HMMT 十一月 2022 · 团队赛 · 第 6 题

HMMT November 2022 — Team Round — Problem 6

[45] A triangle XY Z and a circle ω of radius 2 are given in a plane, such that ω intersects segment XY at the points A, B , segment Y Z at the points C, D , and segment ZX at the points E, F . Suppose that XB > XA , Y D > Y C , and ZF > ZE . In addition, XA = 1 , Y C = 2 , ZE = 3, and AB = CD = EF . Compute AB .

解析

[45] A triangle XY Z and a circle ω of radius 2 are given in a plane, such that ω intersects segment XY at the points A, B , segment Y Z at the points C, D , and segment ZX at the points E, F . Suppose that XB > XA , Y D > Y C , and ZF > ZE . In addition, XA = 1 , Y C = 2 , ZE = 3, and AB = CD = EF . Compute AB . Proposed by: Rishabh Das √ Answer: 10 − 1 Solution: Let d = AB and x = d/ 2 for ease of notation. Let the center of ( ABCDEF ) be I . Because AB = CD = EF , the distance from I to AB , CD , and EF are the same, so I is the incenter of △ XY Z . Let △ XY Z have inradius r . By symmetry, we have XF = 1, Y B = 2, and ZD = 3. Thus, △ XY Z has side lengths d + 3, d + 4, and d + 5. Heron’s Formula gives the area of △ XY Z is p p K = (3 x + 6) ( x + 2) ( x + 1) ( x + 3) = ( x + 2) 3 ( x + 1) ( x + 3) , while K = rs gives the area of △ XY Z as K = (3 x + 6) r. Equating our two expressions for K , we have p p 2 ( x + 2) 3 ( x + 1) ( x + 3) = (3 x + 6) r = ⇒ 3 ( x + 1) ( x + 3) = 3 r = ⇒ ( x + 1) ( x + 3) = 3 r . 2 2 2 2 The Pythagorean Theorem gives x + r = 4, so r = 4 − x . Plugging this in and expanding gives 2 2 ( x + 1) ( x + 3) = 3 4 − x = ⇒ 4 x + 4 x − 9 = 0 . √ √ − 1 ± 10 This has roots x = , and because x > 0, we conclude that d = 10 − 1. 2