iiiilii: iliiilllllliiiiiiilii WENTWORTH-SMITH MATHEMATICAL SERIES ^ELEMENTS OF PROJECTIVE GEOMETRY BY GEORGE HERBERT LING Xy GEORGE WENTWORTH AND DAVID EUGENE SMITH GINN AND COMPANY BOSTON NEW YORK CHICAGO LONDON ATLANTA DALLAS COLUMBUS SAN FKANCISCO COPYRIGHT, 1922, BY GINN AND COMPANY ENTERED AT STATIONERS' HALL ALL RIGHTS RESERVED 722.7 tCbt fltftttntum jprcgg GINN AND COMPANY- PRO- PRIETORS BOSTON U.S.A. PREFACE This work has been prepared for the purpose of providing a thoroughly usable textbook in protective geometry. It is not intended to be an elaborate scientific treatise on the subject, unfitted to classroom use ; neither has it been prepared for the purpose of setting forth any special method of treatment ; it aims at presenting the leading facts of the subject clearly, succinctly, and with the hope of furnishing to college students an interesting approach to this very attractive and important branch of mathematics. There are at least three classes of students for whom a study of the subject is unquestionably desirable ; namely, those who expect to proceed to the domain of higher mathematics, those who are intending to take degrees in engineering, and those who look forward to teaching in the secondary schools. Although the value of the subject to the second of these classes has not as yet been duly recognized in America, European teachers for several decades have realized its useful- ness as a theoretical basis for some of the practical work in this field. For the large number of students belonging to the third class, trigonometry, analytic geometry, and projective geometry are the three subjects essential to a fair knowledge of elemen- tary geometry, and it is believed that the presentation given in this book is such as greatly to aid the future teacher. There is a healthy and growing feeling in America that teachers of secondary mathematics need a more thorough training in the subject matter, even at the expense of some of the theory of education which they now have. This being the case, one of the best fields for their study is projective geometry. iv PREFACE It is recognized that students of projective geometry have usually completed an elementary course in analytic geometry and the calculus, that they have a taste for mathematics which leads them to elect this branch of the science, and that there- fore there may fittingly be some departure from the elementary methods employed in the earlier mathematical subjects. On the other hand, for some students at least, projective geometry is a transition stage to higher mathematics, and the subject should therefore be presented with due attention to the important and recognized principles which must always be followed in the preparation of a usable textbook. It is the belief of the authors that they have followed these principles in such a way as to afford to college students a simple but sufficient introduction to this interesting and valuable branch of geometry. Especial attention has been given to the proper paging of the book, to a clear presentation of the great basal propositions, to the illustrations accompanying the text, to the number and careful grading of the exercises, and to the application of projective geometry to the more elementary field of ordinary Euclidean geometry. CONTENTS PART I. GENERAL THEORY CHAPTER PAGE I. INTRODUCTION 1 II. PRINCIPLE OF DUALITY 15 III. METRIC RELATIONS. ANHARMONIC RATIO .... 21 IV. HARMONIC FORMS 31 V. FIGURES IN PLANE HOMOLOGY 37 VI. PROJECTIVITIES OF PRIME FORMS 41 VII. SUPERPOSED PROJECTIVE FORMS 63 PART II. APPLICATIONS VIII. PROJECTIVELY GENERATED FIGURES 79 IX. FIGURES OF THE SECOND ORDER 101 X. CONICS 115 XI. CONICS AND THE ELEMENTS AT INFINITY .... 143 XII. POLES AND POLARS OF CONICS 151 XIII. QUADRIC CONES 167 XIV. SKEW RULED SURFACES 173 HISTORY OF PROJECTIVE GEOMETRY 181 INDEX . 185 GREEK ALPHABET The use of letters to represent both numbers and geometric magnitudes has become so extensive in mathematics that it is convenient for certain purposes to employ the letters of the Greek alphabet. In protective geometry the Greek letters are used particularly to represent planes and angles. These letters with their names are as follows: A a alpha B beta r y gamma A 8 delta E e epsilon Z c /eta H rj eta e theta I I iota K K kappa A \ lambda M mu N V nu B t xi o o omicron n 7T pi p P rlio 2 a; s sigma T T tau Y V upsilon <5 * phi X X chi * * psi O 0) omega vi ELEMENTS OF PROJECTIVE GEOMETRY PART I. GENERAL THEORY CHAPTER I INTRODUCTION 1. Orthogonal Projection. In elementary geometry the projection of a point upon a line or upon a plane is usually defined as the foot of the perpendicular from the point to the line or to the plane, and the projection of a line is denned as the line determined by the projections of all its points. This simple projection is called orthogonal projection. A FIG. 1 FIG. 2 FIG. 3 Thus, the point A' in Fig. 1 represents the orthogonal projection of the point A upon the line I; the line A{A' 2 in Fig. 2 represents the projection of the line A V A Z upon the line ; and the line k' in Fig. 3 represents the projection of the curve k upon the plane a. 2. Symbols. Projective geometry, like other branches of mathematics, employs special symbols, generally using capital letters to denote points, small letters to denote lines, the first letters of the Greek alphabet, a, /3, 7, 8, , to de- note planes, and the Greek letters (j> and 6 to denote angles. 1 INTRODUCTION 3. Parallel Projection. In a plane a which contains a line p and the points A v A v A y , A n , if a line I is drawn making an angle < with p, each of the lines through A v A v Ay, , A n parallel to I makes with p the angle . The points Jj, Jg ^3* > -^u ^ which these lines inter- sect jt>, are called the projections of ^ ^4 2 , A s , , J ?l upon jw and are said to be found by parallel projection. In space of three dimensions a plane figure A^A^A^ may be projected upon a plane TT by parallel projection. 4. Central Projection. If several coplanar points A v A v A& ' , A n are joined to a point P of their plane, and if these lines are cut by a line jo, the points of intersection of the lines PA V PA V PA 3 , . . ., PA n with the line p are called the projections of A v A v A 3 , ., A n from the center P upon j3 and are said to be found by central projection. In space of three dimensions a plane figure A^A^A Z A^ may be projected centrally upon a plane TT. KINDS OF PROJECTION 3 5. Projection from an Axis. Let A v A v A s , -, A n be points which are not all coplanar. Their orthogonal pro- jections upon a line p may be obtained by drawing a line from each of them perpendicular to p, or by passing a plane through each of them perpendicular to p. The latter method may be generalized by requiring merely that the planes passed through the points shall be parallel to a fixed plane which is not necessarily perpendicular to JP, or by requiring that the planes passed through the points shall pass also through a fixed line p' instead of making them parallel to a fixed plane, p' being different from p. Then the points are said to be projected from an axis, and this second kind of projection is called projection from an axis or axial projection. The projection of points by parallel planes is the limiting case of projection from an axis in which the axis has receded indefinitely. 6. Operations of Projection and Section. The process of finding the projection of a plane figure upon a line or plane consists of two parts. The first part is called the operation of projection and consists in the construction of a figure composed of lines, or of planes, or of both lines and planes, passing through the points and lines of the figure and through the center or axis of projection. These lines and planes are called the projectors of the points and lines of the figure, and constitute the projector of the figure. The second part is called the operation of section and consists in cutting the projector of the figure by a line or plane called the line of projection or the plane of projection. The center or axis of projection, and the line or plane of projection, should be so taken as not to be parts of the figure to be projected. 4 INTRODUCTION Exercise 1. Simple Projections 1. Draw a figure showing the orthogonal projection of a given circle upon a given plane. The figure may be drawn freehand, and at least three cases should be considered : (1) the circle parallel to the plane ; (2) the circle oblique to the plane ; (3) the circle perpendicular to the plane. In (2) consider also the case in which the circle cuts the plane. 2. Draw a figure showing the parallel, but not necessarily orthogonal, projection of a given square upon a given plane. 3. Draw a figure showing the central projection of a given straight line or a given plane curve upon a given line. 4. Draw a figure showing the central projection of a given square upon a given plane. Consider three cases, as in Ex. 1. Consider the case in which P u between the square and the plane as well as the case in which it is not. 5. Draw a figure in which the four vertices of a square are projected from a given axis upon a given line. In projecting from a center P upon a plane TT, describe the projectors and the projections of the following, mentioning all the noteworthy special cases under each : 6. A set of points. 10. Two intersecting lines. 7. A line. 11. Two parallel lines. 8. A triangle. 12. A quadrilateral. 9. A circle. 13. A pentagon. 14. Four points and the lines joining them in pairs. 15. A tangent to a circle at any point. In projecting from an axis p' upon (1) a plane IT and (2) a line p, describe the projectors and the projections of the following, mentioning the noteworthy special cases under each : 16. A set of points. 18. A line not parallel top'. 17. A line parallel to p'. 19. A circle. ELEMENTS AT INFINITY 5 7. Elements at Infinity. Considering central projection only, and supposing the center P and the plane of projec- tion TT to be given, these questions now deserve attention : 1. Does every point of a plane a have a projector? Does it have a projection ? Since every point of the plane a can be joined to P by a straight line, every point of a has a projector; but since this projector may happen to be parallel to TT, a point of a may have no projection. 2. Is every line which passes through P the projector of some point of a ? No ; for certain of these lines may be parallel to a. 3. Does every line of a have a projector ? Does it have a projection ? Consider the answers to Question 1. Draw the figure. 4. Is every plane which passes through P the projector of some line of a ? Consider the answer to Question 2. Draw the figure. Certain exceptional cases have been suggested in con- nection with these questions. Their occurrence is due to the existence of parallel lines and parallel planes, and the difficulty caused by them may be removed as follows : Every straight line is assumed to have one and only one infinitely distant point, and this point is called the point at infinity of the line. Every plane is treated as having one and only one straight line situated entirely at an infinite distance, and as having all its infinitely distant points situated on that line. This line is called the line at infinity of the plane. Space is treated as having one and only one plane situ- ated entirely at an infinite distance, and as having all its infinitely distant points and lines situated on that plane. This plane is called the plane at infinity. 6 INTRODUCTION 8. Illustrations. Let a line c rotate in a plane a about a point C of that plane, and when it has the positions ^i> Cy * let it meet a fixed line p in the points A v A v . Then as long as c is not parallel to p it meets p in one point and only one. As the point of intersection becomes more and more dis- tant, the line c becomes more and more nearly parallel to p. The limiting position of c as the point of intersection recedes infinitely is the line c' through C parallel to p. If, however, the rotation of c is continued ever so little beyond c?', the intersection of c and p is found to be at a great distance in the other direction on p, and as the rotation proceeds farther this point of intersection comes continu- ously back toward A r Hence c' is said to meet p in a point at infinity. If there were several infinitely distant points on jt?, they would with C determine several lines through C parallel to jt?, or several of these points would be common to c 1 and jo, or one or more of these points taken with C would fail to determine a straight line. Apparent conflict with propositions of Euclidean geometry is best avoided by the assumption that every line not situated at an infinite distance has one and only one infinitely distant point. Now consider all points of a plane which are infinitely distant. In elementary geometry we find that the only plane locus met by every line of its plane in one and only one point is a straight line. The locus of infinitely distant points of the plane also possesses this property. Hence this latter locus is called the (straight) line at infinity of the plane. Similarly, the locus of the infinitely distant points in space is called the plane at infinity. PEOJECTOKS AND PROJECTIONS 7 9. Ideas of Projector and Projection Simplified. From the considerations set forth in 7 and 8 it appears that the introduction of the elements at infinity has distinct advan- tages arising out of the fact that, from the new point of view, statements can often be more briefly and more simply made. The greater simplicity is due to the fact that certain cases involved in the questions cease to be exceptional. For example, in dealing with the first question in 7 we now say that every point of a plane a has a projection upon the plane IT ; for if a projector happens to be parallel to TT, it is regarded as meet- ing TT in one point at infinity. It is also clear that we may now say that all the lines through a point P, and lying in a plane determined by P and a line a, consti- tute the projector from the center P of all the points of a. Similarly, we may say that all the lines and all the planes passing through P constitute respectively the projectors from the center P of all the points and all the lines of any plane not passing through P; and that all the planes through any line p form the projector from the axis p of all the points of any line not parallel to p. Exercise 2. Projectors and Projections 1. Draw figures illustrating the four statements in the last three paragraphs above. 2. Consider the truth of the statement that two lines in a plane have one and only one common point. Illustrate the statement by a figure. 3. Do every two planes in a space of three dimensions determine a line ? Explain the statement. 4. In what case do a straight line and a plane fail to determine within a finite distance exactly one point ? 5. In what case do a straight line and a point fail to determine a plane ? 6. Two straight lines which determine a plane determine, without exception, a point. 8 INTRODUCTION 10. The Ten Prime Forms. As fundamental sets of ele- ments we use the following sets, called the ten prime forms : One-Dimensional Forms. 1. The totality of points of a straight line (the base~) is called a range of points, a range, or, less frequently, a pencil of points. The distinction between a line and the totality of its points may be appreciated by considering points as arranged on a line like beads on a string. Similar considerations apply to the other prime forms. 2. The totality of planes through a straight line (the base') is called an axial pencil. This is also called a, pencil of planes or a sheaf of planes. 3. The totality of straight lines in a plane and through a point of the plane is called a flat pencil. In a flat pencil, either the point common to the lines or the plane containing the lines may be regarded as the base of the pencil. The terms range of points, axial pencil, and flat pencil are used for a finite number as well as for an infinite number of elements. Two-Dimensional Forms. 4. The totality of points in a plane (the base) is called a plane of points. 5. The totality of planes through a point (the base) is called a bundle of planes. 6. The totality of lines in a plane (the base) is called a plane of lines. 7. The totality of lines through a point (the base~) is called a bundle of lines. It is also called a sheaf of lines, but because the word sheaf is used in conflicting senses, we shall not use it. Three-Dimensional Forms. 8. The totality of points of three-dimensional space. 9. The totality of planes of three-dimensional space. Four-Dimensional Form. 10. The totality of lines of three- dimensional space. THE TEN PRIME FORMS 9 Exercise 3. The Ten Prime Forms Draw a rough sketch to illustrate each of the following : 1. Range of points. 4. Plane of points. 2. Axial pencil. 5. Bundle of planes. 3. Flat pencil. 6. Plane of lines. Examine each of the following prime forms when the base is at infinity : 7. Flat pencil. 9. Bundle of lines. 8. Axial pencil. 10. Bundle of planes. 11. Find the central projection of a range of points; of a flat pencil ; of a plane of points ; of a plane of lines. 12. Find the plane section of a flat pencil ; of an axial pencil ; of a bundle of planes ; of a bundle of lines. 13. Find the axial projection of a range of points and also of a flat pencil, the axis passing through the base. 14. Find the linear section of an axial pencil and also of a flat pencil, the line of section being in the plane. 15. Investigate the central projection of a bundle of lines ; of the points of space ; of the lines of space. 16. Investigate the plane section of a plane of lines ; of the planes of space ; of the lines of space. 17. Investigate the projection from an axis of a plane of points and also of the points of space. 18. Investigate the linear sections of a bundle of planes and also of the planes of space. 19. Apply each of the four operations to the prime forms not already considered in connection with it. 20. Examine the results of Exs. 11-19 and in each case determine whether to every element of the original figure there corresponds one element and only one element of the resulting figure, and vice versa. 10 INTRODUCTION AI A 11. Classification of Prime Forms. In each of the first three classes of the ten prime forms mentioned in 10 the prime forms of every possible pair are connected by a simple relation. Consider first a range of points A^A^A^ A n on a base p, and consider its pro- jector a 1 2 a 3 a n from a point P exterior to p, this pro- jector being manifestly a flat pencil. By setting up, or arranging, the infinitely many pairs of elements A v a 1 ; A v a 2 ; Ay 3 ; ; A n , a n \ , we find that for every point of the range there is a corresponding line of the flat pencil, and vice versa; and that if two points are nearly coincident, so also are the corresponding lines. Next, make a section of an axial pencil by a plane TT. From the planes a v 2 , a y , a n , ... of the axial pencil and the lines a v a 2 , a y , a n , . planes by the plane TT, infinitely many pairs of elements a v a^ of the section of these a 2 3 may be set up. Evidently for every plane of the axial pencil there is a line of its section (the flat pencil), and vice versa ; similarly for the range and the axial pencil. A similar conclusion may be reached regarding any two prime forms of the second class, and also regarding the two prime forms of the third class. In each case, by the setting up of the pairs, there is established a one-to-one correspondence between the elements of the two forms. This is often written as a 1 1 correspondence. 11 and a(a 2 12. Perspectivity. Certain cases of one-to-one corre- spondence between the elements of prime forms of the same kind should also be noted. For example, in this figure if two transversals p v p[ cut the lines a v a 2 , . ., , of a flat pencil in the points A v A[ ; ^2' 2 ' * * ** H' ?i? * * "9 the ranges A^A^ and A^A'z corre- spond in this w r ay. Similarly, in this figure, if two flat pencils a^a< a' n . are so situated that a v a( ; a a ni a n'-> ' ' ' intersect in the points A v A 2 , , A n , ... of a range, such a correspondence exists. The correspondences in the cases in S 11 re- A A suited from one opera- tion of projection or one of section. In the cases just men- tioned the correspondences resulted from one operation of projection and one of section. All these cases and other similar cases may be brought into one group by means of the following definition : If either of two prime forms can be obtained from the other by means of one operation of projection, or one opera- tion of section, or by means of one operation of each kind, the two forms are said to be perspectively related, or to be in perspective, or to be perspective. The symbol T is often used for " is perspective with." The perspective relation is called a perspectivity. PQ 12 INTRODUCTION Exercise 4. Perspectivity and Projection 1. If the line a' of the plane a' is the projection, from the center P, of the line a of the plane a, then the lines a and a' intersect in a point on the line of intersection of a and a'. 2. If the angle formed by the lines a[ and a' 2 of the plane a' is the projection, from the center P, of the angle formed by the lines t and a. 2 of the plane a, the pairs of lines a 1? a[ ; o. 2 , o a ' intersect in points on the line of intersection of a and a'. 3. If two triangles A 1 A^A 8 and A^A^A^ of the planes a and a' respectively are so situated that the lines A^A\, A 2 A' 2 , and A^Al pass through a common point P, the intersections of the pairs of sides A^A^ A(A' 2 \ A. 2 A 3 , A'. 2 A' a ; A a A lt A' Z A[ are collinear. 4. If two polygons A^A 2 A n and A[A! 2 A' n of the planes a and as.' respectively are so situated that the lines A^A'u A 2 Az, - -, A n A' n pass through a common point P, the intersections of the pairs of sides A^A^ A[A\ ; A^A g , A^A' a ; ; A n A v A' n A[, and the intersections of the pairs of diagonals A t A v A{A\; A t A 4 , A\A[; . . .; A^, A' 2 A' t] - ;A k A m , A' k A' m ; . . . are collinear. It will be noticed that Exs. 1-4 form a related set of problems, as is also the case with Exs. 6-8. 5. If two lines a and a' of the planes a and a' respectively intersect, either may be regarded as the projection of the other from any point exterior to both lines but in their common plane. 6. State and prove the converse of Ex. 2. 7. State and prove the converse of Ex. 3. 8. State and prove the converse of Ex. 4. 9. Given three points on a line a and a point A t not on the line, construct a triangle that shall have A l as a vertex and shall have each of its sides, produced if necessary, pass through one and only one of the three given points. How many such triangles can be constructed ? PERSPECTIVITY AND PROJECTION 13 10. Investigate the problem similar to Ex. 9 in which two given points A l and A 2 of the plane a are to be vertices of the required triangle, and show how to construct the triangle when such a triangle exists. 11. Given the points A l and A[ of two planes a and a' which intersect in a given line a, and given in the plane a a triangle constructed as required in Ex. 9, use Ex. 3 to obtain a triangle in the plane a' that shall have A( as a vertex and shall have sides which, produced if necessary, shall intersect the line a in the points in which this line is cut by the sides of the given triangle in a. How many constructions are possible ? 12. Investigate the cases of Ex. 11 in which a second vertex of one or of each of the triangles is also given. 13. If two triangles A^A^A^ and A[A! 2 A^ in the same plane a are so situated that the lines A^'^ A 2 A' 2) and A S A' S are con- current, the intersections of A t A 2 , A[A 2 ', A 2 A S , A' 2 A' 3 ; A 3 A 1} A' 3 A[ are collinear. Let A^A 2 and A'^A' Z meet in C s , A 2 A S and A' 2 A' S in C v and A 3 A^ and A' Z A\ in C 2 . Take a center of projection P not in the plane a, and project the whole figure upon a plane parallel to the plane PC 3 C^ thus obtaining the line at infinity as the projection of CgC^. Prove that the projection of C 2 is on this line. The development of this problem and similar problems is fully considered in Chapter V. 14. State and prove the converse of Ex. 13. 15. State and prove the proposition of plane geometry which corresponds to Ex. 4. 16. State and prove the converse of Ex. 15. 17. Given three points A I} A^ A s on a line a in a plane a, and three points A[, A 2i A' z on a line ' also in the plane a, find three points A", A", A' a ', not necessarily collinear, into which both sets of three points can be projected. 18. With the same data as in Ex. 17 find three collinear points AC, A 2) A' s ' into which the first two sets of three points mentioned can be projected. 14 INTRODUCTION 19. In Ex. 17 consider also the case in which the lines a and ' are coincident and in which the points A[, A! 2 , A' a are not necessarily all distinct from the points A^ A.^ A 8 . 20. Given three points on a line a, construct a quadrilateral such that the pairs of opposite sides shall intersect in two of the given points, and such that one of its diagonals shall pass through the other point. 21. Assuming the construction asked for in Ex. 20, use Ex. 4 and Ex. 16 to obtain additional quadrilaterals fulfilling the same conditions as the first. Do the other diagonals of these quadrilaterals intersect ? 22. Given two quadrilaterals so constructed as to fulfill the conditions of Ex. 20, the straight lines joining corresponding vertices of these figures are concurrent. 23. If the quadrilateral constructed in Ex. 20 moves so as continuously to fulfill the conditions stated, the other diagonal constantly passes through a fixed point. 24. Show how to find the fixed point mentioned in Ex. 23. 25. In Ex. 20, if the third of the given points bisects the segment joining the other two given points, determine the position of the fixed point mentioned in Ex. 23. 26. Given five points on a line a, construct a quadrilateral such that each of its sides and one of its diagonals, pro- duced if necessary, shall pass through one and only one of the given points. Obtain additional quadrilaterals fulfilling the same conditions. 27. In Ex. 26 investigate the relation that the other diago- nals of any two of the quadrilaterals bear to each other and to the given line a. 28. Extend the problem in Ex. 26 to the case of the pentagon. 29. Given two quadrilaterals so constructed as to fulfill the conditions of Ex. 26, the straight lines joining pairs of corre- sponding vertices of these figures are concurrent. CHAPTER II PRINCIPLE OF DUALITY 13. Principle of Duality. It is now highly desirable to consider a certain important relation between pairs of figures in space, and also between their properties. The nature of this relation, by the use of which the difficulties of the subject may be reduced by almost half, is explained by the Principle of Duality, or the Principle of Reciprocity, which may be stated as follows: Corresponding to any figure in space which is made up of or generated by points, lines, and planes there exists a second figure which is made up of or generated by planes, lines, and points, such that to every point, every line, and every plane of the first figure there corresponds respectively a plane, a line, and a point of the second figure, and such that to every propo- sition which, relates to points, lines, and planes of the first figure, but which does not essentially involve ideas of measure- ment, there corresponds a similar proposition regarding the planes, lines, and points of the second figure, and these two propositions are either both true or both false. The two figures which are related in the manner just described, as well as the two propositions, are said to be dual, reciprocal, or correlative. As a simple illustration of the principle, consider the following : Two points determine a line. Two planes determine a line. Two lines through a point deter- Two lines in a plane determine mine a plane. a point. 