PUMaC 2022 · 几何（B 组） · 第 6 题

PUMaC 2022 — Geometry (Division B) — Problem 6

Let △ ABC be an equilateral triangle. Points D, E, F are drawn on sides AB, BC, and CA respectively such that [ ADF ] = [ BED ] + [ CEF ] and △ ADF ∼ △ BED ∼ △ CEF . The ratio √ [ ABC ] a + b c can be expressed as , where a, b, c, and d are positive integers such that a and d [ DEF ] d are relatively prime, and c is not divisible by the square of any prime. Find a + b + c + d . (Here [ P ] denotes the area of polygon P .)

解析

Let △ ABC be an equilateral triangle. Points D, E, F are drawn on sides AB, BC, and CA respectively such that [ ADF ] = [ BED ] + [ CEF ] and △ ADF ∼ △ BED ∼ △ CEF . The ratio √ [ ABC ] a + b c can be expressed as , where a, b, c, and d are positive integers such that a and d [ DEF ] d are relatively prime, and c is not divisible by the square of any prime. Find a + b + c + d . (Here [ P ] denotes the area of polygon P .) Proposed by Adam Huang Answer: 17 2 2 Assume WLOG that △ ABC has sidelength 1. The similarity condition implies DE + EF = 2 DF , hence ∠ DEF = 90. Angle chasing also yields ∠ BED = 45, so that △ BED, △ CEF are 60-45-75 triangles and △ DEF is a 30-60-90 right triangle. By the Law of Sines applied √ z x z 3 to △ BED and △ CEF , the lengths z = DE and x = BE satisfy = and = sin 60 sin 75 sin 60 √ √ √ √ 2 1 − x 3 − 1 2 6 − 3 2 z 3 . Solving, we find x = and z = . Thus [ DEF ] = and the ratio is sin 75 2 2 2 √ √ 3 / 4 7+4 3 √ [ ABC ] / [ DEF ] = , which reduces to . Our answer is 7 + 4 + 3 + 3 = 17. 2 3 z 3 / 2 2