PUMaC 2023 · 几何（B 组） · 第 4 题

PUMaC 2023 — Geometry (Division B) — Problem 4

Let △ ABC be an isosceles triangle with AB = AC = 7 and BC = 1. Let G be the centroid of △ ABC . Given j ∈ { 0 , 1 , 2 } , let T denote the triangle obtained by rotating △ ABC about j G by 2 πj/ 3 radians. Let P denote the intersection of the interiors of triangles T , T , T . If K 0 1 2 a 2 denotes the area of P , then K = for relatively prime positive integers a, b . Find a + b . b

解析

Let △ ABC be an isosceles triangle with AB = AC = 7 and BC = 1. Let G be the centroid of △ ABC . Given j ∈ { 0 , 1 , 2 } , let T denote the triangle obtained by rotating △ ABC about j G by 2 πj/ 3 radians. Let P denote the intersection of the interiors of triangles T , T , T . If K 0 1 2 a 2 denotes the area of P , then K = for relatively prime positive integers a, b . Find a + b . b Proposed by Sunay Joshi Answer: 1843 Construct the equilateral triangle △ AXY with sidelength 3 such that BC is the middle third of the side XY (with B closer to X , WLOG). Note that T , T , T lie within △ AXY ; they are 0 1 2 simply △ ABC rotated. Let M, N lie on AY, AX such that AM/M Y = 1 / 2 and AN/N X = 1 / 2. Let P = AB ∩ M X , Q = N Y ∩ M X , and R = AC ∩ N Y . Then it is easy to see by symmetry that we see K = 3 · [ P QRG ]. Since P QRG has perpendicular diagonals, its area is 1 1 given by · QG · P R . To compute P R , note by mass points that AP/P B = 3 / 2, hence by similar 2 triangles P R = 3 / 5 · BC = 3 / 5. By mass points, we also have hat AQ is half the height of √ √ √ △ AXY , hence QG = AG − AQ = 3 / 4. Solving for K , we find K = 3 / 2 · 3 / 4 · 3 / 5 = 9 3 / 40. 2 Squaring yields K = 243 / 1600 and our answer is 243 + 1600 = 1843.