HMMT 十一月 2018 · 团队赛 · 第 9 题

HMMT November 2018 — Team Round — Problem 9

[ 60 ] Let A, B, C be points in that order along a line, such that AB = 20 and BC = 18. Let ω be a circle of nonzero radius centered at B , and let ` and ` be tangents to ω through A and C , respectively. 1 2 Let K be the intersection of ` and ` . Let X lie on segment KA and Y lie on segment KC such that 1 2 XY ‖ BC and XY is tangent to ω . What is the largest possible integer length for XY ?

解析

[ 60 ] Let A, B, C be points in that order along a line, such that AB = 20 and BC = 18. Let ω be a circle of nonzero radius centered at B , and let ` and ` be tangents to ω through A and C , respectively. 1 2 Let K be the intersection of ` and ` . Let X lie on segment KA and Y lie on segment KC such that 1 2 XY ‖ BC and XY is tangent to ω . What is the largest possible integer length for XY ? Proposed by: James Lin Answer: 35 Note that B is the K -excenter of KXY , so XB is the angle bisector of ∠ AKY . As AB and XY are ◦ ◦ parallel, ∠ XAB + 2 ∠ AXB = 180 , so ∠ XBA = 180 − ∠ AXB − ∠ XAB . This means that AXB is isosceles with AX = AB = 20. Similarly, Y C = BC = 18. As KXY is similar to KAC , we have that KX XA 20 = = . Let KA = 20 x, KC = 18 x , so the Triangle Inequality applied to triangle KAC gives KY Y C 18 KX x − 1 38 KA < KC + AC = ⇒ 20 x < 18 x + 38 = ⇒ x < 19. Then, XY = AC · = 38 · = 38 − < 36, KA x x so the maximum possible integer length of XY is 35. The optimal configuration is achieved when the radius of ω becomes arbitrarily small and ` and ` are on opposite sides of AC . 1 2