HMMT 十一月 2022 · 团队赛 · 第 8 题

HMMT November 2022 — Team Round — Problem 8

[50] Consider parallelogram ABCD with AB > BC. Point E on AB and point F on CD are marked such that there exists a circle ω passing through A, D, E, F and a circle ω passing through B, C, E, F . 1 2 If ω , ω partition BD into segments BX, XY , Y D in that order, with lengths 200 , 9 , 80, respectively, 1 2 compute BC . 4

解析

[50] Consider parallelogram ABCD with AB > BC. Point E on AB and point F on CD are marked such that there exists a circle ω passing through A, D, E, F and a circle ω passing through B, C, E, F . 1 2 If ω , ω partition BD into segments BX, XY , Y D in that order, with lengths 200 , 9 , 80, respectively, 1 2 compute BC . Proposed by: Albert Wang Answer: 51 Solution: We want to find AD = BC = EF . So, let EF intersect BD at O . It is clear that △ BOE ∼ △ DOF . However, we can show by angle chase that △ BXE ∼ △ DY F : ∠ BEG = ∠ ADG = ∠ CBH = ∠ DF H. This means that EF partitions BD and XY into the same proportions, i.e. 200 to 80. Now, let a = 200 , b = 80 , c = 9 to make computation simpler. O is on the radical axis of ω , ω and its power 1 2 respect to the two circles can be found to be ac bc abc ( a + b + c ) a + = 2 a + b a + b ( a + b ) c ( a + b + c ) 2 However, there is now x for which OE = ax, OF = bx by similarity. This means x = . 2 ( a + b ) Notably, we want to find ( a + b ) x , which is just p √ c ( a + b + c ) = 9 · 289 = 51 . 4