15 16 PRINCIPLE OF DUALITY 14. Assumption of the Principle of Duality. The validity of the principle of duality will not be proved in this book, although it is possible so to formulate the axioms of pro- jective geometry that they are unchanged if everywhere the words point and plane are interchanged, and thus to show this validity. Nevertheless the principle will be applied with great frequency in deriving properties of figures, and in so doing either of two courses may be adopted: On the one hand, it may be assumed that the principle is valid and is capable of a proof which is, of course, entirely inde- pendent of any results obtained by means of the principle itself; on the other hand, the principle may be used simply as the basis of a rule for formulating the dual of any proposition, the rule being justified in every case by a proof of this dual proposition. Of these two courses the latter is not a difficult one, for after the principle of duality has been used to derive the enunciation of the dual proposition, it may be applied to the various steps of the proof of the original proposition to obtain a new set of statements which may be examined to see if they constitute a proof of the dual. In each case it will be found that a proof is secured. The plan has the further advantage that it avoids the feeling of dissatisfac- tion and uncertainty attendant upon making a very general and far-reaching but apparently unjustified assumption, besides which the repeated application of the principle leads to that confidence in its validity which comes from increasing experimental evidence. For this reason the second course has been adopted in this book; and whenever the dual of a proposition is derived by applying the principle of duality, either the proof of the dual is derived by the same means or such derivation is left to serve as an exercise for the student. DERIVATION OF DUAL PROPOSITIONS IT 15. Derivation of Dual Propositions. If a figure or a proposition in the geometry of space is given, the first considerations which enter into the derivation of the dual figures or propositions are the facts that the point and the plane are dual elements, and that every geometric figure may be obtained by using either the point or the plane as the primary generating element. Although neither the point nor the plane has a superior claim over the other to be considered as the primary generating element, it fre- quently happens that the statement of a proposition is so framed as to imply that one or the other of these elements has been so used. For this reason the derivation of the dual of any proposition generally requires more than the mere interchange of the words point and plane. On the other hand, it is true, almost without exception, that propositions which have duals may be so stated that the latter may be found from the former by such interchange. Skill in the derivation of dual figures and propositions is quickly gained, and the examples given in 16 will assist the beginner in acquiring this skill. In general it may be said that the student will find it advantageous to consider, in every proposition he meets, the proposition which results from the method of treatment mentioned above, and then to consider whether the proof of the derived proposition can be obtained from the original proof in the same way. In plane geometry the point and the line are dual ele- ments, and any figure may be regarded as having been gen- erated by either of these elements. In a general way, duals in the plane are derived by interchanging these elements. The principle of duality for threefold space, applied to plane geometry, yields the geometry of the bundle of lines and planes, the line and the plane being dual elements. 18 PRINCIPLE OF DUALITY 16. Examples of Duality, the more simple examples of 1. Point A. 2. Line a. 3. Two points determine a line. 4. Two lines which determine a plane also determine a point. 5. Three points in general de- termine a plane. 6. Several points which lie in a plane. 7. Several lines which lie in a plane. 8. A plane triangle ; that is, three points and the three lines determined by them in pairs. 9. A plane polygon. 10. A range of points. 11. A fiat pencil. 12. A plane of points and a plane of lines. 13. Four points in a plane and the six lines joining them in pairs ; a complete quad- rangle, or four-point. 14. Four lines in a plane and the six points determined by the various pairs of these lines ; a complete quadri- lateral, or four-line. 15. Four points in space and the lines and planes deter- mined by them. The following are a few of duality : 1'. Plane a. 2'. Line a'. 3'. Two planes determine a line. 4'. Two lines which determine a point also determine a plane. 5'. Three planes in general de- termine a point. 6'. Several planes which pass tli rough a point. 7'. Several lines which pass through a point. 8'. A trihedral angle; that is, three planes and the three lines determined by them in pairs. 9'. A polyhedral angle. 10'. An axial pencil. 11'. PL fiat pencil. 12'. A bundle of planes and a fmndle of lines. 13'. Four planes through a point and the six lines of inter- section in pairs ; a com- plete four-fiat. 14'. Four lines through a point and the six planes deter- mined by the various pairs of these lines ; a complete four-edge. 15'. Four planes in space and the lines and 'points deter- mined by them. EXAMPLES OF DUALITY 19 16. Given three collinear points A, B, C, find four points P v P 2 , P 3 , P such that the lines PjP, and P 3 P 4 shall meet in A, the lines P 2 P 3 and P 4 P t shall meet in B, and the lineP 1 P 3 shall pass through C. 17. // the line a" of the plane a' is the projection from the center P of a line a of the plane a, the lines a and a' intersect in a point of the line of intersection of a and a'. Proof. The lines a and a' lie in the plane determined by P and a, and hence they determine a point. Since their point of in- tersection lies in the plane a and also in the plane a', it lies in the line determined by a and a'. 16'. Given three coaxial planes a, ft, y, find four planes TT^ 7r , 7T 3 , ?T 4 such that the lines T^ an d 1r 3 7I 4 shall lie in a, the lines 7T 2 7r 3 and TT^ I shall lie in /3, and the line TT I IT Z shall lie in y. 17'. If the line a' through the point A' is determined by A' and the point of intersection of the plane TT icith the line a through the point A, the lines a and a" lie in a plane which passes through the line A A'. Proof. The lines a and a' pass through the point determined by TT and a, and hence they determine a plane. Since their plane passes through the point A and also through the point A', it passes through the line determined by A and A'. 17. Figures in a Plane and Figures in a Bundle. The Principle of Duality may also be stated for the geometry of figures in a plane, and likewise for the geometry of figures in a bundle. In the first case the point and the line are dual elements, and in the second case the line and the plane. Simple modifications of the statement of the principle in 13 yield the statements for the two cases. Pairs of the examples of 16 may be used to illustrate this fact. For example, the following pairs are duals in the plane : 1, 2 ; 6, 7 ; 8, 8; 9,9; 10, 11; 13, 14. The following pairs are duals in the bundle : 1', 2' ; 6', T ; 8', 8' ; 9', 9'; 10', 11'; 13', 14'. Many other examples of duality will be found as we proceed. 20 PRINCIPLE OF DUALITY Exercise 5. Principle of Duality 1. By means of the Principle of Duality obtain for three- fold space the statement and also the proof of the dual of Ex. 3, page 12. 2. Similarly, find the space dual of Ex. 4, page 12. 3. Derive and prove the space dual of Ex. 13, page 13. 4. Obtain for plane geometry the statement and proof of the dual of Ex. 13, page 13. 5. Derive the space dual of Ex. 4. 6. Verify that in the geometry of the bundle the results of Exs. 3 and 5 are dual. 7. If a proposition in plane geometry or in the geom- etry of the bundle has a dual, but is not self-dual in that geometry, then in the geometry of threefold space it belongs to a set of four propositions each of which is dual with two of the others. 8. If a proposition is self-dual in plane geometry, then the four propositions mentioned in Ex. 7 reduce to two. 9. Can the four propositions of Ex. 7 ever reduce to one proposition ? Discuss in full. 10. Give an example of a self-dual figure in threefold space. 11. State a simple self-dual proposition regarding the figure mentioned in the answer to Ex. 10. 12. Are all propositions regarding the self-dual figure of Ex. 10 themselves self-dual ? Discuss in full. 13. Given three planes passing through a line a of a bundle whose base is A, construct in this bundle a four-edge such that pairs of opposite edges lie in two of the given planes and one of its diagonal lines lies in the remaining plane. Compare this example with Ex. 20, page 14. Notice that when a prop- osition is harder to prove than its dual the proof of the latter may be used to suggest that of the former. CHAPTER III METRIC RELATIONS. ANHARMONIC RATIO 18. Metric and Descriptive Properties. Properties of geo- metric figures are of two sorts : (1) metric, that is, those which relate to the measurement of geometric magnitudes ; (2) descriptive, that is, those which are not metric. Nearly all the propositions of ordinary elementary geometry deal with metric properties, while, speaking generally, those of projective geometry deal with descriptive properties.' In fact, it is possible to exclude almost entirely from projec- tive geometry the consideration of the metric properties of figures. On the other hand, even when the object in view is the study of the descriptive properties of figures, it frequently happens that brevity is secured by the use of metric considerations. For this reason the metric proper- ties of figures will be used freely whenever the nature of the work is such as to make this course advisable. If the student will consider the work which has thus far been done in this book he will see that no statement has been made that depends in any way upon measurement. The lines projected may be of any desired length, the angles may have any desired measure, and the closed figures may have any desired area. It is therefore evident that the work thus far has not been metric in any way. In this chapter, on the other hand, we proceed to establish certain important properties which will prove to be of great service to us in subsequent work. Moreover, as we proceed, it will appear that by virtue of the propositions proved in 26 and 27 the study of these properties is appropriate in connection with the descriptive properties of figures. 21 22 METRIC RELATIONS 19. Relations of Line Segments. In measuring distances along a straight line attention is given to direction as well as to length. One direction along a line is selected as posi- tive, the opposite one being negative. The direction of a line segment is called its sense and is indicated by the order of the end letters, An denoting the segment of a line thought of as extending from A to B and BA the segment of a line thought of as extending from B to A. Evidently, therefore, we have or AB+BA = Q. Having adopted this convention with respect to signs, many identical relations can be proved. For example, A, B, C, , , but not both. 2. (ABCD} approaches AC/BC as a limit as D recedes indefinitely in either direction. Exercise 6. Anharmonic Ratios 1. If (ABCD l ) = (ABCD i ) ) D I and Z> 2 are coincident. 2. Consider Ex. 1 and 22 for the anharmonic ratio (abed). 3. Consider Ex. 1 and 22 for the anharmonic ratio RATIOS OF FOUR POINTS 25 23. Twenty-four Anharmonic Ratios. Corresponding to the order A, B, C, D of four collinear points, there has been defined the anharmonic ratio (ABCIt). There are, however, twenty-four possible orders for these points, that is, the 4 ! permutations of the four letters; and therefore there are twenty -four anharmonic ratios for the four points, as follows : (ABCP), (ABDC), (ACBD}, (ACDB^ (ADBC), (ADCB), (BACV), (BADC), (BCAV), (BCD A), (ED AC), (BDCA), (CABD), (CADB), (CBA&), (CBDA), (CDAB), (CDBA), (DABC), (DACB), (DBAC), (DBCA), (DCAB), (DCBA). But by definition ( 21) AD AC AD.BC while (ABDC} = - : - = - BD BC AC-BD and so In like manner it may be shown that the last eighteen ratios fall into six sets of three each, all those in any set being equal to one of the first six anharmonic ratios. Exercise 7. Anharmonic Ratios Given the three collinear points A, B, C, proceed as follows : 1. Find the collinear point D such that (ABCD)= 7. 2. Find the collinear point D such that (ABCD) = 7. 3. Find the collinear point D such that (ABCD)= k. 4. Prove that (ABCB) = (BADC) = (CDAB) = (DCBA). 5. Determine the several permutations of the four elements A, B, C, D which leave the value of (A BCD) unchanged. 6. Which of the twenty -four ratios are equal to (ADBC) ? 26 ANHARMONIC KATIO 24. Relations of the First Six Ratios. The first six anharmonic ratios given in 23 are also connected by sim- ple relations. If (ABCD)=x, we have the following: 1. (ABCD)=x. 2. (ABDC} = -. AC AD AC-BD lor ( ABCD) = : = Jjr ^ = X , AD AC AD-BC 1 and 3. For ( 19, 2) AB-CD+BC- AD+ CA BD = ; AB CD CA -BD AC AD Therefore ( A CBD) = l-x. 1 4. \-x For (.4 CDB) = , since we have simply interchanged the (A.CBD) last two letters, as in 1 and 2 above. Hence the result follows from 3. 5. X For (A CBD) = 1 x, by 3, where we merely interchange the second and third letters. Hence, by similar reasoning, (ADBC) = l-(A BDC) = 1 - - = i^-1 . nt* 6. x-\ For we found from 1 and 2 that the transposition of the third and fourth letters gave the reciprocal of the original anharmonic ratio, 1 x and so from 5 we have (ADCB) = RELATIONS OF THE RATIOS 27 25. Equality of the Six Expressions. We may now de- termine the values of x for which any pair of the six 1 1 x\ , x expressions x, - , 1 x, , , and are equal, x 1 x x x \ and therefore we may determine the values of these six expressions which correspond to the values of x so found. The results may be put in tabular form as follows : X 1 X 1-x 1 x-1 X 1-X x x-1 1 1 GO cc -1 -1 2 i 2 I J 2 i 2 -1 -1 l+V-3 l_V-3 l+V-3 l+V-3 l-V-3 2 2 2 2 2 2 l_V-3 1+V^3 i + V^~3 l-V-3 l-V-3 1 + V^3 2 2 2 2 2 2 GO 1 1 GO 2 1 -1 -1 * 2 GO GO 1 1 It will be noticed that these values of the six distinct ratios of the 4 ! anharmonic ratios may be classified into three groups as follows: 1. Those in which the values of x are imaginary, the values of all the functions being also imaginary. 2. Those in which the values of x are 1, 0, or oo, the values of the functions being also 1, 0, or oo. 3. Those in which the values of x are 1, |> or 2, the values of the functions being also 1, , or 2. The first group is not concerned with the anharmonic ratio of four real collinear points, and the second does not correspond to the anharmonic ratio of four real points which are distinct. The third group is the only important one for our present purpose, and this will be considered in Chapter IV. PG 28 ANHARMONIC RATIO THEOREM. PRIME FORMS RELATED BY PROJECTIONS AND SECTIONS 26. If two prime forms of the first class are so related, that either may be obtained from the other by a finite number of projections and sections, the anharmonic ratio of any four elements of one is equal to the anharmonic ratio of the corre- sponding four elements of the other. Proof. I. Let either form be obtainable from the other by means of one operation of projection or one of section. 1. Range ABCD and flat pencil abed. From P, the base of the flat pencil, draw PQ perpendicular to p, the base of the range ABCD. Then, equating pairs of expressions for double the areas of the triangles A CP, BCP, ADP, BDP, we have PA.PCsinac = PQ. AC, BC, 'AD, BD. PA - PD sin ad = PQ PB PD sin bd = PQ PA sin ac _ A C PA sin ad 1PB" smbc~~BC' ~PB ' sin bd sin ac sin ad AC ^ AD , ~BC'~ AD JD and whence smbc sinbd BC BD (abed} = (ABCD}. 2. Range ABCD and axial pencil afiyS. From a point P in the base of the axial pencil a/3 Dy X, Y are collinear points, and if (A 1 A Z XY) = (B 1 B 2 XY) = (C 1 C 2 XY) = (D 1 D 2 XY) = -1, it follows that (^ 1 B 1 C 1 7) 1 ) = (J 2 B 2 C 2 Z) 2 ). 13. If A v Ay , A n , X, Fare n -f- 2 collinear points, then 14. If A v A y A 3 , A', F are five coplanar points, and if A 1 (A i A g XYJ denotes the anharmonic ratio of the four lines from A l to Ay A 8 , X, F, it follows that the product of the anharmonic ratios A^A^xr), A^A^XY), A^A^^XY) is 1. 15. Generalize the result found in Ex. 14. 16. If A v Ay A 8 , A t , X, Fare concyclic points, it follows that the anharmonic ratios X (A^i 3 ^ 8 ^ 4 ) and Y(A 1 A^A a A 4 ) are equal. CHAPTER IV HARMONIC FORMS 28. Harmonic Range. When four collinear points A, B, C, D are so situated that (ABCD) = 1, the four points are said to constitute a harmonic range. In the same way we may define a harmonic flat pencil and a harmonic axial pencil. Any one of these three forms is spoken of as a harmonic form, and it follows ( 26) that every form derived from a harmonic form by a finite number of projections and sec- tions is a harmonic form. If three elements of a harmonic form are given, it is evident that the fourth element, called the fourth harmonic to the three, is uniquely determined. Moreover, since the anharmonic ratio is here negative, from the above definition the elements of the first pair, say A and B, separate those of the second pair, say C and D. . AC-BD AC AD That is, since (ABCD) = - 1, -^-^ = - 1, and _ = - _, so that the two ratios have opposite signs. Therefore, either C or D divides AB internally and the other divides it externally. The elements of either of these pairs, A and B, or C and Z>, are said to be conjugates or harmonic conjugates with respect to the other pair. They are also said to be harmonically separated by the elements of the other pair. Since from the definition ( 21) the anharmonic ratios (ABCD) and (CDAB) are equal, it follows that the relation between the pairs of elements A, B and C, D is symmetric with respect to them. 31 32 HARMONIC FORMS Exercise 9. Harmonic Ranges 1. Given three collinear points A, B, C, with C bisecting AB, determine the fourth harmonic D. 2. Consider Ex. 1 when C is the point at infinity. 3. Consider Ex. 1 when A, B, C are any collinear points. 4. Given three concurrent lines a, b, c, with c bisecting the angle ab, determine the fourth harmonic d. Compare Ex. 4 with Ex. 1. 5. Consider Ex. 4 when a, b, c are any concurrent lines. Compare Ex. 5 with Ex. 3. 6. If (A BCD) = 1, the four points A, B, C, D may, by a finite number of projections and sections, be projected into the positions A } B, D, C. 7. If A BCD is a harmonic range, the line segments AC, AB, and AD are connected by the proportion AC:AD = AC -AB:AB-AD. 8. If AB 1 C 1 D 1 and AB^C^D^ are harmonic ranges on differ- ent bases, the lines B^B^ C^C^ D^ are concurrent, and the lines -Bj# 2 , C\.D 2 , C^ are also concurrent. 9. If ^j-Z^C^Dj and A^B^C^D^ are harmonic ranges, and if A^A^ -BjB.,, C^Cg are concurrent at 0, then D^ also passes through 0. 10. If A, B, C, D, 0, P are points on a circle and are so placed that the pencil O(ABCD) is harmonic, the pencil P(ABCD) is also harmonic. 11. If ABCD is a harmonic range, and if is the midpoint of CD, then OC 2 = OA OB. 12. Use the result of Ex. 11 to find a pair of points which shall be harmonic conjugates with respect to two given pairs of collinear points A^ Bj A^, /? 2 . 13. Given four coplanar lines, draw when possible a line which shall cut them in a harmonic range. COMPLETE QUADRANGLE 33 29. Complete Quadrangle. The figure formed by four points in a plane, no three of which are collinear (asP, Q, R, S in the figure below), and the six lines determined by them is called a complete four-point or complete quadrangle. Any two of the six lines of a complete quadrangle which do not intersect in one of the original four points are called opposite sides. The intersections of opposite sides are called diagonal points, and they are the vertices of the diagonal triangle of the complete quadrangle. THEOREM. HARMONIC PROPERTY OF A QUADRANGLE 30. If four collinear points A, B, C, D are so situated that two opposite sides of a complete quadrangle pass through A, two opposite sides pass through B, and the two remaining sides pass through C and D respectively, then (ABCU) = 1. Let P, Q, R, S be the vertices of the complete quadrangle, and let PQ, RS pass through A ; PS, QR through B ; PR through C; and QS through D. Then (ABCD) = Hence and (A BCD) cannot be equal to + 1, since no two points are coinci- dent, as would then be the case. 34 31. Complete Quadrilateral. The figure formed by four lines in a plane, no three of which are concurrent (as p, q, r, s below), and the six points determined by them is called a complete four-side or complete quadrilateral. Any two of the six points of a complete quadrilateral which do not both lie on one of the original four lines are called opposite vertices. The lines determined by pairs of opposite vertices are called diagonal lines, and they deter- mine the diagonal triangle. The student should compare this fig- ure with that of the complete quadrangle in 30, and should notice also the duality suggested by 29 and 31, the dual ele- ments being the point and line. THEOREM. HARMONIC PROPERTY OF A QUADRILATERAL 32. If four concurrent lines a, b, c, d are so situated that two opposite vertices of a complete quadrilateral are on a, two opposite vertices on b, and the two remaining vertices on c and d respectively, then (abed) 1. Let p, q, r, s in the figure above be the sides of the complete quadrilateral, and let p and q, and also r and s, intersect on a ; p and s, and also q and r, intersect on b ; p and r intersect on c ; and and q intersect on d. (abed) = (sqod) = (bacd) = Then Hence and (abed) 2 = I, (abed") = - 1. (abed) Why cannot (abed) = + 1 ? Students should compare this proof, step by step, with that of 30. QUADRANGLES AND QUADRILATERALS 35 Exercise 10. Quadrangles and Quadrilaterals 1. State and prove the converse of 30. 2. State and prove the converse of 32. 3. Two vertices of the diagonal triangle of a complete quadrangle are harmonically separated by the points in which the line determined by them is cut by the remaining pair of opposite sides of the quadrangle. 4. By interchanging certain elements, it is possible to derive 32 from 30 and Ex. 2 above from Ex. 1. Derive a proposition in this way from Ex. 3 and investigate its truth. 5. From 30 derive a theorem respecting the complete four-flat and prove it. In the geometry of space the figure dual to the complete quadrangle is called the complete four-flat, and, similarly, the complete four-edge in Ex. 6 is dual to the complete quadrilateral. 6. As in Ex. 5, from 32 derive a theorem respecting the complete four-edge and prove it. 7. The six points, other than. the diagonal points, in which the diagonal lines meet the sides of a complete quadrangle lie in sets of three on each of four lines. 8. From the result in Ex. 7 prove the existence of a com- plete quadrilateral which has the same diagonal triangle as any complete quadrangle. 9. Prove the plane duals of Exs. 7 and 8. 10. In this figure QS is parallel to AB. Show that PC is a median and is divided harmonically. Consider Z), the intersection of QS and AB, to have moved to infinity. 11. In the complete quadrangle shown in 30 show that A Q PS EC = - A C BS . PQ. 12. As in Ex. 11, show that A Q PS . BD = AD - BS - PQ. 13. Using 30, prove Ex. 12, page 30. 36 HARMONIC FORMS 33. Descriptive Definitions of Harmonic Forms. The har- monic forms might originally have been denned in a purely descriptive fashion based upon the facts just developed. Thus, a harmonic range might have been denned as a set of four collinear points so situated that through each of the first two points there pass two opposite sides of a com- plete quadrangle, and through each of the other two points there passes one of the remaining sides of the quadrangle. Similar definitions might have been given for the har- monic flat pencil and the harmonic axial pencil. These are the definitions which are usually adopted when it is desired to avoid as far as may be possible the use of con- siderations based upon measurement. Exercise 11. Harmonic Forms 1. Given three collinear points A, B, C, construct the fourth harmonic D from the descriptive definition of 33. In the constructions on this page use only an ungraduated ruler. 2. Given three concurrent lines a, b, c, construct the fourth harmonic d from the descriptive definition of 33. 3. Given a line segment AB and an indefinite line parallel to A B, bisect AB. 4. Given a line segment AB, its midpoint C, and any point not in the line of AB } through O draw a line parallel toAB. 5. Given two intersecting lines and the bisector of one of the angles formed by them, construct the bisector of the supplementary angle formed by the lines. 6. Given a line segment AB divided at C in the ratio m : n, construct a point D that divides the segment AB externally in the same ratio. 7. Dualize for space the descriptive definitions of a harmonic range and a harmonic flat pencil. CHAPTER V FIGURES IN PLANE HOMOLOGY 34. Homologic Plane Figures. Further interesting appli- cations of the anharmonic ratio and illustrations of its significance occur in homologic plane figures. Given two figures in a plane, if to every point of one figure there corresponds a point of the other, if to every line of one there corre- sponds a line of the other, if the lines joining corre- sponding points of the two figures are concur- rent, and if the inter- sections of corresponding lines are collinear, the two figures are said to be homologic, or in (plane) ^^^i A B homology. The point in which all lines joining corresponding points are concurrent is called the center of homology, the line which contains all intersections of corresponding lines is called the axis of homology. In the above illustration the two given figures are the triangles ABC, A'E'C', The corresponding points indicated are the three pairs of vertices, but any number of other pairs of points may be chosen. The corresponding lines are AB and A'B', BC and B'C', C'A and C'A'. The center of homology is O, and the axis of homology is o. 37 38 FIGURES IN PLANE HOMOLOGY THEOREM. FIGURES IN HOMOLOGY 35. If two figures are in plane homology, it follows that: 1. All sets of four collinear points consisting of the center of homoloyy, a point on the axis of homology, and two correspond- ing points of the figure have a common anharmonic ratio. 2. All sets of four concurrent lines consisting of a line through the center of homology, the axis of homology, and two corresponding lines of the figure have a common anharmonic ratio. 3. These two common anharmonic ratios are equal. Proof. Let A v A v A s and A[, A' v A' z be corresponding sets of three points of two homologic figures; and let OA^A[, OA Z A' Z , OA 3 A S intersect o in 0^, 2 , 3 respectively ; also let A^A V A[AI meet in P s ; A%A S , A^A' Z meet in ^; and A 3 A V A' Z A[ meet in /. Then it is evident that CONSTANT OF HOMOLOGY 39 36. Constant of Homology. The common value of the two anharmonic ratios found in the theorem of 35 is called the constant of homology for the two figures. Given a center and an axis of homology o, we can construct a figure homologic with any given figure in such a way that to any point A of the given figure there shall correspond any selected point A' on the line OA, or that to any line a of the given figure there shall correspond any selected line a' concurrent with o and a. For the point A' selected there is a value of the constant of homology. Conversely, for any assigned value of the constant of homology, the center and axis being given, one and only one point A' corresponds to any given point A of a given figure. The fact is that when the center, axis, and constant of homology are given, one and only one figure homologic with a given figure can be constructed. Examples of such constructions are given on page 40. There are several notable special cases of the homologic relation. One such case is that of harmonic homology in which the constant of homology is 1. Another case is that in which the axis of homology is the line at infinity. Then all pairs of corresponding lines are parallel, and the figures are similar. In this case the constant of homology (OO^^) becomes OAJOA\, which may be shown to be the ratio of similitude. If in addition the constant is 1, the center is a center of symmetry for the figure composed of the two homologic figures. A third case is that in which the center of homology is a point at infinity. Then the constant is O^A'JO^A^. If also the homology is harmonic, the axis of homology is an axis of symmetry for the figure composed of the two homologic figures. 