HMMT 十一月 2021 · GEN 赛 · 第 9 题

HMMT November 2021 — GEN Round — Problem 9

ABCDE is a cyclic convex pentagon, and AC = BD = CE . AC and BD intersect at X , and BD and CE intersect at Y . If AX = 6, XY = 4, and Y E = 7, then the area of pentagon ABCDE can √ a b be written as , where a, b, c are integers, c is positive, b is square-free, and gcd( a, c ) = 1. Find c 100 a + 10 b + c .

解析

ABCDE is a cyclic convex pentagon, and AC = BD = CE . AC and BD intersect at X , and BD and CE intersect at Y . If AX = 6, XY = 4, and Y E = 7, then the area of pentagon ABCDE can √ a b be written as , where a, b, c are integers, c is positive, b is square-free, and gcd( a, c ) = 1. Find c 100 a + 10 b + c . Proposed by: Eric Shen Answer: 2852 Solution: C D Y X B E A Since AC = BD , ABCD is an isosceles trapezoid. Similarly, BCDE is also an isosceles trapezoid. Using this, we can now calculate that CY = DY = DX − XY = AX − XY = 2, and similarly √ 3 BX = CX = 3. By applying Heron’s formula we find that the area of triangle CXY is 15. 4 Now, note that AC XY 9 [ ABC ] = [ BXC ] = 3[ BXC ] = 3 [ CXY ] = [ CXY ] . CX BX 4 9 Similarly, [ CDE ] = [ CXY ]. Also, 4 CA · CE 81 27 [ ACE ] = [ CXY ] = [ CXY ] = [ CXY ] . CX · CY 6 2 √ 27 Thus, [ ABCDE ] = (9 / 4 + 9 / 4 + 27 / 2)[ CXY ] = 18[ CXY ] = 15. 2