HMMT 二月 2014 · 冲刺赛 · 第 25 题

HMMT February 2014 — Guts Round — Problem 25

[ 17 ] Let ABC be an equilateral triangle of side length 6 inscribed in a circle ! . Let A , A be the 1 2 points (distinct from A ) where the lines through A passing through the two trisection points of BC meet ! . Define B , B , C , C similarly. Given that A , A , B , B , C , C appear on ! in that order, 1 2 1 2 1 2 1 2 1 2 find the area of hexagon A A B B C C . 1 2 1 2 1 2

解析

[ 17 ] Let ABC be an equilateral triangle of side length 6 inscribed in a circle ω . Let A , A be the 1 2 points (distinct from A ) where the lines through A passing through the two trisection points of BC meet ω . Define B , B , C , C similarly. Given that A , A , B , B , C , C appear on ω in that order, 1 2 1 2 1 2 1 2 1 2 find the area of hexagon A A B B C C . 1 2 1 2 1 2 √ 846 3 ′ ′ ′ Answer: Let A be the point on BC such that 2 BA = A C . By law of cosines on triangle 49 √ ′ ′ ′ 2 ∗ 4 4 √ √ AA B , we find that AA = 2 7. By power of a point, A A = = . Using side length ratios, 1 2 7 7 √ 4 √ 2 7+ AA 7 18 1 √ A A = 2 = 2 = . 1 2 ′ AA 7 2 7 Now our hexagon can be broken down into equilateral triangle A B C and three copies of triangle 1 1 1 A C C . Since our hexagon has rotational symmetry, ∠ C = 120, and using law of cosines on this 1 1 2 2 18 30 triangle with side lengths and 6, a little algebra yields A C = (this is a 3-5-7 triangle with an 1 2 7 7 angle 120). √ √ √ 2 6 3 3 846 3 1 18 30 The area of the hexagon is therefore + 3 ∗ = 4 2 7 7 2 49