40 FIGURES IN PLANE HOMOLOGY Exercise 12. Figures in Homology 1. Given a center of homology O, an axis of homology o, and any triangle A^A^A^ construct, homologic with A^A^A^ a triangle that has a vertex A[ at a given position on OA r As to the possibility of this construction, see Ex. 13, page 13. Given a triangle A-^A^Ay the center of homology 0, the axis of homology 0, and the following constants of homology, con- struct the figures homologic with A^A^A^ : 2. 3. 3. -3. 4. -1. 5. 1. 6. In each of Exs. 2-5, if A^A^A^ is the result of the con- struction, use the same center, axis, and constant of homology to con struct the figure homologic with A [A. ^4.^. 7. If there are three coplanar figures f v / 2 , f a , and if for a given center and a given axis of homology two of them, /j and fy are homologic with the constant c a , and if / 2 and / g are homologic with the constant c lt then /j and / g are homo- logic with the constant of homology c z equal to c a - c r 8. Consider carefully Exs. 2- 7 for the case in which O is a point at infinity and the case in which o is the line at infinity. 9. For a given center, axis, and constant of homology con- struct the line corresponding to the line at infinity. This line is called the vanishing line. 10. In Exs. 2-5 determine the vanishing line. 11. Under what conditions is the vanishing line at infinity? Q-iven a circle, the axis of homology o, and the center of Iwmology 0, construct the figure homologic with the circle under each of the following conditions : 12. The vanishing line does not meet the circle. 13. The vanishing line is a secant of the circle. 14. The vanishing line is a tangent to the circle. CHAPTER VI PKOJECTIVITIES OF PRIME FOKMS 37. Projective One-Dimensional Prime Forms. Whenever there exists between the elements of two one-dimensional prime forms a one-to-one correspondence such that, by means of a finite number of operations of projection and section, it is possible to pass simultaneously from all the elements of one prime form to the corresponding elements of the other, the two prime forms are said to be protectively related or to be protective. The correspondence existing between two projective one- dimensional prime forms is called a projectivity. The symbol j- is frequently used for " is projective with." THEOREM. PROJECTIVE PRIME FORMS 38. Prime forms which are projective with the same prime form are projective with each other. Proof. If n^ operations yield a form / 2 from a form f v and if n 2 operations yield / 3 from/ 2 , then Wj + n. 2 operations yield / from / r It is not the purpose of this book to discuss the projectivities of prime forms other than one-dimensional ones, but it may be stated that between the prime forms of higher dimensions there exist relations which have the same general character as those just denned, and that these relations are also called projectivities. A complete study of projective geometry would include the consider- ation of these higher projectivities and of many important geometric propositions relating to them. 41 42 PROJECTIVITIES OF PRIME FORMS THEOREM. PROJECTIVITY OF TRIADS 39. Between two one-dimensional prime forms, each of which consists of three elements in a specified order, there exists a project iinty. Proof. If either of the prime forms is not a range, it is possible by operations of projection and section to obtain from it a range which is projective with it. Hence it is necessary to prove the proposition only for the case in which both prime forms are ranges of three points. This theorem is what was formerly called a lemma, a proposition inserted merely for the purpose of leading up to a fundamental theo- rem ; in this case, the one given in 40. The proof involves the consideration of the three cases below. 1. The ranges may be coplanar and upon different bases. Let the ranges be A 1 B 1 C 1 on the base p l and A 2 B Z C% on the base p y and let both ranges be in the same plane. Draw the line through A l and A 2 , and on it take any points ^ and J^, not coincident with A^ and A 2 respectively. Draw I>B V I^C V J^B y J<7 2 ; and let P^, P^B 2 intersect at B, and let ^<7 r P 2 C 2 intersect at C. Through B and C draw the line p, cutting A^A^ at A. Then range A 1 B 1 C 1 range ABC range J 2 Z? 2 (7 2 . Hence range A 1 B 1 C 1 range 38 PROJECTIVITY OF TRIADS 43 2. The ranges may be coplanar and upon the same base. Let the ranges A^B^C^ and A Z B 2 C 2 be on the same base p. It is here assumed that the points A y B y C 2 may be the same except for order as the points A v B V C v or may be partly or wholly distinct from these points. In any case from a center P, exterior to the base p, project A 2 B 2 C 2 upon a new base p'. Then apply Case 1 to show that the range so obtained on the base p' is pro- jective with the range A^B^C^. It then follows that the range A 2 B Z C Z is projective with the range A 1 B 1 C r 3. TJie ranges may not be coplanar. A 2 B 2 C 2 O 2 ^ Let the bases p 1 and p% of the ranges A^B^ and not be in any one plane. Join any point of the base p l to any point 2 of the base p 2 by the line p. Select three points A, B, C on p. Then, by Case 1, it follows that range A^B 2 C Z range ABC, and range A^B^C^ -^ range ABC. Therefore range A 2 B 2 C 2 -^ range A^C^. From these results the existence of a projectivity follows. PQ 44 PROJECTIVITIES OF PRIME FORMS THEOREM. FUNDAMENTAL THEOREM OF PRIME FORMS 40. Betiveen two one-dimensional prime forms there exists one and only one projectivity in which three elements of one form, in a specified order, correspond to three elements of the other form, also in a specified order. Proof. Let us first consider three special cases. 1. Suppose that the two prime forms are a flat pencil and the range obtained by cutting the pencil by a line. Let the lines of the flat pencil be a, b, c, , I, and let these lines be cut by a line p in the points A,B,C,.'~, L, > '. Then these prime forms are perspective in such a way that a, A ; 5, B ; c, C ; ; I, L ; are pairs of cor- responding elements. Assume that, if possible, a second projectivity exists in which the first three of these pairs of elements correspond, but in which I corresponds to M and not to L. Then, from the perspectivity, and, from the second projectivity, Then (ABCL} = (ABCM), which is impossible unless L = M. Hence the second projectivity cannot exist, and the only projectivity existing between the forms is the perspectivity. FUNDAMENTAL THEOREM 45 2. Suppose that the two prime forms are an axial pencil and the range obtained by cutting the pencil by a line. FIG. 1 FIG. 2 X, Let the planes of the axial pencil be a, /3, 7, (Fig. 1), and let p cut the planes in A, B, C, , L, . A proof similar to that on page 44 should be given by the student. 3. Suppose that the two prime forms are ranges. Let A v B r , ,>'. Similarly, if /' is a range, let the elements 1', 2', 3', 4', 5', ... be A', B', C', D', E', . . . ; but if/' is not a range, let the elements 1', 2', 3', 4', 5', ... be cut by a line in the points A', B', C', D', E' . . .. Between the ranges AB CD and A'B'C'D' . . . there is just one projectivity in which A, A'; B, B' ; C, C' are cor- responding elements. Let Z), D' be corresponding elements. Then form 1234 ...- range ABCD ... - range A'B'C'D' ... - form 1'2'3'4', .... Hence form 1234 ... - form 1'2'3'4' .... Suppose that, if possible, between / and /' there is a second projectivity in which 1, 1'; 2, 2'; 3, 3' are pairs of corresponding elements and 4, 5' are corresponding elements. Then range ABCD-..- form 1234 ..., and form 1'2'3'5' ... - range A'B'C'E' . . .; whence range ABCD . range A'B'C'E' .... Accordingly there would be a second projectivity between the ranges ABC - . and A'B'C' . ., in which A, A'-, B, B' -, C, C' are pairs of corresponding points. This, however, is impossible. Hence the theorem is true in all cases. FUNDAMENTAL THEOREM 47 41. COROLLARY. There is one projectivity and only one projectivity between one-dimensional forms on the same base which makes three distinct elements of a one-dimensional form correspond each to itself. This projectivity makes every element of the form correspond to itself. Exercise 13. Projectivities of Triads 1. If A, B I} B^ C v C 2 are five distinct points on a line, find a set of projections and sections, minimum in number, which connects the triads AB 1 C 1 and AB 2 C 2 . 2. Examine Ex. 1 for the case in which B l coincides with B Z . 3. If A, B, C are three points on a line, find the set of projections and sections, minimum in number, which connects these points with themselves in any selected order. Consider Ex. 3 for each of the six possible orders of the points. 4. If A v A z , B v B v C v C 2 are six distinct collinear points, find a set of projections and sections, minimum in number, which connects the triads A 1 B 1 C 1 and A^C^ 5. If A, B^ B Z , Cj, C 2 are five points, no four of which are coplanar, find a set of projections and sections, minimum in number, which connects the triads AB 1 C l and AB^C 2 . This is a special case of a more general problem. 6. In how many ways can a projectivity between the triads specified in Ex. 5 be established ? 7. If A v A 2 , B I} JS 2 , Cj, C 2 are six points in space, no four of which are coplanar and no three of which are collinear, the triads A 1 B 1 C 1 and A^B 2 C Z are projective. Specify a set of projections and sections which constitutes such a projectivity. 8. Investigate the possibility of establishing a projectivity between the triads specified in Ex. 5 such that any fourth point D l in the plane AB 1 C l shall correspond to any fourth point Z> 2 in the plane AB^C f 48 PROJECTIVITIES OF PRIME FOKMS THEOREM. METRIC CONDITION FOR A PROJECTIVITY 42. If between the elements of two one-dimensional prime forms there exists a one-to-one correspondence such that the anharmonic ratio of every set of four elements of one prime form is equal to that of the set of four corresponding elements of the other prime form, the correspondence is a projectivity. BZ Proof. It is sufficient to consider the case of two ranges because if either of the given prime forms is not a range, it is possible by operations of projection and section to obtain from it a range which is projective with it. From 39 it follows that a projectivity exists between three points A v B V C of the first range and the three corresponding points A y B y C z of the second range. Then if a fourth pair of corresponding points D v D Z on the ranges are found, it follows by hypothesis that Also let range A 1 B 1 C 1 D 1 range A^B^C^D^. Then (4Ai) = OMWA). 26 and so (4AJ) = (4^0^). Accordingly D^ coincides with Z> 2 ; and therefore range A 2 B 2 C 2 D 2 -^ range A^B^D V Thus, any fourth point D^ of the first range has the same corresponding point Z> 2 in the given one-to-one correspond- ence as it has in the projectivity between the ranges which is determined by the triads of points A 1 B 1 C 1 and A^B^C V Hence the correspondence is this projectivity. METRIC CONDITION FOIl A PROJECTIVITY 49 If it is given that two one-dimensional prime forms are projective, it should be noted that it is necessary only that three elements of one of them and the corresponding ele- ments of the other be specified in order that by the con- struction of 39 the whole projectivity may be established ; and tbis construction furnishes a standard method of establishing a projectivity. Associated with the general results of 39-42, several which are special in their nature and application are considered in 43~46. Exercise 14. Projectivity of Prime Forms 1. Given three points A V B I} C 1 of a range and the corre- sponding points A 2 , B 2 , C 2 of a second range projective with the first, the bases being different, construct the point Z> 2 of the second range corresponding to a fourth point D l of the first. 2. Consider Ex. 1 when the point D l is at infinity. 3. Consider Ex. 1 when the ranges are coplanar and D l is their intersection. 4. Obtain simplified constructions for Exs. 1 and 2 when A l and A 2 coincide at the intersection of the ranges. 5. Consider Ex. 1 when the bases are coincident. 6. Consider Ex. 5 when A V B 2 ; A 2 , B l ; C 2 , D l are all pairs of coincident points. 7. Prove the theorem suggested by the result of Ex. 6. 8. If two ranges are projective, to every harmonic range in one of them there corresponds a harmonic range in the other. 9. Assuming that if four points A, B, C, D are properly divided into pairs a common pair of harmonic conjugates with respect to these pairs exists, prove the converse of Ex. 8. 10. Investigate the question of dividing a set of four col- linear points into pairs so that a common pair of harmonic conjugates may exist. 50 PROJECTIVITIES OF PRIME FORMS THEOREM. PROJECTIVE RANGES IN PERSPECTIVE 43. Two projective ranges whose bases are not coplanar are perspective. C 2 Proof. Let A 1 B 1 C 1 L on the base p 1 and A 2 B 2 C Z L 2 on the base p 2 be projective ranges in which A r A z ; B v B Z I C r C z ; are corresponding elements, their bases not being coplanar. On the line A^A^ take a point A, and let C be the point in which the line C-^C^ intersects the plane determined by the point A and the line B^B^. Let the line A C intersect the line S 1 B Z in the point B. Denote by a, ft, 7, , X, the planes determined by the line AB C and the points A r B r C r , L v respectively. The axial pencil afiy ... X is perspective with the range A 1 B 1 C 1 L l and cuts the linejt? 2 in a range pro- jective with the range A l B l C l L 1 . . . Moreover, in the projectivity thus established the points Ay By C 2 correspond to the points A v B v C l respectively. This projectivity is therefore the same as the one that was originally assumed to exist between the ranges ( 40). Hence from the range A^B^C^ on the base jt?j it is possible to pass to the range A Z B 2 C 2 ... on the base p 2 by one operation of projection from the axis ABC and one operation of section by the line p^ Hence the pro- jectivity existing between the ranges is a perspectivity. PROJECTIVE RANGES 51 THEOREM. CONDITION FOR PERSPECTIVE RANGES 44. Two projective ranges on different bases in a plane TT are perspective if and only if the point common to them is self-corresponding in the projectivity. FIG. 1 FIG. 2 Proof. First, let the ranges A 1 B 1 C 1 (Fig. 1) on the base p l and A^B^C Z on the base p z be projective in such a way that X, the intersection of the two ranges, is self- corresponding in the projectivity. Let the lines A^A^ and B^B^ intersect at P, and draw PX. The projectivity is completely determined by the corre- spondence of A v B v X with A v B v X, but this correspond- ence is established by the projection of A v B v X from the center P and the section of the resulting flat pencil by the line p z . Hence the ranges are perspective. Next, let the ranges be perspective (Fig. 2). Let the intersection of p l and p v regarded as a point of the first range, be denoted by X Y Since its correspond- ing point would be found by cutting the projector of X^ from a center P in the plane TT, or from an axis p not in this plane, as the case may be, by the line jt? 2 , it is the intersection X Y That is, X^ is self-corresponding. 52 PROJECTIVITIES OF PRIME FORMS THEOREM. RANGES PERSPECTIVE WITH A THIRD RANGE 45. Two protective (but not perspective) ranges on different bases in a plane are both perspective with the same range on any line that does not pass through the intersection of the bases of the given ranges. 3 Proof. The construction used in Case 1 of 39 estab- lishes the existence of a range which is perspective with each of the two given coplanar ranges. It is now necessary to show that the points P l and P 2 can be so selected that the line AB C shall be any line which does not pass through the intersection of p 1 and p y Let p be any line not concurrent with p 1 and p y Let the intersection of p l and p be D v and that of p z and p be E y Suppose that the point Z> 2 corresponding to D v and the point E l corresponding to E v have been found by the method of Case 1 of 39, or by any other suitable method. The problem now is to choose points P 1 and P 2 such that the line ABC shall coincide with the line p or D^E^ Draw D 2 D V and let this line meet A 2 A 1 in P y Draw EE and let this line meet A and let these lines meet in P v Join P l to 7 2 in 5, C, . . .. PERSPECTIVE RANGES 53 Then range ABCD^ - = range Hence range ABCD^E^ - -^ range But the triad AD^E^ is perspective with the triad A^D^Ey and these triads determine completely the nature of the projecti vity between the range ABCD^E^ and the range Therefore the two ranges A 1 B^C 1 D 1 E 1 on p 1 and J 2 .# 2 C 2 Z> 2 # 2 ... on p 2 are perspective with a range on the given line p. Exercise 15. Projective and Perspective Forms 1. State and prove the dual of 43. State and prove the duals of each of the following : 2. 44, in the plane. 5. 45, in the plane. 3. 44, in space. 6. 45, in space. 4. Ex. 3, in the bundle. 7. Ex. 6, in the bundle. Solve the duals of each of the following : 8. Ex. 1, page 49, in space. 9. Ex. 3, page 49, in the plane. 10. Ex. 3, page 49, in space. 11. Ex. 1, page 49, in the plane, considering the case in which the ranges are coplanar. 12. If three ranges in a plane have concurrent bases and if two of the ranges are perspective with the third, these two ranges are perspective with each other. 13. Three fixed lines p l ,p 2 ,p 3 radiate from A I} one of three fixed collinear points A^ A. 2 , A 3 . A point P a moves on j^, and the lines Af^ A Z P 1 cut p^ p a respectively in P 2 , P 3 . Show that P 2 P 3 has a fixed point. 54 PROJECTIVITIES OF PRIME FORMS 46. Special Case. In the construction described in 39 the only limitation upon the position of the points P and P^ was that P l should not be at A 1 and that P% should not be at A y The necessity for this limitation is due to the complete failure of the construction which would result from the coincidence of the lines PiA v ^-Bp P\C r We shall now consider a special case which is noteworthy because it puts in evidence a line that has important relations to all pairs of corresponding points of the ranges and furnishes a basis for several simple constructions. Let P l be taken at A v and P 2 at A v Then the line p is determined by the intersection of P 1 B 1 (^A Z B 1 ) and (A^B^) and the intersection of ^C 1 (^A^C l ) and It contains likewise the intersections of A Z D V A^ 2 ; A 2 E V A^^\ and so on. It can now be shown that this line p can be located independently of the placing of P l and P% on the particular line A-^A V If the ranges A 1 B 1 C l and A Z B 2 C 2 are projective but not perspective, the point of intersection of p l and p z will not be self-corresponding. Regarding this point as belonging to the range A l B 1 C l , call it X r In the other range there will correspond to X l a point X v Regarding the intersection as a point of range A^B^C Z , call it F 2 . To it will correspond a point Y 1 of the first range. Then A^XZ and A 2 X 1 intersect on the line p. But their inter- section is X. PERSPECTIVE EANGES 55 Similarly, Y v the intersection of A-^Y^ and A^Y V is on the line p. Accordingly the line p is the line determined by the points of the first and second ranges which correspond to the point of intersection of the ranges regarded as a point of the second and first ranges respectively. It follows that the line p is determined by the projectivity between the ranges and bears the same relation to any two correspond- ing points of the ranges that it does to A v A v Accordingly not only do A^B V A^\ A^C V A^\ A^ v A^; - intersect on p, but so do B^C V B^C^ B^D V B^D^', ; C^DV C^D^ C^E V C^E^ ; and so on. If the ranges A^B^ and A^B Z C Z - are perspective, their common point X is self -corresponding. Let p be the line joining X to the intersection of A^B% and A z B r We then see that the flat pencils A^A^B^C^ - X - ) and A^A 1 B 1 C 1 . . . X ) are protective and have a com- mon line A^A V Hence these pencils are perspective, and the axis of perspective is p. Hence, as in the other case, A^B V A^-, A^C V Afy AJ) V A 2 D i; . ; B^Cy B^ ; . . .; C-^Dy C^D 1 all intersect on p. Exercise 16. Perspective Forms 1. If a simple reentrant hexagon has its first, third, and fifth vertices on one straight line of a plane IT and has its second, fourth, and sixth vertices on a . 7 A A second straight line of that plane, the intersections of the first and fourth, the second and fifth, and the third and sixth sides are collinear. 2. State and prove the proposition dual in a plane to Ex. 1. 3. State and prove the proposition dual in space to Ex. 1. 4. State and prove the proposition dual in the bundle to Ex. 3 56 PKOJECTIVITIES OF PRIME FORMS PROBLEM. LINE TO AN INACCESSIBLE POINT 47. To draw a line which shall pass through the inacces- sible intersection of two given straight lines and also through a given point of their plane. P A z B 2 C z Solution. To draw a line which shall satisfy the first condition, let the given lines be p 1 and p y Select a point P in the plane of the lines p l and jt> 2 , but not on either line, and through P pass three lines cutting p l in A v B v C and p 2 in A v B v C r Since the ranges A 1 B 1 C 1 and A 2 B^C 2 are perspective, the pairs of lines A^B^, A^B^, A^C^, A^C-^ B 1 C Z , B^C^ inter- sect on a line p which passes through the intersection of the lines p l and p% ( 46). Now to satisfy the second condition we must so choose the point P that the line p will pass through 0. Through O draw two lines and let them intersect p 1 in A 1 and B l and intersect p 2 in A^ and B y Take P to be the intersection of A V A Z and B^By Through P draw an- other line cutting p 1 in C and p^ in C y Then the line p which is determined by the intersections of two of the three pairs of lines A^B^, A^B^ A^C^, A^C^, B^Cn, B^C l satisfies the two given conditions. PROBLEMS OF CONSTRUCTION 57 PROBLEM. LINE PARALLEL TO A GIVEN LINE 48. Given two parallel lines and a point in their plane, to draw through the point a line parallel to the given lines. The solution is left for the student, who should write out the proof after the method used in 47. The case in 49 is somewhat different. PROBLEM. LINE PARALLEL TO A GIVEN LINE 49. Griven in a plane a point 0, a line p^ not passing through 0, and a parallelogram none of whose sides is known to be parallel to p v to draw through a line parallel to p^. Pi A l Solution. This problem may be reduced to the preced- ing one as follows: Let the adjacent sides a 1 and b 1 of the parallelogram meet p l in A l and B 1 respectively. Let C be any point on the diagonal P^P V and let the lines CA l and CB^ meet a 2 and 5 2 in A 2 and B z respectively. Then the triangles P 1 A 1 B 1 and P Z A 2 B 2 are homologic, and since the intersections of PiA v P%A Z and of PiB v P 2 B 2 are at infinity, the axis of homology is at infinity. Hence A%B 2 is parallel to p r We now have two parallel lines A%B 2 and p v and by means of 48 we can draw through the given point a line parallel to the given line p r 58 THEOREM. SIMILAR RANGES 50. If in two protective ranges the points at infinity are corresponding points, the ratios of all pairs of corresponding segments are the same, and conversely. Proof. Let the points at infinity of two ranges be J^ and <7 2 , and suppose that the points J v A v B v C v of the first range correspond respectively to the points J v A v 2? 2 , (7 2 , of the second range. Then (^(V,) = (^CV/i)' or A 2 C 2 :B^C 2 = A l C^ : B l C r 22 Hence A% C 2 : A l C l = B 2 C 2 : B l C r Similarly, it can be shown that -^22 : ^11 ^-^2^2 * "^1^1 == ^2 2 ' ^\~\ = = J5 2 (7 2 : B l C l = B 2 D 2 : B l D l = =r, where r is independent of the pair of corresponding seg- ments involved. Conversely, suppose that all the pairs of corresponding segments have the same ratio. Let A v A z ; B v B 2 ; C v C 2 ; J v K 2 be pairs of correspond- ing points, t/j being the point at infinity of the first range. Then or A 1 C 1 _ But A l C l : A 2 C Z = S 1 C^ : B^C V Therefore A 2 K Z : B^K 2 = 1, and hence K^ must be at infinity and should be called J v Accordingly 7j and J 2 , the points at infinity of the two ranges, are corresponding points. SIMILAR AND CONGRUENT FORMS 59 51. Similar Ranges. Two projective ranges whose points at infinity are corresponding points are said to be similar. An example of similar ranges is furnished by sections of a flat pencil by parallel lines, or by sections of a flat pencil of parallel rays by any two lines. If similar ranges are on the same base, the point at infinity is a self-corresponding point. If the ratio of corre- sponding segments ( 50) is 1:1, there is no other such point, but in any other case a second point exists. 52. Congruent Ranges. Similar ranges in which the ratio of corresponding segments is unity are said to be congruent. If two similar ranges have parallel bases, their common point at infinity is self-corresponding and the ranges are perspective. 53. Similar and Congruent Pencils. The terms similar and congruent are applied to certain special cases of flat pencils and axial pencils, the mere mention of this fact being sufficient for the present treatment. Two projective flat pencils whose bases are at infinity are said to be similar if linear sections of these pencils are similar ranges. In this and in similar cases the student should draw the figure. _/ Two projeetive flat pencils whose bases are in the finite part of the plane are said to be equal or congruent if every pair of lines of one pencil contains an angle equal to the angle contained by the corresponding pair of lines of the other pencil. Similar and congruent axial pencils are defined in the same way. Flat pencils and axial pencils are said to be proper or improper according as their bases are -or are not in the finite part of space. 60 PROJECTIVITIES OF PRIME FORMS Exercise 17. Projectivities 1. There exist infinitely many sets of projections and sec- tions which connect a prime form with itself. 2. There exist infinitely many sets of projections and sec- tions which connect two projective prime forms. 3. A plane quadrangle is projective with a properly chosen parallelogram. 4. A plane quadrangle is projective with a properly chosen square. 5. A plane quadrangle is projective with every square. 6. Every two plane quadrangles are projective. Specify a set of operations that connects the quadrangles. 7. A triangle and any point in its plane, but not in its perimeter, are projective with any equilateral triangle and its center. 8. Solve the duals of Ex. 6. 9. Two ranges on the same base are projective if they are so related that every pair of corresponding points is a harmonic conjugate with respect to a given pair of points on the base. Apply Ex. 12 on page 30. 10. If J l and K Z are the points at infinity of two projective ranges A l B 1 C l and A 2 B Z C 2 , J 2 and K l being the points corresponding to them, then A l K l A 2 J^ = B l K l B 2 J 2 = -. 