Online Geometry theorems, problems, solutions, and related topics.
Geometry Problem. Post your solution in the comments box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the diagram below to enlarge it.
A property of circles under inversive transformation is quoted here without proof: Let C₁ and C₂ be two circles with radius r₁ and r₂ respectively, with the length of common tangent=l. If under an inversive transformation, they becomes another two circle C₁' and C₂', with radius R₁ and R₂ respectively, and with length of common tangent=L. Then L²/(R₁R₂) = l²/(r₁r₂). ***Let the tangent points of circles A B C D with circle O are P Q R S respectively. Choose any point X on arc AD, perform an inversion with center X. Then circle O becomes a straight line, and circles A B C D becomes circles A' B' C' D'. Let circles A B C D with radius r₁ r₂ r₃ r₄ and circles A' B' C' D' with radius R₁ R₂ R₃ R₄ respectively. Let the image of the tangent points are P' Q' R' S'. Then by Euler's theorem (similar to Ptolemy's theorem), P'Q'×R'S' + P'S'×Q'R' = P'R'×Q'S'P'Q'/√(R₁R₂)×R'S'/√(R₃R₄) + P'S'/√(R₁R₄)×Q'R'/√(R₂R₃) = P'R'/√(R₁R₃)×Q'S'/√(R₂R₄)By the stated property, PQ/√(r₁r₂)×RS/√(r₃r₄) + PS/√(r₁r₄)×QR/√(r₂r₃) = PR/√(r₁r₃)×QS/√(r₂r₄)Hence, PQ×RS + PS×QR = PR×QSa×c + b×d = x×y