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An Elementary Treatise on Pure Geometry; With Numerous Examples

An Elementary Treatise on Pure Geometry; With Numerous Examples

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ISBN10: 1150995300
ISBN13: 9781150995309
Publisher: General Books
Pages: 240
Weight: 0.79
Height: 0.54 Width: 9.01 Depth: 5.98
Language: English
This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1893 edition. Excerpt: ...Ex. 9. If any two segments AA', BB' joining corresponding points in an involution have the same middle point, then all such segments have the same middle point. For the other double point must be n'. Ex. 10. If AP. AP ' = A'P. A'P'. show that the points P and V form an involution in which A and A' are corresponding points; and find the centre and double points. Ex. 11. If any transversal through V (the internal vertex of the harmonic triangle of a quadrilateral circumscribing a conic) cut the sides in AA', BB' and the conic in PP; show that (AA', BB', PP1) is an involution, the double points being V and tfte meet of UW with the transversal. Ex. 12. Through a given point 0 draw a line meeting two conics (or two pairs of lines) in points AA', BB' such that (OAA'BB') = (OA'AB'B). Join 0 to the meet of the polars of 0. Ex. 13. If ABC..., A'B'C... be two homographic ranges on the same line, and if the mates ofP(= Qf) be V and Q, we know that the ranges A and A' and the ranges f and Q have the same common points (E, F say); show that P has the same fourth harmonic for f Q and for EF. (See X. 7. Ex. 4.) We have only to prove that P (= Q/) is one of the double points of the involution determined by P'Q, EF. Now (PQEF)-(P'O'EF) from the first homography = p'PEF) = (PP'FE). Hence P'Q, EF, PP are in involution, i.e. P is a double point of the involution. Ex. 14-With the same data, P and the fourth harmonic of P for PQ generate an involution. 10. A system of coaxal circles is cut by any transversal in pairs of points in involution. For if the transversal cut the circles in AA', BB', CC, ... and the radical axis in 0, then OA. 0A'= OB. 0B'= OC. 0C'=.... Hence A A', BB', CC, ... form an involution of which 0 is the centre. Ex. 1. Give a geometrical...

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