11. If the ranges in Ex. 10 are on the same base, and if 1 is the midpoint of J 9 K l and if # 2 is the point corresponding to O v then 1 X 1 0^ 2 - 0^ . 0/ 2 - Ofc O/ 2 + O& O& = 0. 12. If A l and A 2 in Ex. 11 coincide with A, it follows that 0/ 2 + 0^.0^=0. 13. If P 4 is a fixed point on the side P^P Z of a given triangle J^PjPg, and if a moving line cuts Pf z in P 4 , P 2 P g in Q, and PgPj in R, PjQ and P 2 /e meeting in P, then P 8 P cuts Pf t in a fixed point. PEOJECTIVITIES 61 14. Generalize Ex. 12, page 53. 15. Given A 1 B 1 C 1 -, A^B Z C Z . -, and A s B a C a , three pro- jective but not perspective ranges whose bases p v p 2 , p a are coplanar and concurrent, prove that, if any triad of corre- sponding points is collinear, there exists a range ABC which is perspective with each of the three given ranges. 16. If Pj, P 2 , P 8 in Ex. 15 are the centers from which the three ranges on p v p 2 , p a respectively are projected into the range ABC , each of the sides of the triangle PjP 2 P 3 cuts a pair of the given ranges in points which correspond in one of the projectivities. 17. If in Ex. 16 P moves along the base p of the range ABC , the lines P t P, P 2 P and P 3 P trace on any fourth line of the plane three projective ranges which have one point that is self-corresponding for all three projectivities. 18. If X l} X 2 ; Y v F 2 ; A V A 2 are pairs of corresponding points of two coplanar projective ranges, and if ( 46) X l and F 2 coin- cide at the intersection of the bases, determine the position of B Z which corresponds to any fourth point B 1 of the first range, basing the solution on 46. 19. Solve the plane dual of Ex. 18. 20. Solve the space dual of Ex. 18. In solving the remainder of the problems in this exercise, use only the ungraduated ruler. These exercises are chiefly due to Steiner, one of the founders of the science of projective geometry. 21. Given a segment AB extended to twice its length, divide it into any number of equal parts. 22. In an indefinite straight line given a segment AB divided at C in the ratio of two given whole numbers, draw through any given point a line parallel to the given line. 23. Given two parallel lines p and p', and given onp a seg- ment AB, from any given point on p lay off a segment of p equal to any given multiple of AB. 24. Divide AB in Ex. 23 in the ratio of two given integers. 62 PROJECTIVITIES OF PKIME FORMS 25. Given a parallelogram, divide any segment in its plane in the ratio of two given integers. 26. In a plane, given three parallel lines which cut a fourth line in the ratio of two given integers, draw through a given point a line parallel to a given line. 27. If two parallel line segments which have to each other the ratio of two given integers are given, draw a line parallel to a given third line. 28. Given two parallel lines and a segment divided in the ratio of two given integers, draw through a given point a line parallel to a given fourth line. 29. Given two nonparallel segments, each divided in the ratio of two given integers, draw a line parallel to a given line. 30. Given a square, a point, and a line, all in one plane, draw through the point a line perpendicular to the given line. 31. Given a square and a right angle, both in the same plane, bisect the angle. 32. Given a square and an angle, both in the same plane, construct any multiple of the angle. In each of the following problems, in addition to the data mentioned, one circle fully drawn and its center are assumed to be given. 33. Through a given point draw a line parallel to a given line. 34. Through a given point draw a line perpendicular to a given line. 35. Through a given point draw a line which shall make with a given line an angle equal to a given angle. 36. From a given point draw a line parallel and equal to a given line. 37. Determine the intersections, if any, of a given line and a circle of given center and radius. 38. Determine the intersections, if any, of two circles of given centers and given radii. CHAPTER VII SUPERPOSED PROJECTIVE FORMS 54. Superposed Projective Forms. Hitherto only inci- dental reference has been made to the existence of pro- jective forms on the same base, but the general results obtained apply to them except when the contrary is in- dicated. It is proposed now to consider some special properties of one-dimensional protective prime forms on the same base. Such forms are called superposed forms. In the first place, the existence of superposed projective forms is established if, when two projective ranges are on different bases, as A^B 1 C 1 on the base p 1 and A 2 B Z C 2 - on the base jt) 2 , the second range is projected on the base p 1 from a center P so taken as not to be on the line A-^A^- The result is a range A[B[C[ on the base p l which is pro- jective but not identical with A 1 B 1 C 1 , since A 1 and A[ are not coincident. In the second place, if each of two superposed projective forms is operated upon by section or projection, the result- ing forms are superposed and projective. 63 64 SUPERPOSED PROJECTIVE FORMS THEOREM. EXISTENCE OF SUPERPOSED PROJECTIVE FORMS 55. There exist pairs of projective prime forms which have common bases and which have not all their corre- sponding elements coincident. The proof of this theorem is evident from 54. THEOREM. INVARIANCE UNDER PROJECTION 56. If from two superposed projective prime forms two other superposed prime forms are obtained by projection or by section, the latter forms are projective. The proof of this theorem is evident from 54. 57. Self -corresponding, or Double, Elements. It has been shown ( 39) that three points A v v C of a line are projective with any three points of the same line, even though some or all of the two sets of three are the same ; and it follows from 40 that the correspondence of the three pairs of points establishes for all points of the line a projectivity in which some of the points may coincide with their corresponding points. It is evident that similar con- siderations apply to flat pencils and to axial pencils. Elements of superposed projective prime forms that coincide with those to which they correspond are called self- corresponding elements, or double elements', and the deter- mination of the number of such elements is a problem of importance. Since from superposed flat pencils or axial pencils we may by section obtain superposed ranges whose self-corresponding points are situated on the self- corresponding elements of the pencils, the discussion of this question is substantially the same for all one-dimensional prime forms, and hence will be limited to the case of superposed ranges. SELF-CORRESPONDING ELEMENTS 65 THEOREM. SELF-CORRESPONDING ELEMENTS 58. The number of self -corresponding elements of two dis- tinct superposed projective one-dimensional prime forms is two, one, or none, and all three of these cases occur. Proof. The proof consists of the following four parts : 1. There may be two self -corresponding elements. If A v B v C v C 2 are four points on a line p, the triads A^i ^i an d AA^2 determine a projectivity between dis- tinct ranges on p with A 1 and B v but not with either C l or (7 2 , as self-corresponding points. 2. There may be just one self-corresponding element. On a line p take four points A, B v B v C v and through A pass two lines a and p'. On a take any point 1^. Let the line J^B 2 cut p' in B', the line B'B l cut a in ^, the line J^C l cut p' in C', and the line JFJC" cut p in (7 2 . The triads AB 1 C 1 and AB 2 C 2 determine a projectivity between distinct ranges on p in which the point A is self- corresponding. Moreover, each of these ranges on p is perspective with the range AB'C' on p' and hence, if t/! and C/2 are corresponding points of the range on p v it follows that ^ ?7 2 and P^ U^ must intersect on p'. This would not happen if ZTj and C7" 2 were coincident, except at A. Therefore A is the only self -corresponding point. 66 SUPERPOSED PllOJECTIVE FORMS 3. There may be no self-corresponding element. Taking a range A^B^ ... on a base p, let its projector from a point P, not in the base, be the pencil a^b^ Through P draw the lines a a , 6 2 , c v , making any fixed angle, say 30, with a v b v c v respectively, and let these lines meet the base p in J 2 , B v C 2 , . Then range A^C^ -% pencil ajft - pencil a 2 6 2 J. Because the latter range is perspective with A 1 B l J, the point J at infinity must be self-corresponding ( 44). Therefore the range A 2 B Z J is projective with the range A[B'^ ' J in such a way that the point J at infinity is self-corresponding, and hence these ranges are perspective. The center of perspective of these ranges must be on A[A 2 . Let it be P r Then B 2 must be the intersection of B^ and p. Draw P^P V It will be proved that PP% is parallel to p and p'. If P^ is not parallel to p and p', let it meet p in X 1 and p' in X[. Then range A 1 B 1 X l J= range A[B[ X[ J = range A 2 B 2 . X l ... J. Hence the given ranges have two self-corresponding points X l and J, which is contrary to the hypothesis, and so PP^ is parallel to p. From similar triangles it follows that A[B and hence that Similarly, any two corresponding segments are equal. Hence the ranges are congruent. Compare the above figure with the one in 58. 70 SUPERPOSED PKOJECTIVE FOKMS THEOREM. ANGLE SUBTENDED BY CORRESPONDING POINTS 63. Given two superposed protective ranges having no self-corresponding point, it is possible to find a point at which all pairs of corresponding points in the range subtend equal angles. Proof. Let A-^B^ and A%B 2 C Z be projective ranges on a base p, and let 7j and K 2 be the point at infinity of p. Let ^ and J 2 be the points of the two ranges which correspond to the point at infinity. Bisect K^JI in O r and let 2 of the second range corre- spond to Oj of the first range. Then, J^ and K z being the point at infinity and A v B l being any points of the first range, we have 22 Hence (Oi^i - MO ( OB V if each product is negative, it follows that the point separates every pair of corresponding points. Moreover, if OA^ is longer than OB V it is evident that OA 2 is shorter than OB%, and so any two pairs of corresponding points, as A v A 2 ; B v B v mutually separate each other. In Case 2 by similar reasoning we establish that the involution is hyperbolic with the self-corresponding points equidistant from O, that any two corresponding points lie in the same direction from 0, and that no two pairs of corresponding points mutually separate each other. These metric properties are sometimes used to define elliptic and hyperbolic point involutions. Exercise 19. Point Involutions 1. Choosing two pairs of corresponding points that will determine an elliptic involution, find by construction a third pair of corresponding points and also find the center. 2. Solve Ex. 1 for a hyperbolic involution. 3. Find a pair of corresponding points of an involution of which X (a given point) and J (the point at infinity) are the self-corresponding points. 4. Given one self-corresponding point of an involution and a pair of corresponding points, find by construction the other self-corresponding point. 76 SUPERPOSED PROJECTIVE FORMS THEOREM. LINE INVOLUTION 74. Every involution of lines has one pair of correspond- ing lines at right angles ; if it has two pairs at right angles, all its pairs are at right angles. Proof. Let a v 2 ; b v 2 be pairs of corresponding lines of an elliptic involution on a base P. If a v 2 or b v b 2 are at right angles, the first part needs no proof. If neither of these pairs of lines is at right angles, cut the four lines by any line p in the points A v A v B v B y Describe the circles PA^A^ and PB^B^ meeting again in Q. Let PQ cut p in O, and let the perpendicular bisector of PQ cut p in V. Describe the circle whose center is V and whose radius is VP, or VQ, and let it cut p in C l and C y Then 0^ OC' 2 = OP-OQ = OA l OA 2 = OB 1 . OBy Hence the lines c v or PC V and c v or PC Z , belong to the original involution. Furthermore, they are at right angles, because C^PC^ is a semicircle. For the case of a hyperbolic involution see 70. LINE INVOLUTION 77 Again, if two pairs of corresponding lines are at right angles, these pairs separate each other, and the involution is elliptic. Suppose now that the lines a v 2 and also the lines b v 5 2 are at right angles. Then the segments PA^A Z and PB^B^ are semicircles, and their common chord PQ is at right angles to p at 0. Now if c v and P a A a form a triangle with its vertices on the three given lines ? 17. What modifications of the data in Ex. 15 render it cer- tain that in all positions the lines P^ r P 2 A 2 , and P A form a triangle with its vertices on the same three given lines ? PART II. APPLICATIONS CHAPTER VIII PKOJECTIVELY GENERATED FIGURES 75. Statement of the General Problem. Application of the properties of prime forms will now be made to the study of a problem which is connected with a somewhat wide range of topics in geometry. This problem, to the discussion of which the remainder of the book is devoted, may be stated in these words: To determine the character of all geometric configurations whose generating elements are determined by corresponding elements of two protective one-dimensional prime forms. The problem divides naturally into a number of cases according as the two projective one-dimensional prime forms are any one of the following pairs: 1. Two ranges, considered in 84 and 85. 2. Two flat pencils, considered in 86-89. 3. Two axial pencils, considered in 90 and 95. 4. A range and a flat pencil, considered in 96. 5. A range and an axial pencil, considered in 97. 6. A flat pencil and an axial pencil, considered in 98. 76. Locus of a Point. If a point moves in space subject to a given law, the figure consisting of all the points with which the moving point may coincide, and of no others, is called the locus of the point. 79 SO PROJECTIVELY GENERATED FIGURES 77. Envelope of a Plane. If a plane moves in space subject to a given law, the figure which is tangent to all the planes with which the moving plane may coincide, and to no others, is called the envelope of the plane. 78. Envelope of a Line. If a line moves in a plane sub- ject to a given law, the figure which is tangent to all the lines with which the moving line may coincide, and to no others, is called the (plane) envelope of the line. 79. Generation of a Figure by a Line. If a line moves in space subject to a given law, the figure consisting of all the lines with which the moving line may coincide, and of no others, is said to be generated by the moving line. 80. Order of a Figure. The greatest number of points of a figure that lie on a line which is not entirely in the figure is called the order of the figure. Thus a circle may be met by a line in two, one, or no points. Consequently the order of the circle is two. Similarly, the order of a straight line is one. The order of a plane is also one, while that of a sphere is two. 81. Class of a Figure in Space. The greatest number of tangent planes of a figure in space which pass through a line that does not have all the planes through it tangent to the figure is called the class of the figure. Thus a sphere may have tangent to it two, one, or no planes which pass through a straight line, and hence the sphere is of class two. 82. Class of a Figure in a Plane. The greatest number of tangent lines which can be drawn to a plane figure from any point in its plane is called the class of the figure. Thus, of the lines in a plane which pass through a given point two, one, or none may be tangent to a given circle, and hence the circle is of class two. DEFINITIONS 81 83. Dual Elements. From the point of view of the Prin- ciple of Duality the following are corresponding elements : 1. In geometry of three dimensions, a point on a figure and a plane tangent to a figure. It can also be shown that a line on a figure is self-dual. 2. In geometry of the plane, a point on a figure and a line tangent to a figure. 3. In geometry of the bundle, a line on a figure and a, plane tangent to a figure. Exercise 21. Preliminary Definitions 1. What is the envelope of a system of tangents to a given circle ? What is the dual of a circumscribed polygon ? 2. What is the order and what is the class of the projector of a circle from a point not in its plane ? The student may consult the chapters on higher plane curves in texts on elementary analytic geometry, such as the one in this series, and deter- mine the orders and the classes of the curves considered there. 3. In Ex. 2 what is the order and what is the class of any plane section of the figure ? 4. Find the surface generated by a line so moving as to be constantly parallel to and at a given distance from a given line, and state the order and the class of this surface. 5. In Ex. 4 consider the various plane sections of the sur- face, stating the order and the class of each. 6. Find the locus in space of a point which so moves as to be constantly at a given distance from the nearest point of a given line segment, stating the order and thq class of the locus. 7. Find the order and the class of the plane sections of the figure obtained in Ex. 6. 8. Find the envelope of a plane which so moves as to be constantly at a given distance from the nearest point of a given line segment, stating the order and the class of the locus. 82 PROJECTIYELY GENERATED FIGURES THEOREM. RANGES WITH A COMMON ELEMENT 84. The envelope of the line which so moves as always to contain corresponding points of two coplanar projective ranges is of the second class, unless the ranges are superposed and without a self-corresponding point, in which case the envelope is the common base. If the ranges are perspective, the enve- lope consists of two points, one of which is common to the ranges. If the ranges are not perspective and not superposed, the envelope is tangent to the base of each range at that point of it which corresponds to the point common to the ranges ichen this common point is regarded as belonging to the other range. Proof. The two projective ranges referred to in 75 may or may not have one common element, and in this theorem we consider two ranges having such an element. In this case the ranges must evidently be coplanar. The element determined by two corresponding points of the ranges is a straight line, and hence in this case the problem is that of determining the envelope of a line which so moves as always to contain two corresponding points of two coplanar projective ranges. These ranges maybe (1) superposed ; (2) not superposed but perspective; (3) neither superposed nor perspective. 1. Let the ranges be superposed. Then all pairs of corresponding but not self-corresponding points determine the common base of the ranges ; and the lines through any self-corresponding point are infinitely many. The only figure that has all these lines and no others as tangents consists of the one or two self-corresponding points or, if there is no self -corresponding point, consists of the common base of the ranges. RANGES WITH A COMMON ELEMENT 83 2. Let the ranges be not superposed but perspective. Then all lines determined by distinct corresponding points pass through the center of perspective, and every line through the point common to the ranges joins that point to its corresponding point, that is, to itself. Hence the envelope consists of two points ; namely, the center of perspective and the point common to the ranges. 3. Let the ranges be neither superposed nor perspective. To the common point X v or F 2 , there correspond in the ranges two points X 2 and Y r Since the base p l of the first range joins Y 2 to Y v the enve- lope is tangent to p ; similarly, it is tangent to p y Consider now F x and F 2 , two corresponding points nearly co- incident with X l and X 2 respec- tively. The line v joining them is nearly coincident with p 2 . If, now, Fj approaches coincidence with X v v and F 2 approach p 2 and X y But if a moving tan- gent to a curve approaches a fixed tangent as a limiting line, the intersection of these two approaches the point of contact of the fixed tangent as a limiting point. Hence in this case Xj is the point of contact of p 2 with the envelope. Similarly, Y l is the point of contact of p r Finally, the class of the envelope is two ; for two tan- gents to the envelope, namely, p^ and A^A V pass through a point A^. If through any point there should pass three tangents to the envelope, would be a center of perspec- tive for the ranges ; but the ranges, are not perspective. 84 PROJECTIVELY GENERATED FIGURES THEOREM. RANGES WITH No COMMON ELEMENT 85. The lines which join corresponding points of two pro- jective ranges that have no common point are the intersections of corresponding planes of two projective axial pencils which have no common plane. Proof. If two projective ranges have no common ele- ment, their bases cannot meet, and therefore the ranges cannot be coplanar. Let A 1 B 1 C\ - and A 2 B 2 C 2 be projective ranges on the bases p l and p z which are not coplanar. Let a v /8j, y v be the planes determined by the line p 1 and the points A v B v (7 2 , respectively, and let 2 , /3 2 , 7 2 , be the planes determined by the line p 2 and the points A v B v (?! respectively. Then we have axial pencil a l ^ l ^ l -^ range A 2 B 2 C 2 axial pencil a 2 # 2 7 2 .... Also the line A t A 2 is the intersection of the planes a l and 2 . Moreover, if the axial pencils had a common plane, the ranges along their axes would both be in this plane. But this is contrary to hypothesis. Hence the proof is complete. As a consequence of this theorem the discussion of the figure which is generated by these two projective ranges may be deferred until we consider the figure which is generated by projective axial pencils that have no common plane ( 95). RANGES AND PENCILS 85 THEOREM. COPLANAR FLAT PENCILS 86. The locus of the point which so moves as always to be common to two corresponding lines of two coplanar protective flat pencils is of the second order, unless the pencils are super- posed and without a self-corresponding line, in which case the locus is the common vertex. If the pencils are perspective, the locus consists of two straight lines, one of which is common to the pencils. If the pencils are not superposed, the locus con- tains the base of each pencil and at each of these points is tangent to the line of the corresponding pencil which corre- sponds to the common line of the pencils when this common line is regarded as belonging to the other pencil. Proof. Two flat pencils may or may not have a common base. In the former case the common base may be a plane containing both pencils or a point which is the vertex of both pencils. We shall now deal with the first of these subcases, the second being discussed in 87. Essentially, then, the theorem involves the problem of finding the locus of the intersection of two coplanar pro- jective flat pencils. It is evident that these pencils may be (1) superposed ; (2) not superposed but perspective ; (3) neither superposed nor perspective. 1. Let the pencils be superposed. In this case the flat pencils have in common not only their planes but also their vertices. The points common to the corresponding lines include in any case the common vertex and also include all points of any self -corresponding lines of the pencil. Hence in this case the locus is the one or two self-corresponding lines of the pencils or, in case there is no self -corresponding line, the common point of all the lines of the pencil. 86 PKOJECTIVELY GENERATED FIGURES 2. Let tlie pencils be not superposed but perspective. In this case the pencils have a self-corresponding line, all the points of which are in the locus. In addition the intersections of pairs of corresponding lines are in the locus. But these intersections are on a straight line, and accordingly the locus is a pair of straight lines. 3. Let the pencils be neither superposed nor perspective. Since the common line is not self-corresponding, let it be called x l and y In the two pencils it has corresponding to it the lines y l and x v The base I{ of the first pencil, being the intersection of the lines y l and y v is on the locus. Similarly, the base P 3 of the second pencil is on the locus. Consider now a line v^ that is nearly coincident with x v and the corresponding line v 2 that is nearly coincident with x v The point Voi the locus determined by v 1 and v 2 is nearly coincident with P v If, now, the point V approaches P i along the locus, the line z> 2 , or VP V approaches x y But the limit- ing position of a secant VP^ as V approaches P^ is the tan- gent to the locus at P v Hence the locus is tangent to the line :r 2 at P Y Similarly, it follows that the line y 1 is tangent to the locus at P r RANGES AND PENCILS 87 Finally, the order of the locus is two; for any line of the first pencil, as a v meets the locus at I{ and at A, the intersection of a x and its corresponding line a 2 . Moreover, if any line o cuts the locus in three points, the triads of lines of the pencils which would intersect in these points would be perspective, and so would the complete pencils. But the pencils are not perspective. The theorem is, therefore, completely proved. Exercise 22. Ranges and Pencils 1. Two fixed lines AO^R and CO^D are each perpendicular to 0^0^ and a moving line cuts them in P X and P 2 so that the ratio Of^: a P 2 is a constant. Find the envelope of the moving line. Consider the case in which O 1 P 1 and O g P 2 are equal. Compare the ranges traced by P x and P 2 . 2. Two fixed lines intersect at right angles at O, and a moving line cuts them at equal distances from O. Find the envelope of the moving line. 3. Examine Ex. 1, substituting the condition that O^P^ and # 2 P 2 maintain a constant difference. 4. Examine Ex. 2, substituting the condition that the dis- tances from maintain a constant difference. 5. Examine Ex. 2, substituting the condition that one dis- tance exceeds a given multiple of the other by a fixed amount. 6. Two lines revolve at the same angular velocity in opposite senses about the fixed points O l and 2 respectively. Initially they make angles of 90 and 45 respectively with the line O^O^. Find the locus of their intersection. 7. Consider Ex. 6, substituting the condition that initially the lines coincide. 8. Examine Exs. 6 and 7 on the assumption that the lines revolve in the same sense. 88 PROJECTIVELY GENERATED FIGURES THEOREM. FLAT PENCILS WITH A COMMON VERTEX 87. The envelope of the plane, which so moves as always to contain corresponding lines of two protective flat pencils that are not coplanar but have a common vertex, is of the second class. If the flat pencils are perspective, the envelope consists of tivo straight lines, one of which is common to the pencils. If the pencils are not perspective, the envelope is a surface tangent to the plane of each flat pencil along the line which corresponds to the common line of the pencil when this common line is regarded as belonging to the other flat pencil. All the planes and the surface generated pass, through the common vertex of the pencil. Proof. Since each pair of corresponding lines of the given flat pencils determines both a point and a plane, we are concerned with the problem of finding the aggre- gate of elements, either points or planes, determined by corresponding lines of two projective flat pencils which have a common vertex. The first of these cases is trivial. It may here be assumed that the flat pencils are not coplanar, as the case of coplanar pencils has just been discussed. From the point of view of loci the points determined by corresponding lines either will be the common vertex alone or, if the common line happens to be self-corresponding, will be this common line itself. FLAT PENCILS 89 On the other hand, any two corresponding lines deter- mine a plane which passes through the common vertex. If we cut the whole configuration by a plane that does not pass through the common vertex, we obtain two pro- jective coplanar ranges and the lines joining corresponding points. This section of the envelope is then one of the figures described in 84. Consequently the envelope sought is the projector, from the common base of the two flat pencils, of one of these figures. If the flat pencils are perspective, this projector consists of two straight lines through the common vertex. If the flat pencils are not perspective, the projector is a conic surface and is tangent to the plane of each flat pencil. This is evident from the fact that the plane of either pencil is determined by the common line of the pencils and re, that one of its own lines which corresponds to the common line. The plane of this pencil has the line x in common with the envelope. Through the common line of the pencils there pass the two planes of the pencils, and these are tangent to the envelope. If through any line there should pass more than two tangent planes, the given pencils would be perspective. 