HMMT 十一月 2024 · GEN 赛 · 第 9 题
HMMT November 2024 — GEN Round — Problem 9
题目详情
- Let ABCDEF be a regular hexagon with center O and side length 1 . Point X is placed in the interior ◦ of the hexagon such that ∠ BXC = ∠ AXE = 90 . Compute all possible values of OX .
解析
- Let ABCDEF be a regular hexagon with center O and side length 1. Point X is placed in the interior ◦ of the hexagon such that ∠ BXC = ∠ AXE = 90 . Compute all possible values of OX . Proposed by: Ethan Liu, Isabella Zhu, Pitchayut Saengrungkongka √ 1 7 Answer: , 2 7 Solution 1: A B X 1 O F C X 2 E D Point X is the intersection of circles with diameter AE and BC . Thus, there are two possible inter- section points. Since AC ⊥ BE , the first point, X , is the intersection of AC and BE , from which we 1 1 can see OX = as our first answer. Let X be the other intersection point. 1 2 2 1 Let M be the midpoint of BC and N be the midpoint of AE . Then M X = N O = and M O = 2 2 √ 3 N X = , so OX M N is an isosceles trapezoid. By law of cosine on △ OM N , we have 2 2 2 p 2 2 ◦ M N = OM + ON − 2 · OM · ON cos 150 r √ √ √ √ 2 2 3 1 3 1 3 7 = + + 2 · · · = . 2 2 2 2 2 2 Moreover, by Ptolemy’s theorem, 2 2 1 OX · M N = M O − N O = . 2 2 √ 7 Combining the previous two equations gives OX = . 2 7 Solution 2: Recall the first paragraph of the previous solution that X = AC ∩ BE is the first point. 1 Thus, the second point X is the Miquel point of cyclic quadrilateral ACBE . 2 By a well-known property of Miquel point, if Y = AB ∩ CE , then Y and X are inverses with respect 2 to the circumcircle of ABCDEF . Thus, OX · OY = 1. 2 ◦ One can compute OY as follows: from triangle BCY , we get that BY = BC/ sin 30 = 2. Thus, by power of point, √ 2 2 OY − 1 = Y B · Y A = 2 · 3 = 6 = ⇒ OY = 7 , √ 7 implying OX = . 2 7