HMMT 二月 2014 · 几何 · 第 9 题

HMMT February 2014 — Geometry — Problem 9

Two circles are said to be orthogonal if they intersect in two points, and their tangents at either point of intersection are perpendicular. Two circles ! and ! with radii 10 and 13, respectively, are externally 1 2 p tangent at point P . Another circle ! with radius 2 2 passes through P and is orthogonal to both ! 3 1 and ! . A fourth circle ! , orthogonal to ! , is externally tangent to ! and ! . Compute the radius 2 4 3 1 2 of ! . 4

解析

Two circles are said to be orthogonal if they intersect in two points, and their tangents at either point of intersection are perpendicular. Two circles ω and ω with radii 10 and 13, respectively, are externally 1 2 √ tangent at point P . Another circle ω with radius 2 2 passes through P and is orthogonal to both ω 3 1 and ω . A fourth circle ω , orthogonal to ω , is externally tangent to ω and ω . Compute the radius 2 4 3 1 2 of ω . 4 92 Answer: Let ω have center O and radius r . Since ω is orthogonal to ω , ω , ω , it has equal i i i 3 1 2 4 61 2 power r to each of them. Thus O is the radical center of ω , ω , ω , which is equidistant to the three 3 1 2 4 3 sides of △ O O O and therefore its incenter. 1 2 4 For distinct i, j ∈ { 1 , 2 , 4 } , ω ∩ ω lies on the circles with diameters O O and O O , and hence ω i j 3 i 3 j 3 r r r 130 r 92 2 1 2 4 4 itself. It follows that ω is the incircle of △ O O O , so 8 = r = = = ⇒ r = . 3 1 2 4 4 3 r + r + r 23+ r 61 1 2 4 4 Comment: The condition P ∈ ω is unnecessary. 3