88. Flat Pencils having no Common Base. The discus- sion of the case of projective flat pencils that have a com- mon line but are not coplanar and do not have a common vertex can be given quickly. No intersection of corre- sponding lines can occur except on the common line. The common line may be self-corresponding, in which case any point on it is common to corresponding lines, and any plane through it contains corresponding lines. If the common line is not self-corresponding, there is in each pencil a line corresponding to it. With these lines it deter- mines two points, the vertices of the pencils, and two planes, the planes of the pencils. 90 PROJECTIVELY GENERATED FIGURES 89. Projective Flat Pencils having no Common Element. Figures generated by means of projective flat pencils hav- ing no common line are so simple as not to demand special study. They will be noticed in passing, but no formal theorem regarding them need be stated. Since pencils which lie in the same plane or in differ- ent planes that intersect in a line of the pencils have a common element, it follows that the pencils under consider- ation lie in planes whose intersection does not pass through the vertex of either pencil. Upon this line the pencils determine ranges which may be either identical or distinct. If the ranges are identical, their common base is the locus of the intersections of corresponding lines of the pen- cils. Likewise each pair of corresponding lines determines a plane through the line joining the bases of the flat pencils ; and the envelope of these planes is this line. Hence the figure determined is either the line determined by the ver- tices of the pencils or the line determined by the planes of the pencils, according to the point of view. If the superposed ranges mentioned above are not iden- tical, the only pairs of corresponding lines of the flat pen- cils whicli determine elements are those which meet in the self-corresponding points of the superposed ranges. These determine two, one, or no points on the line common to the planes of the pencils, or two, one, or no planes through the line joining the vertices. Hence, from the point of view of loci, the figure generated by means of the projective flat pencils consists of two, one, or no points ; and from the point of view of envelopes, the figure consists of two, one, or no planes. Though neither of the configurations obtained has any special interest for us, it is clear that they conform in a general way to the type of figures which we obtain in the other cases. FLAT AND AXIAL PENCILS 91 THEOREM. AXIAL PENCILS WITH A COMMON PLANE 90. The surface generated by the line which so moves as always to be contained in corresponding planes of two axial pencils that have a common plane is of the second order unless the axial pencils are superposed and are without self- corresponding planes, in which case the surface degenerates into the common axis of the pencils. If the axial pencils are perspective, the surface consists of two planes, one of which is common to the pencils. If the pencils are neither perspective nor superposed, the surface contains the axis of each pencil and along each of these axes is tangent to the plane which corresponds to the common plane of the pencil when this com- mon plane is regarded as belonging to the other pencil. The generating line continually passes through the intersections of the axes of the pencils. Proof. Two projective axial pencils may have or may not have a common element. In this theorem we con- sider only the former case. Evidently the axes of the two pencils are coplanar. Then the pencils may be (1) super- posed ; (2) not superposed but perspective; (3) neither superposed nor perspective. 1. Let the axial pencils be superposed. Then all pairs of corresponding planes intersect in the axis, but any line in a self-corresponding plane may be regarded as common to two coincident corresponding planes. The surface generated by the lines common to corresponding planes consists, therefore, of one or two self-corresponding planes if there are such planes or, if there are no such planes, this surface degenerates into a line, the common axis of the pencil. This last case is of minor importance only. 92 PROJECTIVELY GENERATED FIGURES 2. Let the axial pencils be not superposed but perspective. The axes of the pencils intersect in a point O which lies in every plane of both pencils. Then the line deter- mined by any pair of corresponding planes passes through this point. Moreover, if a plane is passed so as not to con- tain this point, it cuts the axial pencils in coplanar per- spective flat pencils, and the locus of the intersections of corresponding lines of these pencils is two straight lines, one of which is the line through the bases of the flat pencils ( 86). Consequently the surface generated by the inter- sections of corresponding planes of the axial pencils is the projector of the two straight lines from the point ; that is, the surface is two planes, one of which is the common plane of the axial pencil. 3. Let the axial pencils be neither superposed nor perspective. As in the previous case, the axes p 1 and p% intersect in a point through which passes every line determined by corresponding planes of the axial pencils. A plane which does not pass through cuts the axial pencils in coplanar projective but not perspective flat pencils whose bases are P l and P 2 , the locus of whose intersections is described by 86. The surface generated by the lines common to cor- responding planes of the axial pencils is the projector from the point of the locus just mentioned. EULED SURFACES 93 91. Regulus. Any three straight lines, no two of which are coplanar, are met by infinitely many straight lines which, taken as an aggregate, are said to form a regulus. The three given lines are called the directrices of the regulus, and the lines which meet the directrices are called the generators of the regulus. REGULUS QUADRIC SURFACE SKEW QUADRIC RULED SURFACES 92. Quadric Surface. The aggregate of the points of the lines of a regulus constitute a surface called a quadric surface, and the generators of the regulus are also called the generators of this surface. There are quadric surfaces which are not constituted in this way. 93. Ruled Surface. A surface generated by the move- ment of a straight line is called a ruled surface. 94. Skew Ruled Surface. A ruled surface in which no two consecutive generators intersect is called a skew ruled surface. For a full discussion of skew ruled surfaces see 187. 94 PROJECTIVELY GENERATED FIGURES THEOREM. AXIAL PENCILS WITH No COMMON ELEMENT 95. The lines of intersection of two protective axial pencils which have no common element generate a skew ruled surface of the second order in which lie the bases of the pencil. Every section of this surface ly a plane through a generating line is a pair of straight lines. All other plane sections are curves of the second order. Proof. The proof may be divided into five parts : 1. The ruled surface is skew. Each generator intersects each axis. If two generators intersect, they determine a plane which contains two points and therefore all points of each axis. This plane must then be a common element of the pencils, which is con- trary to hypothesis. Hence no two generators intersect, and so the surface is skew. 2. The bases of the two axial pencils lie in the surface. Through any point A of the base of either axial pencil there pass all the planes of that pencil and one plane a of the other. Hence A is on the line determined by the plane a and its corresponding plane. Hence the surface contains every point of the base of either pencil. 3. Every section of the surface by a plane which does not pass through a generating line is a curve of the second order. Any plane TT which does not contain a generating line cuts the axial pencils in two projective flat pencils. If these flat pencils were perspective, their common line would be self-corresponding and the plane TT would cut two planes of the axial pencils in their common line, that is, in a gen- erating line ; and this is contrary to hypothesis. Then ( 86) the section is a curve of the second order. AXIAL PENCILS 95 4. Every section of the surface by a plane through a gen- erating line is a pair of straight lines, one of which is the generating line and the other of which meets every one of the generating lines mentioned above. Let the surface be generated by the projective axial pencils cc 1 /3 1 y 1 and a 2 /? 2 7 2 Any plane that is an element of one of these axial pencils intersects the surface in two straight lines, of which one line is the base of its pencil, and the other line is the line of intersection of this plane with the plane that corresponds to it in the other pencil. Now let a plane TT that does not belong to either axial pencil be passed through a generating line a which is deter- mined by the planes a and 2 of the axial pencils. This plane cuts the axial pencils in projective flat pencils a\c^ - and a5 2 P*> ' ' '*Pn on the respective cones are such that each plane PkPk+i contains the line of contact of the cones C^, C^ +1 . If o v o n are any fixed lines lying on C lt C n respectively, find the sur- face generated by the intersection of the planes o^p^ o n p n . 20. As a train is running along a straight track at a uniform rate, an automobile moves, also at a uniform rate, down a hill along a straight road which passes beneath the railroad. Find the figure generated by the line joining two fixed points, one on the train and the other on the automobile. 21. Initially a plane cuts two fixed intersecting planes per- pendicularly in the lines o l and 2 . As it moves it cuts these planes in the lines p v p 2 , which cut o l and o 2 at their intersection and always make equal angles with them. Find its envelope. Examine the flat pencils traced by p l and p 2 . 22. If A BCD is a regular tetrahedron, and a line so moves as always to intercept on AB and CD equal distances from .1 and C, find the surface generated by the line. 23. Consider Ex. 22 if the line continually divides AB and CD proportionally. 24. A sloping telephone wire and an electric-light pole cast upon the side of a house shadows Avhich intersect. If the wind causes the source of light to swing in a straight line, find the path traced by the intersection of the shadows. 25. Consider Ex. 24 if the source of light swings in a circle which intersects the wire and the pole. 26. Two fixed lines o l and 2 intersect and pierce a plane o> at the points O l and O 2 . Two planes TT I and 7r 2 revolve about o t and o 2 respectively so that their intersections with w describe equal angles in the same time. Find the nature of the envelope of the intersections of ir l and 7T 2 . CHAPTER IX FIGURES OF THE SECOND ORDER 100. Purpose of the Discussion. The results obtained in the preceding chapter may be given a more general as well as a more compact and symmetric form. It will be noted that these results relate to three types of figures, namely, figures in a plane ; figures in a bundle, or conical figures ; and figures in space. These three types of figures will be considered separately. 101. Plane Figures. It has already been shown that the figure obtained as the envelope of the line joining corre- sponding points of coplanar projective ranges is a curve of the second class, and the one obtained as the locus of the intersections of corresponding lines of projective flat pencils is a curve of the second order. Whether all curves of the second class and all curves of the second order may be generated in this way, whether the ranges and flat pen- cils which give rise to the curves in question have special situations relative to those curves, and what relation, if any, exists between curves of the second class and those of the second order, are questions whose answers will exhibit clearly the importance and generality of the results obtained. These questions will now be discussed, but for the sake of brevity the treatment will be limited, the parts of the argument which are omitted being indicated. The student should not lose sight of the omission, and he should later seek to complete the argument which answers the questions. 101 102 FIGUKES OF THE SECOND ORDER 102. Generalization of Results. By a line of reasoning which will not be given here the following can be proved : Every curve of the second class is the envelope of the lines joining corresponding points of two coplanar protective ranges whose bases are tangent to the curve. The application of the Principle of Duality for the plane to this result immediately leads to the further result : Every curve of the second order is the locus of the inter- sections of corresponding lines of two coplanar projective Jlat pencils whose bases are on the curve. Accordingly it is clear that the developments in the preceding chapter relate to all curves of the second class and to all of the second order. Consider the second question suggested in 101. We have already seen that the bases of the projective ranges by means of which the curves of the second class were obtained are tangent to these curves at certain of their points. It will be shown ( 106) that any two tangents to a curve of the second class may be taken as the bases of projective ranges such that the given curve is the envelope of the lines joining corresponding points of these ranges. Correspondingly, it may be shown that any two points on a curve of the second order may be taken as bases of projective flat pencils such that the given curve is the locus of the intersections of corresponding lines of these pencils ; and because to most students the idea of the locus of points is more familiar than that of an envelope, the latter proposition will be established first. The proofs of these statements regarding curves of the second order depend upon an auxiliary proposition which will now be stated and proved. This theorem will later ( 120) be generalized. GENERALIZATION OF RESULTS 103 THEOREM. AUXILIARY PROPOSITION 103. If P v P s , P 4 , P % are four points on a curve of the sec- ond order which is the locus of the intersections of correspond- ing lines of the two coplanar protective flat pencils whose bases are P l and Jj?, then the pairs of lines P^, P 4 P 5 ; determine collinear points. Proof. By hypothesis the points on the curve in ques- tion determine projective flat pencils whose bases are P l and P^. These flat pencils may be denoted by P 1 (^P 2 P S P^P^ and P^P 2 P,P 4 P^. Cut these pencils by the lines P S P 4 and P. 2 P S respectively, and let the resulting ranges be Q^^z^Q^ an d P 2 P S R 4 R 6 . These ranges are not only projective but also have a self-corresponding element P A . Hence they are perspective, and the lines P^Q^ (or P^P^, PR (or PP^), and $ 6 .# 6 pass through 0, the center of perspective. Therefore the points 6 , _R 6 are collinear with the inter- section of P 1 P 2 and PP 5 . But Q Q is the intersection of and PP V and R is the intersection of P 2 P 3 and Hence the proposition is proved. By means of the above theorem the desired proposition, known as Steiner's theorem, may now be proved. 104 FIGURES OF THE SECOND ORDER THEOREM. STEINER'S THEOREM 104. Every curve of the second order is the locus of the intersections of coplanar protective flat pencils whose bases are any two points of the curve. Proof. Let P v P b be the bases of the protective flat pencils which generate the curve, and let P v -?* ^e be anv three fixed points on the curve. Let -?3 be any point moving along the curve and occu- pying successive positions, as P%, P s ',. It will be shown that the flat pencils generated by the moving lines P^P Z and P^ are projective. For each position of P 3 the pairs of lines 7^, P^P 5 ', P^P 3 , P 6 P Q ; P Z P, P^ determine as collinear the fixed point O and the points # 6 > -#6 ( 103 )- As P% moves, ^ 6 and R 6 move along the fixed lines I^ and P 5 P 6 and trace ranges on them. Then flat pencil ^(^' - - 0^ ran g e ^e^e -(on = range ^ 6 ^-..(on = flat pencil JJ(Jg# . . .). Accordingly the curve is the locus of the intersections of corresponding lines of the projective flat pencils whose bases are P 2 and P, any two points of the curve. STEINER'S THEOREM 105 THEOREM. AUXILIARY PROPOSITION 105. If tj, t SJ t, 6 are four tangents to a curve of the second class which is the envelope of the lines joining cor- responding points of two coplanar protective ranges whose bases are ^ and 6 , then the pairs of points tfo t 4 t 5 ; t z t 3 , t 5 t 6 ; < g 4 , t^ determine concurrent lines. This theorem is the dual of the theorem of 103 and leads to the dual of Steiner's theorem. It will later ( 121) be generalized into a highly important proposition. The proof is left for the student. THEOREM. DUAL OF STEINER'S THEOREM 106. Every curve of the second class is the envelope of the lines joining corresponding points of the coplanar protective ranges whose bases are any two tangents to the curve. It is particularly important that by means of the Principle of Duality, or otherwise, the student should follow out in detail the proof of this proposition, as well as the proof of the proposition which immediately precedes it ( 105). There is not much difficulty in obtaining the steps of the proof as duals of the corresponding steps of the proof in 104, but the figure and the verification of the various steps of the argument in connection with this figure require close attention on the part of the student. 107. Relations between Curves of the Second Order and Curves of the Second Class. The two sorts of plane curves which have been obtained can now be shown to be iden- tical. In other words, it can be shown that every curve of the second order is of the second class, and conversely. Only one of these proofs will be given, since the other can be derived from it by means of the Principle of Duality. In this proof use will be made of a limiting case of the prop- osition in 103, in which two of the four arbitrarily chosen points on the curve have moved up to coincidence with the bases of the pencils, and this will first be established. 106 FIGURES OF THE SECOND ORDER THEOREM. INSCRIBED QUADRANGLE 108. If P 3 , PQ are two points on a curve of the second order which is generated by two coplanar protective flat pen- cils whose bases are P l and P 6 , the tangents at P l and P^ the tangents at P 3 and P Q , and the pairs of opposite sides 7^, P^i P S P^ PPI of the inscribed quadrangle P^P^ inter- sect in collinear points. Proof. In the figure of 103 let P% and P move along the curve so as to approach P 1 and P 5 as limiting points. Then P^ and P 4 P 5 approach the tangents to the curve at P l and P 6 respectively, and approaches the intersection of the tangents at P l and P 5 as a limiting point. The hexagon P^P^P^P^ approaches the quadrangle P^P^PQ, together with the tangents at P l and P b . During the motion of the points and lines the intersec- tions of the pairs of lines J^, P^P b \ P^, P & P 6 ; P Z P, P^ remain collinear, and the limiting positions which they approach are collinear. Since P r P 5 are not special points on the curve ( 104), it follows that the tangents at ^3, P$ meet on the same line. Then the theorem of 103 takes the form of this theorem. This proposition will be used as auxiliary to the proof of the identity of curves of the second order with those of the second class. STEINER'S THEOREM 107 Exercise 24. Steiner's Theorem and Related Theorems 1. In the theorem of 103, if P v P 2 , P 4 , P g are fixed points, and if P and P so "move that the intersection R. of P P , P P o 62 3" 56 moves on a fixed line through the intersection of P P , P P , then the intersection Q of P.P., P.P. moves on the same line, O O *' D 1 / and Q 6 , R 6 trace on this line two superposed projective ranges. 2. In Ex. 1 find the self-corresponding points of the super- posed projective ranges. 3. Prove the proposition which is related to the theorem of 108 as Ex. 1 is related to the theorem of 103. 4. Solve the problem which is related to Ex. 3 as Ex. 2 is related to Ex. 1. 5. If t s , t 6 are two tangents to a curve of the second class which is the envelope of the lines joining corresponding points of two coplanar projective ranges whose bases are t v t & , the points of contact of t v t & , the points of contact of t a , t G , and the pairs of opposite vertices t^, t 6 t 6 ; t a t & , t 6 t t of the circumscribed quadrilateral /// 6 determine concurrent lines. 6. A variable hexagon ^^^s^^s^e inscribed in a curve of the second order so moves that P.P., P.P.: P P , P f P. la 7 4 5 7 23 7 56 always intersect in fixed points and R respectively. Find the locus of the intersection of P a P 4 , P 6 P^ 7. Prove the duals of Exs. 1, 3, and 6 for the plane. 8. Solve the dual of Ex. 2 for the plane. 9. By an argument independent of that given in this chap- ter prove Steiner's theorem for the special case of the circle. 10. By an argument independent of that referred to in 106 prove the theorem stated there for the special case of the circle. 11. P 1? P 2 , P 8 , P 4 , P 5 are five fixed coplanar points no three of which are in a straight line; find the locus of a point P which so moves that the intersections of the pairs of lines PJP V P 4 P 5 ; P 2 P 3 , P 5 P ; P 3 P 4 , PP l are constantly collinear. 108 FIGURES OF THE SECOND ORDER THEOREM. IDENTITY OP CURVES 109. Every plane curve of the second order is of the second class, and conversely. N K. Proof. When a curve of the second order is given, the pencils of lines drawn from two of its points to all of its points are projective. From any three pairs of corre- sponding lines all additional pairs can be obtained by the method of 39. In particular, since the tangent at either of the two points corresponds to the line joining the two points, it can be drawn by the same method; and there- fore, when the whole curve is given, the tangents at as many points as may be desired can be drawn. Let a curve of the second order be given, and select on it any three points P r P v P y Draw P^P V Pf y P^P Z and the tangents MP^N, NP^L, and LP^M. The projectivity is determined by the triads P^M, P^P V P^ and P^P V P^L, P%P y Consider the curve as the locus of a moving point P. It will be shown that as P moves along the curve the tangent at P so moves as to meet the tangents at P l and P z in corresponding points X and Y of two projective ranges. IDENTITY OF CURVES 109 Let 0, K be the intersections of the sides P^, P Z P and %P Z of the quadrangle PPJ^ inscribed in the given curve. For this inscribed quadrangle, M is the intersection of tangents at opposite vertices and 0, K are intersections of pairs of opposite sides, and hence it follows from 108 that the points M, 0, K are collinear. By means of a like reasoning the points 0, K, Y are proven collinear. Hence the points M, 0, Y are collinear. Similarly, the points L, 0, X may be proved collinear. Since JJ, P z , P z are fixed points, the tangents at these points are also fixed. As the point P moves, so do the lines P^P, Pf, P A P, XY, P Z O, LO, MO, and the points 0, X, Y. The lines LO, MO generate perspective flat pen- cils with bases at L and M, and the points X and Y trace on the lines MI^N and LP^N ranges perspective with these pencils and hence projective with each other. Therefore, as P moves, the tangent at P so moves as always to meet the tangents at P l and P% in corresponding points of two projective ranges. The given curve is the envelope of the tangent at P and, by 104, is of the second class. The converse of this proposition being also its dual for the plane, its proof is also the dual of the above argu- ment. The theorem is therefore established. 110. Conic. A plane curve which is of the second order and second class can be shown to be a curve ordinarily called a conic section or a conic. The proof will not be given, but, independently of its other uses, the word conic will be employed to designate any curve of the second order. The use of properties of conies not deduced from the definition here given will be avoided. A conic section, or conic, is a curve of the second order and second class. 110 FIGURES OF THE SECOND ORDER THEOREM. ORDER AND CLASS OF SURFACES (IN THE BUNDLE) 111. Every surface (in the bundle) that is of the second order is of the second class, and conversely. Proof. A comparison of the problems whose discussion led to 84 and 90 shows that these theorems deal with figures and state results that, for threefold space, are dual. Similarly, the theorems of 86 and 87 are dual. It was indicated, but not proved, that 84 is true for all plane curves of the second class and that 87 is true for all plane curves of the second order. Correspondingly, 90 and 87 can be shown to apply to all surfaces (in the bundle) that are of the second class and all surfaces (in the bundle) that are of the second order. Moreover, the iden- tity of the curves of the second order with curves of the second class having been established in 109, the identity of these surfaces of the second order with those of the second class may be established by reasoning dual to that of 109. This involves the derivation of an auxiliary theorem dual for space to that of 103 and one that is dual to the limiting case of the latter as worked out in 108. The student will find that the principal difficulty is connected with the drawing of appropriate figures for these cases. In this way the classes of figures numbered 3 and 4 in 99 are shown to be identical. 112. Quadric. Since a surface (in the bundle) of the second order and second class may be thought of as gener- ated by the motion of a line that always passes through a fixed point, it is said to be a conic surface or a cone ; and on account of its order and class it is called a quadric conic surface. A quadric conic surface or quadric cone is a surface (in the bundle) of the second order and second class. ORDER AND CLASS OF SURFACES 111 THEOREM. SKEW RULED SURFACES 113. Every ruled surface (not in the plane or in the bundle) of the second order is of the second class, and conversely. Proof. In 85 and 95 certain ruled surfaces of the second order that exist in threefold space but not in the plane or in the bundle were found to be generated by means of both projective ranges and projective axial pencils. It can further be shown that all ruled surfaces of the second order can be so generated. Moreover, a complete discussion of these surfaces from the point of view of both methods of generation would have led to results similar to those obtained for the conies and for the quadric cones. Among other things it would have appeared that the planes which pass through a point P not on the sur- face, and are tangent to the surface, would generate a quadric cone, and that of these planes not more than two, but in some cases two, would pass through a line that contains P. Hence these surfaces are of the second class. The converse may also be proved. The discussion indicates that the class of figures numbered 5 in 99 is self-dual in threefold space. 