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PUMaC 2022 · 几何（B 组） · 第 7 题

PUMaC 2022 — Geometry (Division B) — Problem 7

专题
Discrete Math / 离散数学
难度
L3
来源
PUMaC

题目详情

  1. Let △ ABC be a triangle with AB = 5, BC = 8, and, CA = 7. Let the center of the A -excircle be O , and let the A -excircle touch lines BC , CA , and, AB at points X , Y , and, Z , respectively. Let h , h , and, h denote the distances from O to lines XY , Y Z , and, ZX , respectively. If 1 2 3 2 2 2 m h + h + h can be written as for relatively prime positive integers m, n , find m + n . 1 2 3 n
解析
  1. Let △ ABC be a triangle with AB = 5, BC = 8, and, CA = 7. Let the center of the A -excircle be O , and let the A -excircle touch lines BC , CA , and, AB at points X , Y , and, Z , respectively. Let h , h , and, h denote the distances from O to lines XY , Y Z , and, ZX , respectively. If 1 2 3 2 2 2 m h + h + h can be written as for relatively prime positive integers m, n , find m + n . 1 2 3 n Proposed by Sunay Joshi Answer: 2189 Let a, b, c denote the lengths of sides BC, CA, AB , and let r denote the radius of the A - A 2 2 2 C B A r sin r sin r cos A A A 2 2 2 excircle. We claim that h = , h = , and h = . We begin with 1 2 3 s − b s − c s 1 1 2 h . Computing the area of △ OXY in two ways, we find h · XY = r sin XOY . Since 1 1 A 2 2 C XY = 2( s − b ) cos and ∠ XOY = C , solving the equation for h yields the desired formula. 1 2 By symmetry, this implies the expression for h . For h , we compute the area of △ Y OZ in 3 2 1 1 A 2 two ways to find h · Y Z = r sin Y OZ . Since ∠ Y OZ = π − A and Y Z = 2 s sin , solving 2 A 2 2 2 the equation for h yields the desired formula. 2 Having established the above, we now compute each of h , h , h . By the Law of Cosines, 1 2 3 1 1 11 A 2 √ cos A = , cos B = , and cos C = . By the half-angle formulae, it follows that cos = , 7 2 14 2 7 √ B 1 C 3 √ sin = , and sin = . Next, since r ( s − a ) = K , Heron’s formula implies that A 2 2 2 2 7 s ( s − b )( s − c ) 2 r = = 75. Putting everything together, we find that A s − a √ √ √ 2 / 7 3 / (2 7) 1 / 2 2175 2 2 2 2 2 2 2 h + h + h = 75 · ( ) + ( ) + ( ) = 1 2 3 10 3 5 14 This gives an answer of m + n = 2189.