Substituting these values, we have, ΠΑ 2 vol. AHBCD = (2r2+2r2+2rr') +3πA (R2 — LM2). But since ALM and ABK are right-angled triangles, we have, R2— LM2 = AL2= | AB2 = — (BK2+AK2)= {} [A2+(~~~')']; COR. If the segment has but one base, as vol. ABED, the radius r - O, and we have, vol. ABED = 17r2A+1πA3. MODERN GEOMETRY. BOOK IX.* ANHARMONIC RATIO. I. IF A, B, C, and D are four points in a straight line, the ratio AC AD CB DB is called the anharmonic ratio of the four points. This ratio is also expressed by [ABCD], which gives the A order in which the ratio is written. B The line AB is considered positive, and BA negative. (See p. 2.) II. The value of the anharmonic ratio of four points is not changed when two of the points are interchanged, provided the other two are also interchanged. for Thus [ABCD]=[BADC] = [CDAB]=[DCBA]; AC AD BD BC CA CB DB DA BC AC This book is designed for those students who have time and desire to learn the elements of modern geometry in addition to what has, until recently, been given in our text-books of elementary geometry. The following authors are recommended to those who wish to pursue the subject further, - Mulchay, Townsend, Salmon, Rouché and Comberousse, Chasles, Poncelet, Poinsot, and Steiner. Any other changes in the order of points will change the value of the ratio. The four letters may be written in twentyfour different orders; and hence, from four points in a line, six different anharmonic ratios may be formed. (Write the six anharmonic ratios, and show that three are the reciprocals of the other three.) III. AC AD is positive when the two The anharmonic ratio points C and D are both between A and B, or both without; and negative, when one is between, and the other without. If ABCD is the order of the points in the line, and AB is supposed to be cut by C and D, or AD is supposed to be cut by B and C, the anharmonic ratios are positive. If AC is supposed to be cut by B and D, the anharmonic ratio is negative. The reciprocals of these have the same signs. Hence, of the six anharmonic ratios of four points, four are positive, and two are negative. That is, if two points coincide with each other, the anharmonic ratio takes one of the values, ∞, 0, +1. AC AD = 1. (See Harmonic Proportion.) IV. If one of the points, as C, bisects AB, or is at an infinite distance, AC = CB, and the anharmonic ratio AC AD = DB also the five remaining anharmonic ratios reduce to simple ratios. DEF. 1. A system of lines passing through a common point is called a pencil. Each line is called a ray; and the common point, the vertex. Theorem. If a pencil of four rays is cut by any transversal, the anharmonic ratio of the points of intersection is constant for all positions of the transversal. HYPOTH. O-ABCD is a pencil B d A d B D cut by the transversals ABCD and A'B'C'D'. TO BE PROVED. [ABCD]=[A'B'C'D′]. PROOF. Draw Bd and B'd' || OA. Then ▲ OAC ~ cBC, and ▲ OAD ~ hence dBD (III., 22): Dividing these equations member by member, we have, SCHOLIUM. The anharmonic ratio is the same when the transversal cuts one or more of the rays on the other side of the vertex. (Draw the figure, and apply the same demonstration.) DEF. 2. The anharmonic ratio [ABCD] of the four points in the transversal is also called the anharmonic ratio of the pencil, which may be expressed by the form [0-ABCD]. COR. 1. If two pencils are mutually equiangular, they have the same anharmonic ratio. B COR. 2. If the intersections of the rays of two pencils are in the same straight line, they have the same anharmonic ratio. COR. 3. If two pencils have the same anharmonic ratio, it does not follow that they are mutually equiangular. VI. Problem. Given three rays, the anharmonic ratio, and the relative position of the fourth ray, of a pencil, to find the exact position of the fourth ray. Given OA, OB, OC, and [ABCD]; also OD lies to the right of OC. Find the exact position of OD. B A B (5), which fixes the position of d and of the ray OdD. COR. D is a fixed point in the transversal ABC when the anharmonic ratio [ABCD] is given. Hence, also, the ray OD is fixed. From this follow the two corresponding theorems, VII. and VIII. |