114. Summary. We have now shown that the config- urations whose generating elements are determined by corresponding elements of two projective one-dimensional prime forms are as follows: 1. All plane curves of the second order and second class (conies). 2. All conic surfaces of the second order and second class (quadric cones). 3. All ruled surfaces of the second order and second class not in the plane or in the bundle. 112 FIGURES OF THE SECOND ORDER Exercise 25. Review 1. State and prove the dual for space of 103. 2. State and prove the dual for the bundle of Ex. 1. 3. Compare the dual for space of the theorem in 105 with the result of Ex. 2. 4. Give the dual for the plane of the statement and proof of Ex. 11, page 107. 5. Give the dual for space of the statement and proof of the theorem in 108. 6. Compare the space dual of the result of Ex. 5, page 107, with the dual for the bundle of the result of Ex. 5, above. 7. Derive the dual for space of Ex. 1, page 107. 8. Establish three projectivities between flat pencils which shall lead to the generation of conies having respectively two, one, and no points of intersection with any given straight line. In this connection consider 58. 9. Consider Ex. 8 for the case in which the given straight line is the line at infinity of a given plane. The student will observe that the solution of this problem establishes the existence of conies with two, one, and no points at infinity. 10. Establish three projectivities between ranges which shall lead to the development of conies having respectively two, one, and no tangents whose points of contact are on any straight line. 11. Consider Ex. 10 for the case in which the given straight line is the line at infinity of a given plane. 12. Prove the dual for space of Ex. 6 on page 107. 13. Solve the dual for the plane of Ex. 8. 14. Consider Ex. 13 for the case in which the given point is at infinity in a given direction. 15. Solve the dual for space of Ex. 8. REVIEW EXERCISES 113 16. Five concurrent lines, no three of which are in any one plane, all lie on one conic surface of the second order. 17. Prove the proposition regarding five parallel lines which corresponds to Ex. 16. 18. Establish three projectivities between axial pencils which shall lead to the generation of conic surfaces having their vertices at infinity and having as right sections curves with two, one, and no points at infinity respectively. 19. Establish between two given ranges which are not in the same plane a projectivity such that if the surface gener- ated is cut by any given plane in the finite part of space, the section shall be two straight lines. 20. In a bundle TT I} ?r 2 , 7T 8 , 7r 4 , ?r 5 are five fixed planes, no three of which are coaxial. Find the envelope of a plane IT which moves so that the planes determined by the intersec- tions of TTj, ?T 2 and 7T 4 , 7r 5 , of 7T 2 , ?T 8 and 7T 5 , TT, and of 7T 3 , ?T 4 and TT, TT I are constantly coaxial. 21. Consider Ex. 19 for the case in which the given plane is at infinity. 22. Given a plane and two projective axial pencils which have no common element, establish between other pencils a projectivity which shall lead to the generation of a surface that shall be the projector from a given center of the intersec- tion of the given plane and the surface generated by the given axial pencils. 23. Derive the dual of Ex. 22 for space. 24. Given in a plane three nonconcurrent bases p lt p^ p a passing through the points A I} A 2 , A Z respectively, specify three projectivities which connect ranges on the bases p lf p z ', P^P 9 'i p s ,p 1 respectively, and which are such that the three conies that they determine shall coincide and also be tangent to the three lines p 1} p^, p a at A v A^ A & respectively. 25. Examine Ex. 24 for the case in which the three points A., A a , A, are collinear. I/ &' o 114 FIGURES OF THE SECOND ORDER 26. In Ex. 24 select such lines p lt p^, p s and such points Aj, A 3 , A a that the conic generated shall be a circle. 27. Given in a bundle three non-coaxial planes TT V 7T 2 , 7r 3 passing through the lines a 1? 2 , 8 respectively, specify three projectivities which connect flat pencils in the planes TT V 7T 2 ; 7T 2 , 7T 8 ; 7r 8 , TT I respectively, and which are such that the three conic surfaces they determine shall coincide and shall be tangent to the planes TT^ 7T 2 , 7r g along the lines a v a. 2 , a 3 . 28. Solve the dual of Ex. 27 for the bundle. 29. Consider Ex. 19 for the case in which the section by the plane at infinity is to be two straight lines. 30. Establish such a projectivity between two given axial pencils, not in the same bundle, that if the surface generated is cut by any given plane in the finite part of space, the section shall be circular. 31. Find two axial pencils, not in the same bundle, between which such a projectivity may be established that the corre- sponding surface generated shall be cut by a given plane in a given circle of that plane. 32. Given two protective axial pencils, not in a bundle, pass a plane which shall cut the surface generated by them in two straight lines. 33. Given three bases p lt p^ p 9 in space, no two of which intersect, specify three projectivities between ranges on the bases p v /> 2 ; p. 2 ,p s ', p s , p l respectively, such that the three skew ruled quadric surfaces determined by them shall coincide. 34. Solve the dual for space of Ex. 33. 35. Develop completely the proof of the theorem corre- sponding to the theorem of 109 for the case of figures of the second order in the bundle. 36. For the case of figures of the second order in three- fold space describe accurately the figure for the theorem corresponding to that of 109, and outline the proof. CHAPTER X CONICS 115. Determination of Conies by Certain Conditions. Some of the more important properties of the curves and sur- faces to which attention has been drawn in the preceding chapters will now be deduced. The conies will be dealt with much more fully than the other figures because of their more frequent application and also because, after their properties have been set forth, the corresponding properties of the quadric cone may be obtained by means of the Principle of Duality. Chapters X-XII are devoted to the conies. In Chapter XIII there is given a discussion of quadric cones, this being confined to a few topics in addition to those suggested by the developments obtained for the conies. Notwithstanding this limitation, students should give due attention to these properties of quadric cones. In Chapter XIV will be found a brief introduction to the study of the properties of skew quadric ruled surfaces. A thorough study of these figures may well be deferred until the student has an opportunity to approach the sub- ject from the point of view of analytic geometry also, when a comparison of the analytic and synthetic treatments will heighten the interest. The first body of facts to be established relates to sets of data which completely determine conies. It constitutes the important theorem stated in 116, which consists of six simple propositions. 115 116 CONICS THEOREM. A CONIC DETERMINED 116. In a plane there is one and only one conic which has one of the following properties : 1. It passes through five given points, no four of which are collinear. 2. It passes through four given points, no three of which are collinear, and at any one of these points is tangent to a given line which passes through this point but not through any other of the given points. 3. It passes thrmigh three given points, not collinear, and at each of two of these points is tangent to a given line which parses through this point but not through any other of the three given points. 4. It passes through two given points, and at each of these is tangent to a given line which passes through that point but not through the other given point, and in addition is tangent to a third given line which is not concurrent with the other two given lines. 5. It passes through a given point and is tangent at that point to a given line through the point, and is tangent to each of three other given lines so situated that of four given lines no three are concurrent. 6. It is tangent to five given lines, no four of which are concurrent. Without doubt the student will generally use in his work an abridgment of this statement. The longer statement given above may be regarded as an interpretation of the shorter one in 117, making clear the meaning of the determination of a figure by means of certain data. The student will find that very frequently in geom- etry this abridged form of statement is used in the sense expressed more fully by the other one. Occasional expansions of shorter statements into the corresponding longer ones are well worth the attention of the student. A CONIC DETERMINED 117 THEOREM. ALTERNATIVE STATEMENT OF 116 117. A conic is determined by any one of the following sets of elements that are associated with it : 1. Five of its points. 2. Four of its points and the tangent at one of these points. 3. TJiree of its points and the tangents at two of these points. 4. Three of its tangents and the points of contact on two of these tangents. 5. Four of its tangents and the point of contact on one of these tangents. 6. Five of its tangents. Proof. We shall deal with the cases in the above order. 1. A conic is determined by jive of its points. Let P v Py, P y P, P^ be five points of a plane, no four of them being collinear. Join P l and P 2 to P Z , P^ P^. The triads of lines P^, P^, P^ and P^P y P^, P^ determine a projectivity between the flat pencils whose bases are P V P V and hence they determine a conic through the five points. Any conic through these points could be generated from the projectivity determined by the same triads ( 104) and would be the same as the one mentioned. If three of the five points are situated on a line /, the other two points should be taken as the bases of the pencils. In this case the conic consists of the line I and the line through the bases. 118 CONICS 2. A conic is determined by four of its points and the tangent to it at one of these points. Let P r 1%, P%, P be four points of a plane, no three of them being collinear, and in the plane let ^ be any line which passes through P l but which does not pass through any other of the four given points. Draw from P l the lines .7^, I^P, and draw from P 2 the lines JZ, P Z P 3 , P&. Consider P l and P 2 as bases of flat pencils. The lines t v P^ ; P^, P 2 P A ; P^P, P 2 P being taken as corresponding, one and only one projectivity is thereby established between the flat pencils. This projectivity determines one conic passing through the points P v P v P%, P 4 and having the line ^ as its tangent at P r No other conic can fulfill these conditions, since in that case the conic would also be generated from the projectivity just mentioned, and hence this conic would coincide with the first one. Therefore the second statement is proved. If three of the four points other than the one at which the tangent is given are on a line /, the conic consists of the given tangent and the line /. If P l and two only of the other points are collinear, no conic is determined. A CONIC DETERMINED 119 3. A conic is determined by three of its points and the tangents to it at two of these points. Let PV PV P be three noncollinear points of a plane, and in the plane let ^ and t z be lines which pass through P l and P 2 respectively but through no other of the three given points. Draw the lines P^, P^, and P^P r The triads t v P^, P^ and P^P V t 2 , P^P^ determine a pro- jectivity between the pencils whose bases are P l and P%, and this projectivity determines a conic passing through P v P%, P and tangent at P 1 and P i to ^ and 2 respectively. As in the other cases, it may be shown that there is only one such conic. Hence the third statement is proved. If the three points are on a line I and if one of the given tangents is I, the conic consists of / and the other tangent. If neither of the tangents coincides with Z or both tangents coincide with I, the conic may be thought of as the line I taken twice. 4. A conic is determined by three of its tangents and the points of contact of two of these tangents. Since Nos. 3 and 4 are dual in the plane, the proof of No. 4 follows at once. 5. A conic is determined by four of its tangents and the point of contact of one of these tangents. Since Nos. 2 and 5 are dual in the plane, the proof of No. 5 follows at once. 6. A conic is determined by five of its tangents. Since Nos. 1 and 6 are dual in the plane, the proof of No. 6 follows at once. 120 . CONICS 118. Construction of Conies. The statements of 117 establish the existence of conies that fulfill certain condi- tions, and suggest but do not solve the problem of con- structing the conies under these various conditions. The solution of this problem can be based upon the notion of projectivity involved in 117, but it can also be based upon two very celebrated theorems which will be considered on page 121. After these theorems have been proved, the problems of the construction of conies will be treated from both points of view. Before considering these two theorems, however, it will be found necessary to make some extension of the common notion of a hexagon with which the student is familiar from elementary geometry. 119. Hexagon. If any six coplanar points are taken in a given order, the figure formed by the lines through all pairs of successive points, as well as through the first and last points, is called a hexagon. As in the ordinary case of the hexagon, the first and fourth, the second and fifth, and the third and sixth sides are called opposite sides. Thus, in the above figures the pairs of opposite sides are -P X P 2 , p p . p p p p . p p p p 1 i' 5 > 28' 58' 84' 61* In each of the above figures the diagonals from P l are P^P S , P X P 4 , and PjP 6 . A similar generalization applies to each of the other polygons. It thus appears that opposite sides of a quadrilateral may intersect and that a diagonal may lie wholly outside a polygon. THEOREMS OF PASCAL AND BEIANCHON 121 THEOREM. PASCAL'S THEOREM 120. If a hexagon is inscribed in a conic, the three intersections of the three pairs of opposite sides are collinear. Proof. This is the theorem of 103 with the restriction upon P v P b removed by Steiner's theorem ( 104). Unless a cross hexagon is taken, the figure is usually very large. The proposition is due to Blaise Pascal (1623-1662). THEOREM. BRIANCHOITS THEOREM 121. If a hexagon is circumscribed about a conic, the three lines joining the three pairs of opposite vertices are concurrent. Proof. This is simply a generalization of 105. The student should write out the proof of this theorem. The proposition is due to Charles Julien Brianchon (1785-1864). 122 CONICS 122. Pascal Line. The line containing the points of inter- section of the three pairs of opposite sides of a hexagon in a conic is called the Pascal line of the hexagon. 123. Brianchon Point. The point of concurrence of the three lines joining the opposite vertices of a hexagon about a conic is called the Brianchon point of the hexagon. 124. Converses of the Theorems of Pascal and Brianchon. The converses of the theorems of Pascal and Brianchon can be established as in the exercise below, and each of them may then be given a different interpretation. Thus, if six coplanar points are chosen and joined to form a hexa- gon, a conic passes through any five of them. Does it pass through the sixth point ? It does if and only if the three points of intersection of the pairs of opposite sides are collinear. Hence Pascal's theorem and its converse imply the necessary and sufficient conditions for the passing of a conic through six given coplanar points. Brianchon's theorem can be interpreted in a corresponding fashion. Exercise 26. Theorems of Pascal and Brianchon 1. State and prove the converse of Pascal's theorem. 2. If two pairs of opposite sides of a hexagon inscribed in a conic are parallel, the other two opposite sides are parallel. 3. A hexagon is to be inscribed in a conic in such a way that a given line shall be its Pascal line. Determine the maximum number of sides of the hexagon that may be given, and solve the problem. 4. Solve Ex. 3 for the case when the given line is at infinity. 5. State and prove the converse of Brianchon's theorem. 6. Circumscribe a hexagon about a given conic in such a way that a given point shall be its Brianchon point, as many of the vertices of the hexagon as possible being given in advance. THEOREMS OF PASCAL AND BKIANCHOX 123 125. Limiting Cases of the Theorems of Pascal and Brian- chon. There are several limiting cases of the theorems of Pascal and Brianchon which have useful applications and which require mention at this point. They arise out of approach to coincidence of vertices of an inscribed hexagon of a conic and also of sides of a circumscribed hexagon. Let P^P^P^ be a hexagon inscribed in a conic. If 7 approaches P l along the conic, the line P^ approaches the tangent ^ at the point P v and the hexagon approaches the figure composed of the pentagon P^P^P^ and the tangent ^ to the conic at P r The pairs of opposite sides are t v 7^7? ; 7^, P 6 P & ; P S P 4 , P Q P r These pairs determine collinear points. Similarly, P 2 may approach P 1 and either P approach P s or P 5 approach P, yielding an inscribed quadrilateral and tangents to the conic at two of the vertices of the quad- rilateral. A third case is that in which P^ approaches P v P approaches P%, and P 6 approaches P b . In each case the propriety of extending Pascal's theorem, and others, to limiting cases in which two distinct elements are allowed to become coincident is left for the student's consideration. In the case of a circumscribed hexagon, if one side approaches coincidence with a second, their point of in- tersection approaches a limiting position at the point of contact of the second side. There arise out of the approach of sides to coincidence a number of limiting cases of Brianchon's theorem which can be worked out and which will be needed from time to time. Other limiting cases of these propositions are those in which the points or lines of the figures are not all in the finite part of the plane. For example, one or two vertices of the Pascal hexagon and one side of the Brianchon hexagon may be at infinity. Coincident elements and infinitely distant elements may be present in one hexagon. 124 CONICS PROBLEM. CONIC THROUGH FIVE POINTS 126. Griven Jive points in a plane, no four of them being collinear, construct the conic which is determined by them. Solution. This problem admits of two simple solutions. 1. Method based on a projectivity. Let the given points be P v 1%, P A , P, P 6 . Then in any chosen direction from any point, as P v there can be found another point of the conic which is not collinear with two of the others. Let the chosen direction be along the line p v and let the point to be found be called P. Draw P 1 P S , P^, P^, P 2 P S , P^ P 2 P 5 . The triads of lines JJJ, JJJJ, P^ and P 2 P 3 , P 2 P t , P^P b determine the projectivity between two flat pencils which generate the required conic. The point P is the intersection of p 1 and its correspond- ing line of the pencil whose base is P% ; and it may be found by the method used in 39, Case 1. Draw this line and produce it to meet p v thus determining P. By varying the position of the line p 1 any number of additional points of the conic may be found. Evidently it is not feasible to obtain all the points of the conic by this method, nor is the method convenient in practice. In this respect it is similar to the method of plotting in analytic geometry. CONIC THROUGH FIVE POINTS 125 2. Method based on Pascal's theorem. Let the given points be P r J, J, p^ p & . As before, an- other point P can be found on a chosen line p l that passes through any one of the points, as P r Q: The given points P v P v P z , P^ P b and the point P, which is to be found, are the vertices of a hexagon inscribed in the conic determined by the five given points. Then PJ> V J^>; JJ, 1>P; PJ> V PP l intersect on the Pascal line of this hexagon. Of these six lines, P 5 P is not given and PP l is the given line p r The Pascal line is determined by the intersection of P^P V P 4 P & and that of PJ^, p v Draw the Pascal line and let it meet P Z P Z in Q y Draw Q^P y This line Q^P b must coincide with the line P b P and must intersect the line p 1 in the required point P. Since every line through JJ determines a point on the conic, it is possible to locate any number of points. This method, like the first one, bears a certain resemblance to the method of plotting in analytic geometry. From the point of view of convenience it is decidedly superior to the first method. The student will observe that, since the conic is of the second order, the line p 1 cuts it in one and only one point other than P v and also that in either solution of the prob- lem the use of the ruler alone is sufficient. 126 CONICS PROBLEM. FOUR POINTS AND A TANGENT 127. Given four points in a plane, no three of them col- linear, and a line passing through one and only one of these points, construct the conic which passes through the given points and at one of them is tangent to the given line. Solution. As was the case in 126, there are two simple methods of construction. In each method any number of additional points of the conic may be found by determining where the conic would be cut by lines which pass through one of the points. 1. Method based on a projectivity. Let the given points be P^ P%, P 4 , P b , and let the given line be 2 passing through the point P%. Draw any line p l through the point P r Join P l to each of the points P 2 , P, P b , and join P 2 to each of the points P 4 , P y The triads P^, P^, P^ and t v P^, P^P b determine a projectivity between the flat pencils whose bases are P v P%, and the required conic is the locus of the intersections of corresponding lines of the projective pencils. The line through P% which corresponds to p l of the pencil whose base is P l can be determined by the method used in 39, Case 1, and P, the intersection of this line with p v is the point required. By varying the position of p } any number of points of the conic may be found. FOUK POINTS AND A TANGENT 2. Method based on Pascal's theorem. 127 Let P be the point on p 1 that is to be found. It is determined if the direction of P-P can be determined. The pentagon P^P^P is inscribed in the required conic, and the line 2 is tangent to the conic at P%. The intersections of P^, P^P b ; 2 , JP; and P^P V PP i (or jt?j) are on the Pascal line. Produce P^ and PP b to meet at Q v and produce P^ and p to meet in Q y Draw the Pascal line. Let t z meet this line in Q 2 ; join P 5 and 2 . Then the lines P 5 Q% and P 5 P are coincident and the intersection of P 5 Q 2 and j^j is the required point P. PROBLEM. THREE POINTS AND Two TANGENTS 128. Given in a plane three noncollinear points and two lines, each of which passes through one and only one of the given points, construct the conic which passes through the three given points and at each of two of them is tangent to the given line through that point. The solution is left for the student. It should be effected by two methods, as in the two preceding theorems. As in the other prob- lems the second method is to be preferred for practical reasons. An appreciation of the superior convenience of the second method is best secured by making the actual construction necessary for find- ing by the first method the line through P z which corresponds to p t of the first pencil. 128 CONICS PROBLEM. THREE TANGENTS AND Two POINTS 129. G-iven three nonconcurrent lines in a plane, and on each of two of these lines a point which is not on any other of the three, construct the conic which is tangent to each of the given lines and has each of the given points as the point of contact of the given line on which it lies. Of what problem is this the dual? The solution is left for the student. PROBLEM. FOUR TANGENTS AND ONE POINT 130. Given four lines in a plane, no three of them con- current, and a point on one but not on two of them, construct the conic which is tangent to each of these lines and has the given point as the point of contact of the given line on which it lies. Of what problem is this the dual? The solution is left for the student. PROBLEM. FIVE TANGENTS 131. Given five lines in a plane, no four of them con- current, construct the conic which is tangent to each. Of what problem is this the dual? The solution is left for the student. PROBLEM. CONSTRUCTING A TANGENT 132. Given five or more points of a conic, construct the tangent to the conic at any one of these points. The solution is left for the student. PROBLEM. FINDING A POINT OF CONTACT 133. G-iven five or more tangents to a conic, determine by construction the point of contact of any one of these tangents. The solution is left for the student. PROBLEMS OF CONSTRUCTION 129 Exercise 27. Problems of Construction 1. If two project! ve flat pencils generate a circle, they are congruent. 2. Using the result in Ex. 1, find any number of additional points of a circle when three of its points are given. 3. Find any number of points of a circle when two of its points and the tangent at one of them are given. 4. Solve the problem in 126 when one of the five points is at infinity in a given direction. 5. Solve the problem in 126 when two of the points are at infinity in given directions. 6. Solve the problem in 127 when the given line is at infinity. 7. Solve the problem in 127 when one of the four, points is at infinity in a given direction. 8. Solve the problem in 128 when one of the points is at infinity in a given direction and the tangent at that point is given to be the line at infinity. 9. Solve the problem in 129 when the two given points are at infinity. 10. Solve the problem in 131 when one of the five given lines is the line at infinity. 11. Solve the problem in 132 when the point at which the tangent is to be constructed is at infinity in a given direction. 12. Solve the problem in 133 when the given tangent whose point of contact is to be found is the line at infinity. ^13. If a parallelogram is inscribed in a conic, the tangents to the conic at the vertices form a parallelogram circumscribed about the conic. 14. If P v P 2 , P 8 , P 4 , P 5 are fixed points and P moves on the conic determined by them, find the envelope of the Pascal line of the hexagon PffJPf^P. 130 CONICS THEOREM. INVOLUTION ON COMPLETE QUADRANGLE 134. If a straight line cuts all the sides of a complete quadrangle but does not pass through any vertex, it cuts the three pairs of opposite sides of the quadrangle in conjugate points of an involution. Proof. Let a straight line p cut the pairs of opposite sides of the complete quadrangle whose vertices are P v P 2 , P s , P in A, A'; B, B'\ C, C"; and let J/, N, be the diagonal points of the quadrangle. Then range ABA'Cj^ flat pencil j^ range MP B A'P 4 - flat pencil P l - range AC'A'B'. range ABA'C^ range A'CAB. range AC'A'B' x range A'CAB. But Hence 23 Accordingly, A, A' ; B, B'; (7, C' are conjugate points of an involution on p. 65 This theorem is auxiliary to, and is in fact a special case of, an interesting and important theorem which was first established by the French geometer Girard Desargues (1593-1662). Desargues's theorem offers another line of approach to some of the preceding constructions and to other similar problems. In particular, on page 133, it is applied to the solution of 126. DESARGUES'S THEOREM THEOREM. DESARGUES'S THEOREM 131 135. If a complete quadrangle is inscribed in a conic, and if a straight line cuts the conic in two points distinct from each other and from the vertices of the quadrangle, these two points form a conjugate pair of the involution of points on the line, which is determined by the intersections of the line with the pairs of opposite sides of the quadrangle. Proof. Let the complete quadrangle P^P 2 P Z P^ be inscribed in a conic, and let a line p which does not pass through any of the four vertices cut the conic in P, P' and the pairs of opposite sides in A, A'; B, B'; C, C'. Then range PBP'A-ft&t, pencil P^PP^P'P^ -flat pencil P z (PP^P'P^ - range PA'P'B'. But range PA'P'B'- range P'B'PA'. 