SECTION V. ANHARMONIC AND HARMONIC RATIO. DEFINITIONS. 1. THE anharmonic ratio of four points is the quotient obtained by dividing the ratio of the distances of the first two from the third, by the ratio of the distances of the first two from the fourth. Thus, the anharmonic ratio of the A four points A, B, C, D is AC AD ÷ B C D 2. If the anharmonic ratio of four points be equal to unity, they are said to be in harmonic ratio. Thus, if AC AD 1, or AC AD A, B, C, D are harmonic points, which agrees with Def. 5, page 245. 3. A pencil is a system of straight lines diverging from a point. 4. A ray is one of these diverging lines. 5. The vertex of the pencil is the point from which they diverge. Proposition 14. Theorem.-The anharmonic ratio of a system of points is not changed by interchanging two of the points, provided the other two points be also interchanged. Hence, also, [ABCD]=[BADC] = [CDAB] = [DCBA]. Scholium.—There are twenty-four ways in which these four letters may be arranged; hence there are six different anharmonic ratios which may be formed from them. Proposition 15. Theorem. If a pencil of four rays be cut by any two transversals, the anharmonic ratio of the four cutting points of one transversal is equal to the anharmonic ratio of the four corresponding points of the second. Scholium.-The anharmonic ratio of a pencil of four rays is the anharmonic ratio of four points cut from the rays by any transversal. The angles of a pencil are the six angles formed by the four rays with each other, taken two and two. Hence, the values of the anharmonic ratios of two pencils are equal, when the rays make equal angles with each other, each to each. The anharmonic ratio of a pencil is represented by [0 - ABCD]. Proposition 16. Theorem. If four points be taken in the circumference of a circle, and lines be drawn to any other point in the circumference, the anharmonic ratio of the pencil is the same, whatever the position of the fifth point. For the angles at O and O' are equal; hence (Prop. 15, Sch.), [Ó-ABCD]-[0-ABCD]. Preposition 17. Theorem.-If the centre of a circle be joined with the points of intersection of four fixed tangents with a fifth tangent, the anharmonic ratio of the pencil formed is the same, whatever be the position of the fifth tangent. Let four tangents touch the circle in A, B, C, D, and intersect a fifth tangent in E, F, G, H; then will [0-EFGH] be the same, whatever the position of EFGH For the angle EOK is one half AOK, and FOK is one half BOK; therefore EOF is one half the fixed angle AOB, and is therefore constant, whatever the position of EFGH The same may be proved of FOG and GOH. Hence the anharmonic ratio of the pencil is the same, whatever the position of EFGH. E Proposition 18. K Problem.-To divide a straight line harmonically in a given Scholium.-If any point not in AB be taken, and a pencil of four rays drawn through A, C, B, D, it is said to be a harmonic pencil. Proposition 19. Problem. To find the locus of all points whose distances from two given points are in a given ratio. Let A and B be the points, and let M: N be the given ratio. Divide (Prop. 18) AB harmonically in C and D; on CD describe a circle. The circumference of this circle will be the required locus. Take any point, P, and F D E join PA, PC, PB, PE, and PD. Because AB is divided harmonically in C and D, or, hence, or, AD AC: BD : BC, AE+ EC: AE-EC :: CE+EB : CE- EB ; Therefore the sides about the common angle E of the triangles AEP, BEP are proportional; hence the triangles are similar, and the angle BPE is equal to A. But P is any point of the circumference; hence the circumference is the required locus. |