PUMaC 2020 · 几何（B 组） · 第 3 题

PUMaC 2020 — Geometry (Division B) — Problem 3

Let γ and γ be circles centered at O and P respectively, and externally tangent to each other 1 2 at point Q. Draw point D on γ and point E on γ such that line DE is tangent to both 1 2 circles. If the length OQ = 1 and the area of the quadrilateral ODEP is 520 , then what is the value of length P Q ?

解析

Let γ and γ be circles centered at O and P respectively, and externally tangent to each other 1 2 at point Q. Draw point D on γ and point E on γ such that line DE is tangent to both 1 2 circles. If the length OQ = 1 and the area of the quadrilateral ODEP is 520 , then what is the value of length P Q ? Proposed by: Ollie Thakar Answer: 64 Let r be the radius OQ of γ and s the radius P Q of γ . 1 2 It is a well-known theorem that angle ∠ EQD is right, and the length of the hypotenuse ED √ is 2 rs. Call a = EQ and b = DQ. Call x the measure of angle ∠ DQO. Then, ∠ EQP has measure 90 − x. Furthermore, triangles DOQ and EP Q are isosceles, so a = 2 s sin x and b = 2 s cos x. 2 r 2 2 2 Since triangle EQD is right, we have a + b = ED , which gives us sin x = and r + s s 2 cos x = . r + s The area A of quadrilateral ODEP is given by the sum of the areas of triangles DOQ, EP Q, and EDQ, so: √ 1 1 1 2 A = ( s cos x )(2 s sin x )+ ( r sin x )(2 r cos x )+ 4 rs sin x cos x = ( r + s ) sin x cos x = ( r + s ) rs. 2 2 2 √ We are given that r = 1 and then that (1 + s ) s = 520 , which can be solved as s = 64 by inspection. 1