23 Therefore range PBP'A- range P'B'PA'. Hence P, P'; A, A'; B, B' are conjugate points of an involution on p determined by the pairs A, A' ; B, B'. The involution formed by the intersections of p with the pairs of opposite sides of the quadrangle P^P^ is also determined by the pairs of points A, A'; B, B'. Accordingly, P, P' are conjugate points of the involution of points determined by the intersections of the line p with the pairs of opposite sides of the quadrangle 132 CONICS 136. Restatement of Desargues's Theorem. It should be noted that many conies pass through the four points JJ, P 2 , P%, P 4 and that to each of such conies Desargues's theorem applies. Moreover, it should be remembered that the pairs of lines P^P V P Z P; P^, P%P 3 are degenerate conies through the four points. Hence Desargues's theorem is capable of restatement as follows : The infinitely many conies, including pairs of lines, which pass through four given coplanar points, no three of which are collinear, determine on any line which intersects them (but does not pass through any one of the points) infinitely many pairs of points of an involution. 137. COROLLARY. If the involution determined by the conies is hyperbolic, two of the conies which pass through the four points touch the straight line; if it is elliptic, no conic through the four points is tangent to the straight line. Exercise 28. Application of Desargues's Theorem 1. What sort of involution is determined upon a side of the diagonal triangle of the quadrangle mentioned in the theorem of 135 ? 2. Test the validity of the proof of Desargues's theorem when it "is applied to a line through one of the given points, say P 2 . Given four points in a plane, no three of which are col- linear, show how to draw a straight line subject to each of the following conditions : 3. There shall be two conies passing through the four points and tangent to the line. 4. There shall be one conic passing through the four points and tangent to the line. 5. There shall be no conic such as described in Ex. 4. DESAKGUES'S THEOREM 133 PROBLEM. CONIC THROUGH FIVE POINTS 138. Given five points, no four of which are collinear, con- struct the conic which is determined by them. Solution. Let us consider two cases. 1. No three of the five given points are collinear. Let the five points be P v P^ P 3 , P, P 5 . Draw , P 4 P 6 , P 6 P V PrP v and any transversal p l through P v It is now required to find the point P in which this line again cuts the conic determined by the five given points. The points A^, A%, B^, B^, in which jt^ is cut by the lines P 2 P 3 , PJ^i fsP^ P 5 P-2-> determine an involution in which P 1 and the required point P are corresponding points. Hence the point P can be determined from the pro- jectivity between the ranges in the involution. For in- stance, the three known points A v P> v A 2 and the required point P are projective with the four known points A v 7? 2 , A v P r Various special devices for finding P based upon the method of 39, Cases 1 and 2, can be found. By varying the position of the line p any number of points on the conic can be found. 2. Three of the five given points are collinear. In this case the required conic is a pair of straight lines, one the line through the three points and the other the line through the other two points. 134 CONICS PROBLEM. POSITION OF SELF-CORRESPONDING ELEMENTS 139. Given two superposed protective one-dimensional prime forms, construct the position of the self -corresponding elements. Solution. If the superposed prime forms are not ranges, it is possible by operations of projection and of section to obtain from them two superposed projective ranges. Hence it is necessary to solve the problem only for the case in which the prime forms are ranges. Let A 1 B 1 C l and J 2 J5 2 C 2 be two triads of corresponding points of superposed projective ranges on a base p. Describe any circle coplanar with the line p. Join any point P of the circle to each of the six given points, and let these lines cut the circle again in A[, B' v C l and A' z , J5 2 , C' 2 . Join A( to A' v B'v (7 2 , and A' z to C{. Then flat pencil A' 2 - flat pencil -range A^B^ ^ range A^C^ x flat pencil P - flat pencil A{ But the flat pencils A^A\B\C ) 53 ), A{ ) have a self -correspond ing element, and hence are perspective. Let X' be a point, if there be any, in which p 1 ', the axis of perspectivity, cuts the circle, and let PX' meet p at X. In the four flat pencils previously mentioned the corre- sponding lines are A(X', PX', A'^X', PX', and hence in the superposed pencils whose bases are at P, PX' is a self- corresponding line. Therefore X is a self-corresponding point of the ranges on p. Conversely, it is true that, corresponding to each self- corresponding point of the ranges on the line p, there is an intersection of the line p' with the circle. Hence, to find the self -corresponding points on p we join P to the intersections of p' and the circle, and produce these lines to intersect p. There may be no, one, or two intersections with p, and each of these intersections is a self-corresponding point. 140. Constructions of the Second Order. All constructions made before 139 were effected wholly by the use of straight lines, and at every stage the results were uniquely determinate ; that is, all the problems had one and only one solution. If the solutions of the problems analogous to these constructions are effected by the methods of analytic geometry, it is found that only equations of the first degree are used. For this reason these and similar problems are said to be of the first order. Beginning with 141, attention will be given to prob- lems whose solutions by the method of analytic geometry would involve the use of at least one equation of the second degree, as in 139. Correspondingly, each construction will require the use (at least once) of a curve of the second order, and for simplicity the circle will be taken. The problem in 139 furnishes a basis for others, and hence has been deferred from its most natural place, which was in connection with the treatment of superposed projective forms in Chapter VII. 136 CONIC S PROBLEM. INTERSECTIONS OF A LINE AND A CONIC 141. Given any of the sets of elements mentioned in 117 as determining a conic, construct the intersections (if there are any) of the conic with a given straight line p in its plane. Solution. If the set of elements is not five points, by means of 127-133 find five points P r P 2 , P 3 , P 4 , P & on the conic. Join any two of the points, as P 1 and J^, to P s , P, P b , and let these lines meet the line p in P^ P{, Pj and J>", JF>", J>". These triads determine superposed protective ranges on p. By the method of 139 find the self -corresponding points (if there are any) of these ranges. Since these self -corresponding points are common to corresponding lines of the protective flat pencils whose bases are -7J and P 2 , they are on the conic. Moreover, they are the only points of p which are on the conic. There may, therefore, be two, one, or no intersections. PROBLEM. TANGENTS FROM A POINT 142. Given any of the sets of elements mentioned in 117 as determining a conic, construct the tangents (if there are any) to the conic from a given point P in its plane. Solution. If the set of given elements is not five tangents to the conic, by means of 127-133 find five tangents j, 2 , 3 , 4 , 5 to the conic. The tangents t s , 4 , t & cut t v t 2 in triads of points which, being joined to P, determine a projectivity between flat pencils whose base is P. Find the self-corresponding lines of these pencils. Any such line passes through P and also joins corresponding points of the projective ranges on ^, 2 which serve to generate the conic. Hence the line is tangent to the conic. The number of these lines is two, one, or none. PROBLEMS OF CONSTRUCTION PROBLEM. FOUR POINTS AND A TANGENT 137 143. Construct a conic which shall pass through four given points, no three of which are collinear, and shall be tangent to a given line that does not contain any of the points. Solution. Let the given points be P v P^, P s , P 4 , and let the given line be t r Let ^ cut the lines P^, P 3 P^ in A v A z and cut the lines P^, P%P S in B r B z . Find the self-corresponding points of the involution on ^ which is determined by these pairs of points. Through the four points and any self-corresponding point P 5 construct a conic ( 138). This conic is tangent to ^ at P y For if it cuts j in a second point P Q , then the point P & is not self-corresponding, and this is contrary to fact. Hence, for every self -corresponding point of the involu- tion on t^ one conic can be constructed. There may be no, one, or two self-corresponding points ( 139). Hence no, one, or two conies may be constructed. THEOREM. FOUR POINTS AND A TANGENT 144. The number of conies which pass through four given points, no three of tvhich are collinear, and are tangent to a given straight line which does not pass through any of the points, is none, 'one, or two. The proof is left for the student. 138 CONICS PROBLEM. FOUR TANGENTS AND A POINT 145. Construct a conic ivhich shall be tangent to each of four given straight lines, no three of which are concur- rent, and which shall pass through a given point exterior to the lines. This problem is the dual of 143 and may be solved as such. The solution is left for the student. Likewise, a theorem dual to 144 results from the proof of the construction. 146. Special Case of Desargues's Theorem. To complete a set of constructions which include 126, 129-133, 143, and 145, two others are necessary, and these are given in 148 and 149. In order to solve these two problems special cases of Desargues's theorem (and its dual) may be used. Instead of the four distinct points of the conic con- sidered in Desargues's theorem, let the first and second points move up to coincidence, and also let the third and fourth points move up to coincidence. Then the line join- ing the first and second points and that joining the third and fourth points become tangents to the conic. Also the lines joining the first and third points, the second and fourth points, the first and fourth points, and the second and third points move into coincidence upon the chord of contact of the two tangents mentioned. THEOREM. SPECIAL FORM OF DESARGUES'S THEOREM 147. Two straight lines and the conies which are tangent to them at two given points intersect a given line that does not pass through either of these points in pairs of points of an involution, one of the self-corresponding points of which is the intersection of the given line with the chord of contact. The proof is left for the student. PEOBLEMS OF CONSTRUCTION 139 PROBLEM. THREE POINTS AND Two TANGENTS 148. Construct a conic which shall pass through each of three given noncollinear points and be tangent to each of two given lines that do not pass through any of the points. Solution. Let the points be P v P 2 , J^, and let the lines be t v f 2 . Let t v t 2 cut the line P^ in A v A 2 and the line P%P A in B, B z . We shall first find the points of contact of t v < 2 with the conic. Find the self-corresponding points (if there are any) of the involutions determined by P v P% and A v A% and by PL, P 3 and B v B 2 respectively. Let the line through M v one of the first of these, and M v any one of the second, cut 2 in P and ^ in P y Pass a conic through P v and tangent to j and 2 at P 5 and P 4 respectively ( 128). Since the point corresponding to P% in the involution on A^A^ is completely determined by Jfj and the pair of points A v A v it follows that this conic must pass through P l ( 147), for the involution determined on P^ by conies tangent at P and P 5 to 2 and ^ is also determined by a self-corresponding point and the pair of points A v A v Similarly, this conic can be shown to pass through P 3 . Hence, for every possible pair of points P v P B one conic may be constructed. There may be no, one, two, or four pairs of points, as P 4 , P, ( 144), and for each a conic may be constructed. 140 CONICS THEOREM. THREE POINTS AND Two TANGENTS 149. The number of conies which pass through three given noncollinear points and ivhich are tangent to two given lines that do not contain any of the points is none, one, two, or four, as the case may be. The student should write out the proof, which is essentially that of 148. PROBLEM. THREE TANGENTS AND Two POINTS 150. Construct a conic which shall be tangent to three given nonconcurrent lines and shall pass through two given points which are exterior to the lines. The student should write out the solution, which is simply the dual of that of 148. Exercise 29. Review 1. If the sides of an angle of constant size rotating about a fixed vertex intersect respectively two fixed lines, the line joining these intersections envelops a conic. f--/f i_i ^jA^^n 2. Two vertices of a variable triangle move along two fixed lines, and the three sides respectively pass through three fixed collinear points. Find the locus of the third vertex. 3. Consider Ex. 2 for the case in which the three fixed points are not collinear. 4. If two triangles are in plane homology, the intersections of the sides of one triangle with the noncorresponding sides of the other lie on a conic. 5. State Pascal's theorem for the case in which the first and second, the third and fourth, and the fifth and sixth vertices have become coincident. 6. The complete quadrilateral formed by four tangents to a conic, and the complete quadrangle formed by their four points of contact, have the same diagonal triangle. KEVIEW EXERCISES 141 7. If a variable quadrangle P 1 P 2 P g P 4 inscribed in a conic has as fixed points P V P 2 , and the intersection of Pfy PyP^ the other vertices of its diagonal triangle move along the same fixed straight line. 8. If P a , P 5 are fixed points on a given conic, and if P is a moving point, as P moves along the conic the Pascal line of the hexagon, consisting of the triangle PP 8 P 5 and the tangents to the conic at the points P V P g , P 5 , envelops a conic. 9. If P r P g , P 4 are fixed vertices of a complete quadri- lateral whose fourth vertex P moves along a given conic through Pj, P g , P 4 , all the vertices of the diagonal triangle trace straight lines and all the sides pass through fixed points. 10. If the points P 2 and P 3 trace superposed projective ranges on the base AB of a fixed triangle ABC, if P l is a fixed point not on any side of the triangle, if P^ meets A C in P 4 , and if P.P. meets EC in P,, find the locus of P. the inter- 13 6' ' section of .4 P. and BP.. 5 4 11. In Ex. 10 find the envelope of P 4 P g . 12. State Desargues's theorem for the case in which a pair of the four given coplanar points become coincident. 13. State Desargues's theorem for the case in which two pairs of the given coplanar points become coincident. 14. Three sides, AB, AD, CD respectively, of a variable quadrangle inscribed in a given conic pass through three given points of a line. Find the envelope of BC. 15. Extend Ex. 14 to the case of a simple inscribed polygon having 2 n sides. 16. From the data of 127 construct, by means of 147, tangents at additional points of the conic. 17. Prove the dual of 149, namely, that the number of conies which can be constructed under the conditions of 150 is none, one, two, or four. 18. Solve the dual of 139 for the plane. 142 CONICS 19. If the lines p l and p 2 are drawn through the vertices Pj and P 2 respectively of a given quadrangle, the conies which pass through the vertices of the quadrangle determine perspec- tive ranges on p v and p^. 20. If the lines p l and p z are drawn through the vertex P l of a given quadrangle, the system of conies which pass through the vertices of the quadrangle determine protective ranges on p l and p^. 21. State and prove the dual of Ex. 19 for the plane. 22. Construct a conic which shall pass through two given points Pj and P 2 , shall be tangent to a given line t a at the point P 8 , and shall be tangent to a second given line t t . Apply Ex. 12 for the line 4 . Find the self-corresponding points of the involution. 23. Construct a conic which shall be tangent to a given line tfj at the point P 1? to t z at P 2 , and to t a . 24. Consider the problem of 141 for the case in which the given line is the line at infinity. 25. Consider the problem of 143 for the case in which the given line is the line at infinity. 26. Consider the problem of 148 for the case in which one of the given lines is the line at infinity. 27. Solve the dual of Ex. 22 for the plane. 28. Construct a triangle which shall be inscribed in a given triangle and have its sides pass through three given points. Observe that if a triangle has two vertices, as required, but not the third, the sides through the latter cut a side of the given triangle in corresponding points of superposed protective ranges. 29. Construct a triangle which shall be inscribed in a given conic and have its sides pass through three given points. 30. If a conic can be described through the six vertices of two given triangles, another conic can be described which shall be tangent to the six sides of the two given triangles. CHAPTER XI CONICS AND THE ELEMENTS AT INFINITY 151. Classification of Conies. In the discussion of conies in the preceding chapter no classification was made, nor was any account taken of the fact that on certain occasions the term straight line may mean " straight line at infinity " and the term point may mean " point at infinity." These considerations can be associated very advantageously. ELLIPSE PARABOLA HYPERBOLA In projective geometry, conies are classified by means of their relations to t^e line at infinity. This line, like any other, may intersect a conic in no, one, or two points, and hence conies are divided into three classes as follows : 1. Ellipses, or conies that do not intersect the line at infinity. 2. Parabolas, or conies that intersect the line at infinity in one point (or are tangent to the line at infinity). 3. Hyperbolas, or conies that intersect the line at infinity in two distinct points. While these conies are familiar to the student from his work in analytic geometry, the study of conies will now be considered from a different point of view. 143 144 CONICS AND THE ELEMENTS AT INFINITY 152. Elements at Infinity. In the interpretation of the results already obtained, in so far as the ellipse is con- cerned, it will be seen that the expressions point on the curve and tangent to the curve always mean a point and a line in the finite part of the plane. On the other hand, in connection with the parabola, one and only one tangent, and one and only one point of the curve (the point of contact of that tangent), may be taken to be at infinity. In the case of the hyperbola there are two points on the curve which are at infinity, but the line at infinity is not a tangent. At each of the infinitely distant points of the curve there is, however, a tangent which has no infinitely distant point except its point of contact. It follows that in the cases of the parabola and hyperbola the interpretations of the theorems of Pascal and Brianchon and of similar theorems obtained by the methods of pro- jective geometry vary according as all or only part of the elements are assumed to be in the finite part of the plane. In the light of the procedure indicated, the results which have been obtained are capable of restatements which vary for the three types of conies, but which have a great interest, because they bring these results into clearer relation to those obtained by the methods of analytic geometry. 153. Asymptote. A line, not the line at infinity, which is tangent to a conic at an infinitely distant point is called an asymptote. In this figure a is an asymptote. Every hyperbola has, then, two asymp- totes, and the other conies have none, though sometimes the parabola is said to have the line at infinity as an asymptote. This latter form of statement is convenient when geometry is treated algebraically, but it will not be adopted in this text. ELEMENTS AT INFINITY 145 154. Special Interpretations. As indicated in 152, each of the results that have been derived for the conies should be examined for interpretations based upon the rela- tions of the elements at infinity to the three types of conies. The great variety of results that can be obtained prevents a systematic and detailed reexamination in this place of all the theorems and constructions that have been derived. A few of these will be obtained, but for the most part their derivation must be left to the student, a work which will prove both interesting and profitable. Of the elements (points and lines) which determine a conic not more than two points and not more than one line may be at infinity, except in the limiting case of coincident points or coincident tangents. The existence of one infinitely distant point on a conic determines that the curve is not an ellipse, and the existence of two such points determines that the curve is a hyperbola. Similarly, the tangency of the line at infinity to the curve determines it to be a parabola. On the other hand, when all the given determining elements are in the finite part of the plane, the conic may prove to be of any one of the three types. The determi- nation of the character of the conic of which certain ele- ments are given is a particularly interesting case. It is the problem of 141 as modified in Ex. 24, page 142. In view of what is said above, we shall now restate the important theorem of 116. In each case the student should draw the figure and satisfy him- self that the statement is correct and that it is a special case of one of the corresponding statements in 116 and 117. He should also supplement the results here set forth by the others which can be obtained if a thorough examination of the theorem is made for its various interpretations. 146 CONICS AND THE ELEMENTS AT INFINITY THEOREM. A CONIC DETERMINED 155. 1. Tliere is one and only one conic (parabola or hyperbola) ivhich passes through four points in the Jinite part of a plane and one infinitely distant point in a speci- fied direction. This follows from the theorem stated in 116, No. 1. 2. There is one and only one hyperbola which passes through three points in the Jinite part of a plane and has given direc- tions for its asymptotes. This follows from the theorem stated in 116, No. 1. 3. There is one and only one parabola which passes through three noncollinear points in the finite part of a plane and has its infinitely distant point in a given direction. This follows from the theorem stated in 116, No. 2. 4. There is one and only one hyperbola which passes through any point in the finite part of the plane and has two given straight lines as asymptotes. This follows from the theorem stated in 116, No. 3. 5. There is one and only one hyperbola which has two given lines as asymptotes and is tangent to a third line which is not parallel to either of the others. This follows from the theorem stated in 116, No. 4. 6. There is one and only one parabola which is tangent to each of three nonconcurrent lines lying in the finite part of the plane and has its infinitely distant point in a given direction. This follows from the theorem stated in 116, No. 5. 7. There is one and only one parabola which is tangent to any four lines of a plane, no three of which are concurrent and no two of which are parallel. This follows from the theorem stated in 116, No. 6. PASCAL'S THEOREM 147 THEOREM. SPECIAL INTERPRETATION OF PASCAL'S THEOREM 156. The chords from a point on a hyperbola to each of two other points on the hyperbola intersect the lines through these two points parallel to one of the asymptotes, the inter- sections being collinear with the intersection of the tangents at the two points. The proof of this theorem is included in the proof given in 157. THEOREM. FURTHER INTERPRETATION OF PASCAL'S THEOREM >157. The chords from a point on a parabola to each of two other points on the parabola intersect the lines from these two points to the infinitely distant point of the curve, the inter- sections being collinear with the intersection of the tangents at the two points. Proof. These two theorems are closely related, being obtained by applying to such conies the case of Pascal's theorem in which two pairs of vertices coincide. If a conic is known to have one point at infinity, it may be either a hyperbola or a parabola. Consider a hexagon inscribed in a conic in such a way that the first and second vertices coincide, the fourth and fifth vertices coincide, and the sixth vertex is at infinity. We may also assume that the curve is a hyperbola or that it is a parabola. If it is a hyperbola the sides of the hexagon which intersect at the infinitely distant point are parallel to the same asymptote. In either case one pair of opposite sides is a pair of tangents. The two theorems considered above are merely state- ments of Pascal's theorem for the two cases described, the terms used being appropriate in connection with these two kinds of conies. 148 CONICS AND THE ELEMENTS AT INFINITY THEOREM. SPECIAL INTERPRETATION OF BRIANCHON'S THEOREM 158. Given jive tangents to a parabola, the line parallel to the first tangent and concurrent with the third and fourth tangents cuts the line parallel to the fifth tangent and concur- rent with the second and third tangents on the line joining the intersection of the first and second tangents to that joining the fourth and fifth. Proof. In Brianchon's theorem ( 121), simply let one of the tangents be the line at infinity, and the proof follows at once. Pascal's theorem and Brianchon's theorem have a large number of special interpretations. Of these we have space for only the three given in 156-158. They have been selected not because of their intrinsic importance but because they indicate the method of procedure. 