PUMaC 2022 · 几何（A 组） · 第 8 题
PUMaC 2022 — Geometry (Division A) — Problem 8
题目详情
- Let △ ABC have sidelengths BC = 7, CA = 8, and, AB = 9, and let Ω denote the circumcircle of △ ABC . Let circles ω , ω , ω be internally tangent to the minor arcs BC , CA, AB of Ω, A B C respectively, and tangent to the segments BC, CA, AB at points X, Y, and, Z , respectively. BX CY AZ 1 Suppose that = = = . Let t be the length of the common external tangent AB XC Y A ZB 2 of ω and ω , let t be the length of the common external tangent of ω and ω , and let A B BC B C t be the length of the common external tangent of ω and ω . If t + t + t can be CA C A AB BC CA m expressed as for relatively prime positive integers m, n , find m + n . n (Write answers on next page.) 1 Name: Team: Write answers in table below: Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 2
解析
- Let △ ABC have sidelengths BC = 7, CA = 8, and, AB = 9, and let Ω denote the circumcircle of △ ABC . Let circles ω , ω , ω be internally tangent to the minor arcs BC , CA, AB of Ω, A B C respectively, and tangent to the segments BC, CA, AB at points X, Y, and, Z , respectively. BX CY AZ 1 Suppose that = = = . Let t be the length of the common external tangent AB XC Y A ZB 2 of ω and ω , let t be the length of the common external tangent of ω and ω , and let A B BC B C t be the length of the common external tangent of ω and ω . If t + t + t can be CA C A AB BC CA m expressed as for relatively prime positive integers m, n , find m + n . n Proposed by Sunay Joshi Answer: 59 BX 1 2 2 Let k = = . First, we show that t = (1 − k ) a + k b +(1 − k ) kc . Let t , t , t denote the AB A B C BC 3 length of the tangent from A, B, C to ω , ω , ω , respectively. By Casey’s Theorem applied A B C to circles ( A ) , ( B ) , ω , ( C ), we find that c · CX + b · BX = a · t . Since CX = (1 − k ) a A A and BX = ka , solving yields t = kb + (1 − k ) c . By symmetry we find t = kc + (1 − k ) a . A B Applying Casey’s Theorem to circles ( A ) , ( B ) , ω , ω , we find c · t + AY · BX = t · t . A B AB A B Since AY = (1 − k ) b and BX = ka , solving yields the claimed expression for t . AB 2 2 2 2 By symmetry, we therefore have t = (1 − k ) b + k c +(1 − k ) ka and t = (1 − k ) c + k a +(1 − BC CA 2 2 2 k ) kb . Summing yields t + t + t = ((1 − k ) + k +(1 − k ) k )( a + b + c ) = ( k − k +1)( a + b + c ). AB BC CA 1 7 56 Plugging in k = and a = 7, b = 8, c = 9, we find · 24 = and hence an answer of 56+3 = 59. 3 9 3 4