159. Special Constructions. On account of the elements at infinity the problems which were considered in Chap- ter X may also be given special statements for certain cases. Thus a point at infinity may be specified by its direction ; and since a hyperbola is determined by means of any three of its points in the finite part of the plane and its two points at infinity, it is determined by the three points mentioned and the directions of the two points at infinity. These latter directions are also the directions of the asymptotes. In actual constructions certain special situations arise. Thus, drawing a line to a given infinitely distant point is the same as drawing a line parallel to a given line. To effect this with the ungraduated ruler it is necessary to have additional data, as in the exercises on pages 99 and 100. Problems involving considerations of this sort will be considered in 160-162. PROBLEMS OF CONSTRUCTION 149 PROBLEM. CONSTRUCTION OF THE HYPERBOLA 160. Given in a plane three noncollinear points and two pairs of parallel lines, each pair having a direction different from those of the lines joining the three points and also different from that of the other pair, construct a hyperbola through the three given points and having asymptotes parallel to the two given pairs of lines. The student should write out the solution, making appropriate modifications of the methods employed in 126 and 138. PROBLEM. DETERMINATION OF A CONIC 161. Given a set of elements sufficient for the determination of a conic, determine the nature of the conic and the directions of its infinitely distant points (if there are am/). Among the constructions of the second order the construction in 141 deserves attention in this connection, and this problem is one of its special forms. If elements sufficient for the determination of the conic are given, the finding of the intersections of the conic with the line at infinity includes determining whether the conic is an ellipse, a parabola, or a hyperbola. As in the original case, if five points of the conic are not given they may be found by construction. Let them be P v P 2 , P 3 , P t , P & . The triads of lines P^g, P^, PiP & and P 2 P 8 , P 2 P 4 , P Z P S deter- mine the projectivity by means of which any number of additional points of the conic may be found. A difficulty now arises in following the original construction, because the line p is at infinity. The triads of points of the super- posed projective ranges on this line that are determined by the triads of the flat pencil are now not available from the point of view of construction by the ruler. If, however, the possibility of drawing lines parallel to all given lines is assumed, the resulting difficulty disappears. For the purpose of drawing the necessary parallels, the compasses must be used more freely than in the construction in 141. With this difference the construction follows as before, and the student should write out the solution in full. 150 CONICS AND THE ELEMENTS AT INFINITY PROBLEM. CONSTRUCTION OF THE PARABOLA 162. Gf-iven four points in a plane, no three of them col- inear, construct the parabola which passes through these points. By 144 the number of such parabolas is none, one, or two. The solution of this problem, which is based on that of 143, is left for the student. Exercise 30. Elements at Infinity As suggested on pages 144-148, investigate with respect to the elements at infinity the following cases already considered : 1. 120. 5. 128. 9. Page 142, Ex. 19. 2. 121. 6. 149. 10. Page 142, Ex. 22. 3. 126. 7. Page 141, Ex. 10. 11. Page 142, Ex. 28. 4. 127. 8. 147. 12. Page 142, Ex. 29. In Exs. 1-12 practice in special interpretation, not the finding of im- portant results, is the object. 13. Using the compasses only once arid the ruler, find five points in the finite part of the plane which shall determine an ellipse other than a circle. 14. Consider Ex. 13 for the case of a parabola. 15. Consider Ex. 13 for the case of a hyperbola. 16. In the finite part of the plane find four points through which no parabola passes. 17. In the finite part of the plane find four points through which two parabolas pass. 18. Find four points through which both a circle and a parabola pass. 19. On a straight line p passing through a given point P lt find a point P such that through P^ P and two other given points P 2 , P s there pass (1) two parabolas ; (2) one parabola ; (3) no parabola. In each case indicate all possible positions of P. CHAPTER XII POLES AND POLARS OF COXICS 163. Polar of a Point. In 165 it will be proved that if three lines, concurrent at a point 0, cut a conic in A v A% ; B v J5 2 ; C v C v the harmonic conjugates A, B, C of O with respect to A ly A^\ S v B be the polar plane of o with respect to the cone. In o> there is one line, and there may be many lines, perpendicular to o. Through o and a line o' in to perpendicular to o pass a plane TT, and let I and I' be the lines in which the plane TT cuts the cone. Then the lines o and o' are conjugates with respect to I and Z', and, being perpendicular to each other, they bisect the angles formed by I and I'. Hence, through any line o there is one plane which cuts the cone in lines that form an angle of which a is the bisector. If and only if the line o is perpendicular to its polar plane o>, all planes through o cut the cone in lines that make an angle of which o is the bisector. In this case the line o is an axis of symmetry with respect to the cone. Manifestly every axis of symmetry is perpendicular to its polar plane. Also, a line o' parallel to o, an axis of symmetry, cuts a cone in two points, and the segment joining these points is bisected by the polar plane, since the axis of symmetry, the line joining the vertex to the in- tersection of the plane o> with o', and the lines I and I 1 are harmonic. Hence the polar plane of an axis of symmetry is a plane of symmetry. It can be shown that every quadric cone has one axis o l of symmetry and a plane o> l of symmetry which is per- pendicular to it. The plane Wj cuts the cone in a conic that has two principal axes which we may call o 2 and o y Of the lines Oj, 2 , o 3 each pair is conjugate to the third line and determines the polar plane of this line. Moreover, each of these polar planes is perpendicular to its polar line, and hence to the other two polar planes. There are, there- fore, three axes of symmetry, perpendicular eaoh to each, and three planes of symmetry, perpendicular each to each. CYLINDERS 171 186. Cylinders. Hitherto it has been assumed that the axes of the generating axial pencils intersect in the finite part of space. If, however, the axes of the generating axial pencils are parallel, then all the generating lines are parallel to them, and the vertex of the surface is at infinity. In this case the surfaces generated are called cylinders. HYPERBOLIC CYLINDER PARABOLIC CYLINDER ELLIPTIC CYLINDER Cylinders are classified with reference to their relation to the plane at infinity. The plane at infinity may cut the cylinder in two, one, or no straight lines. In these cases a section perpendicular to the generating lines of the cylinder is a hyperbola, a parabola, or an ellipse respectively; and the cylinder is said to be hyperbolic, parabolic, or elliptic, as the case may be. In the case of a cylinder the plane at infinity and cer- tain of its lines are in the bundle to which the cylinder belongs. The plane at infinity has a polar line which is an axis of symmetry and is called the axis of the cylinder. It can be shown that certain planes through this axis and all planes perpendicular to it are planes of symmetry. In the case of the parabolic cylinder the polar line of the plane at infinity is at infinity and lies in the cylinder. 172 QUADRIC CONES Exercise 34. Quadric Cones 1. Prove the theorem regarding quadric cones which corre- sponds to Steiner's theorem regarding conies. 2. Every conic surface of the second class is of the second order. Prove also the converse. 3. The hexahedral angle whose faces are determined by the six pairs of alternate edges of another hexahedral angle which is inscribed in a quadric cone has its faces tangent to a quadric cone. 4. Inscribe in a quadric cone a trihedral angle whose three edges shall be in three given planes. 5. If a variable simple four-flat so moves as always to be circumscribed about a given quadric cone, while three of its edges move each in one of three fixed coaxial planes, then the fourth edge moves on a fourth fixed plane coaxial with the three given cones. 6. State the properties of polar lines and planes of quadric cones corresponding to those of poles and polars of conies which are given in 166 and 167. 7. Find the points of intersection of a given straight line with a quadric cone of which five determining elements are also given. 8. In a bundle a v (3^ y t and a 2 , /8 2 , y 2 are two sets of fixed coaxial planes. Two planes JT I and 7T 2 so move that the lines determined by 7T X and a 1? 7T 2 and 2 ; TT I and /3 15 7r 2 and /?.,; TT and yj, 7r 2 and y 2 lie in three coaxial planes. Find the surface generated by the line common to TT I and 7r 2 . 9. In a bundle the edges of a trihedral angle, whose planes are , /?, y and which is self-polar with respect to a given quadric cone, determine with any line o the planes a', /8', y'. If the polar plane of o is w, the pairs of lines determined by o> with a and a', by o> with ft and ft 1 , and by w with y and y' form a pencil in involution. CHAPTER XIV SKEW RULED SURFACES 187. Skew Ruled Surfaces. The third set of figures which were projectively generated was found to consist of the ruled surfaces of the second order that are not conic. Before discussing these we shall consider the classification of ruled surfaces in general. One classification of ruled surfaces is based upon the law governing the motion of the generating line. At any instant the motion of this line may be a revolution about one of its own points or it may be a displacement by virtue of which the line immediately ceases to intersect its present position. In the former case it is sometimes said that every pair of consecutive generators intersect, and in the latter case it is said that no two consecutive generators intersect. Surfaces generated in the first way are called developable surfaces, and those generated in the second way are called skew surfaces. Cones and cylinders are examples of developable surfaces, but they are of a special type, inasmuch as each of their generators inter- sects every other one. Likewise, skew ruled surfaces of the second order and second class are special in character, for no generator intersects any other of the set, even though they be not consecutive. Generators usually cut other generators of the set if the latter are not consecutive. In this chapter only a few specially interesting facts regarding the surfaces of the second order and second class will be established. 173 174 SKEW KULED SUKFACES THEOREM. SECOND SET OP GENERATING LINES 188. Every skew ruled surface generated by the intersec- tions of corresponding planes of two projective axial pencils has also a second set of generating lines whose relation to the surface is similar to that of the first set. Each member of either set of generators intersects no others of its own set, but intersects every one of the other set. Through every point of the surface there pass two generators, one of each set. Proof. Every plane through any generating line a of the surface cuts the surface along the line a and also along a second line 1? and nowhere else. Moreover, the line a l cuts each of the generating lines that have been noted. The infinitely many planes through a cut the surface in infinitely many lines a l5 2 , 3 , , each of which cuts every one of the generators ; and every point of the surface lies on one and only one of the new lines. No two of these lines intersect ; for if they did, all the generators would lie in the plane determined by them. SECOND SET OF GENERATING LINES 175 Consider now the two sets of points B v B v B%, and C v (7 2 , C 3 , in which the lines a v a 2 , a 3 , intersect two other generators b and c. These sets of points are the intersections of the gener- ators b and c with the planes of the axial pencil whose base is a and whose planes pass through the lines a v 2 , 3 , ; and consequently they constitute perspective (but not coplanar) ranges. Then the axial pencil whose base is b and whose planes pass through C v C v C 8 , and the axial pencil whose base is c and whose planes pass through B v B v B s , -, being perspective respectively with the ranges C^C^C^ and B-^B^B^ , are projective with each other. Corresponding planes of these axial pencils intersect in the lines a v a v 3 , ., which are therefore generating lines of a skew quadric ruled surface. The latter skew ruled surface must coincide with the original one, since the lines j, a 2 , a 3 , contain all the points of the original ruled surface, and no others. Hence these lines must constitute a second set of generators for that ruled surface. 189. COROLLARY. The lines of either set of generators determine projective ranges on any two lines of the other set. 190. Conjugate Reguli. Two reguli which are related as are the two in 188 are called conjugate reguli. The theorem of 188 may be restated as a corollary to this definition as shown below. 191. COROLLARY. Every skew ruled surf ace of the second order carries two conjugate reguli. Each line of either regulus intersects no lines of its own regulus, but intersects each line of the other regulus. Through every point of the surface there pass two lines, one from each regulus. 176 SKEW RULED SURFACES THEOREM. DETERMINATION BY GENERATORS 192. Given three straight lines no two of which are coplanar, there exists one and only one skew quadrie ruled surface of which each of these lines is a generator. Proof. Let a, 6, c be three straight lines no two of which are coplanar. Through each point of a one line and only one can be drawn to meet all three of the lines. Let p v p z , and p 3 be three lines which meet a, b, c. Then p 1 together with the three lines a, 6, c, and p 2 together with the same lines, determine triads of planes of axial pencils whose bases are p l and p 2 . These triads determine a projectivity between the pencils, and this projectivity determines one skew quadrie ruled surface of which a, 6, c are generators. Any two corresponding planes of the projective axial pencils intersect in a line that meets p 1 and p 2 . This line also meets /> 3 ; for if p s meets a, 6, c in A z , -Z? 3 , C 3 respec- tively, the two axial pencils above mentioned are each per- spective with the same range on jt? 3 , the perspectivities being determined by the correspondence of A%, B%, C 3 to the planes p^ jOjJ, p^c in the one case, and the planes j(? 2 a, p^b, p. 2 c in the other. It follows that corresponding planes cut p s in the same point, and hence their line of intersection cuts p y Accordingly, the surface is uniquely determined by the three generators a, 6, c. 177 Hyperbolic Paraboloid, the conjugate reguli being formed by rods. Certain sections are parabolas, other sections are hyperbolas. The curvature of the surface is not secured by the pressure of one set of rods upon the other Hyperboloid of One Sheet, the conjugate reguli being formed by straight rods. The surface in the neighbor- hood of its center of symmetry is shown. Horizontal sections are ellipses, vertical sections are hyperbolas 178 SKEW RULED SURFACES 193. Skew Quadric Ruled Surfaces Classified. These sur- faces are classified according to the nature of their sections by the plane at infinity. As has been shown, any plane section of one of these surfaces is a conic and degenerates into two straight lines if the cutting plane contains a generator. The surfaces are, therefore, of two sorts: 1. Hyperbolic paraboloids, or those whose intersections with the plane at infinity are pairs of generators. 2. Hyperboloids of one sheet, or those whose intersections with the plane at infinity are nondegenerate conies. It may be noted that if a hyperbolic paraboloid is regarded as generated by the joining lines of corresponding points of two protective ranges, the points at infinity of the ranges are found to be corresponding points. Hence in this case (and in this case only) the ranges are similar. Accord- ingly, if corresponding points of two similar (but not coplanar) ranges are connected by threads, a good model of a hyperbolic paraboloid may be constructed. To exhibit both sets of generating lines it is better to use a quadrilateral A BCD hinged at two opposite vertices, as B and D, so that the triangles ABD, CBD can be adjusted to lie in different planes. Congruent ranges can be taken on AB and CD and also on EC and DA. Cor- responding points can be joined by strings, and in this manner an excellent model can be constructed with very little trouble. Directions for constructing a string model of the hyperboloid of one sheet are not so easily given. The existence of such a surface is evident, since it is generated by the lines joining corresponding points of projective ranges which are not coplanar and not similar. CLASSIFICATION 179 Hyperbolic Paraboloid, the conjugate reguli being formed by straight rods. The surface near the vertex, or saddle point, is shown. Certain sections through the vertex are parabolas, others are hyperbolas Hyperbolic Paraboloid, the conjugate reguli being formed by strings, ff every string in either set is cut, the strings in the other set retain their positions 180 SKEW KULED SURFACES Exercise 35. Skew Ruled Surfaces 1. A regulus is determined by two nonintersecting lines and three noncollinear points, no two of which are coplanar with any of the lines. 2. If a regulus contains a line at infinity, the conjugate regulus also contains a line at infinity. 3. Determine three pairs of quadric ruled surfaces which have in common two given noncoplanar lines and also respec- tively no, one, and two generators of the other set. 4. Determine a quadric ruled surface which contains two given noncoplanar lines and a given point exterior to them. How many such surfaces are there ? Find any additional lines which are common to such surfaces. 5. If four generators of a regulus cut one generator of the conjugate regulus in a harmonic range, they cut every generator of the conjugate regulus in a harmonic range. Four generators of a regulus which have the property mentioned in Ex. 5 are called harmonic generators. 6. Given any three lines in space, no two of which are coplanar, find a fourth line which, with the three given lines, constitutes a set of harmonic generators of a regulus. 7. If a line so moves as constantly to intersect each of two noncoplanar lines and also to remain parallel to a given plane, the line generates a hyperbolic paraboloid. 8. If a range and a flat pencil which do not lie in the same plane or in parallel planes are projective, and if from each point of the range a line is drawn parallel to the correspond- ing line of the flat pencil, these parallel lines all lie on -a hyperbolic paraboloid. 9. The locus of the harmonic conjugates of any point with respect to a ruled surface is a plane. 10. The lines (or planes) of any bundle which are tangent to a quadric ruled surface generate a quadric cone. HISTORY OF PROJECTIVE GEOMETRY The history of geometry may be divided roughly into four periods: (1) The synthetic geometry of the Greeks, including not merely the geometry of Euclid but the work on conies by Apollonius and the less formal contri- butions of many other writers; (2) the birth of analytic geometry, in which the synthetic geometry of Desargues, Kepler, Roberval, and other writers of the seventeenth century merged into the coordinate geometry of Descartes and Fermat ; (3) the application of the calculus to geom- etry, a period extending from about 1650 to 1800, and including the names of Cavalieri, Newton, Leibniz, the Ber- noullis, L'HQpital, Clairaut, Euler, Lagrange, and D'Alem- bert, each one, especially after Cavalieri, being primarily an analyst rather than a geometer ; (4) the renaissance of pure geometry, beginning with the nineteenth century and characterized by the descriptive geometry of Monge, the projective geometry of Poncelet, the modern synthetic geometry of Steiner and Von Staudt, the modern analytic geometry of Pliicker, the non-Euclidean hypotheses of Lobachevsky, Bolyai, and Riemann, and the laying of the logical foundations of geometry, a period of remarkable richness in the development of all phases of the science. It is in this fourth period that projective geometry has had its development, even if its origin is more remote. The origin of any branch of science can always be traced far back in human history, and this fact is patent in the case of tliis phase of geometry. The idea of the projection 181 182 HISTORY of a line upon a plane is very old. It is involved in the treatment of the intersection of certain surfaces, due to Archytas, in the fifth century B.C., and appears in various later works by Greek writers. Similarly, the invariant prop- erty of the anharmonic ratio was essentially recognized both by Menelaus in the first century A.D. and by Pappus in the third century. The notion of infinity was also famil- iar to several Greek geometers, so that various concepts that enter into the study of projective geometry were com- mon property long before the science was really founded. One of the first important steps to be taken in modern times, in the development of this form of geometry, was due to Desargues, a French architect. In a work on conic sections, published in 1639, Desargues set forth the founda- tion of the theory of four harmonic points, not as done today, but based on the fact that the product of the dis- tances of two conjugate points from the center is con- stant. He also treated of the theory of poles and polars, although not using these terms. In 1640 Pascal, then only a youth of sixteen, published a brief essay on conies setting forth the well-known theorem that bears his name. The descriptive geometry of Monge is a kind of pro- jective geometry, although it is not what we ordinarily mean by this term. He was a French geometer of the period of the Revolution, and had been in possession of his theory for over thirty years before the publication of his " Geometric Descriptive " (1795). It is true that certain of the features of this work can be traced back to De- sargues, Taylor, Lambert, and Frezier, but it was Monge who worked out the theory as a science. Inspired by the general activity of the period, but following rather in the steps of Desargues and Pascal, Carnot treated chiefly of the metric relations of figures. In particular he investigated HISTORY 183 these relations as connected with the theory of transver- sals, a theory whose fundamental property of a four- rayed pencil goes back to Pappus, and which, though revived by Desargues, was set forth for the first time in its general form by Carnot in his " Geometric de Posi- tion " (1803), and supplemented in his " Theorie des Transversales " (1806). In these works Carnot introduced negative magnitudes, the general quadrilateral, the general quadrangle, and numerous other similar features of value to the elementary geometry of today. Projective geometry had its origin somewhat later than the period of Monge and Carnot. Newton had discovered that all curves of the third order can be derived by central projection from five fundamental types. But in spite of this the theory attracted so little attention for over a century that its origin is generally ascribed to Poncelet. A pris- oner in the Russian campaign, confined at Saratoff on the Volga (1812-1814), " prive," as he says, " de toute espece de livres et de secours, surtout distrait par les malheurs de ma patrie et les miens propres," Poncelet still had the vigor of spirit and the leisure to conceive the great work, "Traite des Propriete's Projectives des Figures," which he published in 1822. In this work was first made promi- nent the power of central projection in demonstration and the power of the principle of continuity in research. His leading idea was the study of projective properties, and as a foundation principle he introduced the anharmonic ratio, a concept, however, which dates back to Menelaus and Pappus, and which Desargues had also used. Mobius, following Poncelet, made much use of the anharmonic ratio in his " Barycentrische Calcul" (1827), but he gave it the name Doppehchnitt-Verhaltnisa (ratio bi8ectionalis~), a term now in common use under Steiner's abbreviated 184 HISTORY form Doppelverhaltniss. The name anharmonic ratio or anharmonic function (rapport anharmonique, or fonction anharmonique^) is due to Chasles, and cross-ratio was sug- gested by Clifford. The anharmonic point-and-line prop- erties of conies have been elaborated by Brianchon, Chasles, Steiner, Dupin, Hachette, Gergonne, and Von Staudt. To Poncelet is due the theory of figures homologiques, the per- spective axis and perspective center (called by Chasles the axis and center of homology), an extension of Carnot's theory of transversals, and the cordes ideales of conies, which Pliicker applied to curves of all orders. Poncelet also discovered what Salmon has called " the circular points at infinity," thus completing and establishing the first great principle of modern geometry, the principle of continuity. Brianchon (1806), through his application of Desargues's theory of polars, completed the foundation which Monge had begun for Poncelet's theory of reciprocal polars (1829). Steiner (1832) gave the first complete discussion of the projective relations between rows, pencils, etc., and laid the foundation for the subsequent development of pure geom- etry. He practically closed the theory of conic sections, of the corresponding figures in three-dimensional space, and of surfaces of the second order. With him opens the period of special study of curves and surfaces of higher order. His treatment of duality and his application of the theory of pro- jective pencils to the generation of conies are masterpieces. Cremona began his publications in 1862. His elementary work on projective geometry (1875) is familiar to English readers in Leudesdorf s translation. The recent contribu- tions have naturally been of an advanced character, seek- ing to lay more strictly logical foundations for the science, and in this line the American work by Professors Veblen and Young is noteworthy. INDEX PAGE Anharmonic ratio . . 21, 24, 183 Asymptote 144 Axial pencil 8, 96 Axial projection 3 Axis, of a cylinder .... 171 of homology 37 of projection 3 of symmetry 170 Base 8 Brianchon point 122 Brianchon theorem . . 121, 148 Bundle . 8 Center, of a conic 163 of homology 37 of involution 74 of projection 2 Central projection ..... 2 Class of a figure 80 Classification, of conies . . . 143 of prime forms .... 10 of projectivities .... 66 Congruent elements .... 72 Congruent pencils 59 Congruent ranges .... 69,68 Conic 109, 115, 143, 156 Conjugate reguli 175 Conjugates ... 31, 72, 157, 164 Constant of homology ... 39 Constructions of second order 135 Correlative propositions . . 15 Cylinder 171 PAOR Desargues's theorem . . , . 131 Descriptive properties ... 21 Developable surface .... 173 Diagonal lines 34 Diagonal points 33 Diagonal triangle . . . . 33, 34 Diameter 163 Directrix 93 Double elements 64 Duality 15, 81, 162 Elements at infinity ... 5, 144 Ellipse 143 Elliptic cylinder 171 Elliptic projectivity .... 66 Envelope 80 Figures of second order . . 101 Flat pencil . . 8, 50, 89, 90, 96, 97 Four-point 33 Four-side 34 Fundamental theorem ... 44 Generation of a figure ... 80 Generator 93 Harmonic conjugates . .31, 151 Harmonic form 31, 36 Harmonic homology .... 39 Harmonic pencil 31 Harmonic range 31 Hexagon 120 Homology 37 185 186 INDEX PAGE Hyperbola 143 Hyperbolic cylinder .... 171 Hyperbolic involution ... 73 Hyperbolic paraboloid ... 178 Hyperbolic projectivity ... 66 Hyperboloids 178 Infinity 5, 144 Involution 72, 74 Line at infinity 5 Line involution 76 Locus 79 Metric properties 21 One-dimensional forms ... 8 One-to-one correspondence . 10 Opposite sides 33, 120 Opposite vertices 34 Order, of a construction . . 135 of a figure 80 Orthogonal projection ... 1 Parabola 143 Parabolic cylinder 171 Parabolic projectivity ... 66 Parallel projection 2 Pascal line 122 Pascal's theorem . . . 121, 147 Pencil 8 Perspectivity 11 Plane, at infinity 5 of points 8 of symmetry 170 Plane figures 101 Point at infinity 5 Point involution 75 Polar 151, 154 Polar reciprocal 162 PAGE Pole 151, 155 Prime forms 8 Principal axes 164 Principal diameter .... 164 Principle of duality . . 15, 162 Projection 1 ? 7 Projectivity 41, 66 Projector 3, 7 Quadrangle 33 Quadric 110 Quadric cone 167 Quadric surface . . . 93, 110, 178 Quadrilateral 34 Range of points 8, 96 Reciprocity 15 Regulus 93, 175 Relation, of angles .... 23 of anharmonic ratios . . 26 of segments 22 Ruled surface 93 Section 3 Self-conjugate triangle . . . 158 Self-corresponding elements . 64 Self -polar triangle . . . . . 158 Sense 22,23 Sheaf 8 Similar figures 39 Similar ranges 69 Similitude, ratio of .... 39 Skew surface . . 93, 111, 173, 178 Steiner's theorem 104 Superposed forms 63 Symbols 1, 11, 41 Ten prime forms 8 Three-dimensional forms . . 8 Two-dimensional forms . 8 Qfi THE LIBRARY UNIVERSITY OF CALIFORNIA Santa Barbara THIS BOOK IS DUE ON THE LAST DATE STAMPED BELOW. 3 1205 00571 4801 UC SOUTHERN REGIONAL LIBRARY FACILITY A 000 81 1 785 5 ill 'fiitlHni