HMMT 十一月 2022 · THM 赛 · 第 9 题

HMMT November 2022 — THM Round — Problem 9

Alice and Bob play the following “point guessing game.” First, Alice marks an equilateral triangle ABC and a point D on segment BC satisfying BD = 3 and CD = 5. Then, Alice chooses a point P BQ BP on line AD and challenges Bob to mark a point Q ̸ = P on line AD such that = . Alice wins if QC P C BP and only if Bob is unable to choose such a point. If Alice wins, what are the possible values of for P C the P she chose?

解析

Alice and Bob play the following “point guessing game.” First, Alice marks an equilateral triangle ABC and a point D on segment BC satisfying BD = 3 and CD = 5. Then, Alice chooses a point P BQ BP on line AD and challenges Bob to mark a point Q ̸ = P on line AD such that = . Alice wins if QC P C BP and only if Bob is unable to choose such a point. If Alice wins, what are the possible values of for P C the P she chose? Proposed by: Pitchayut Saengrungkongka √ √ 3 3 3 Answer: , 1 , 3 5 Solution: First, if P = A then clearly Bob cannot choose a Q . So we can have BP : P C = 1. Otherwise, we need AP to be tangent to the Apollonius Circle. The key claim is that AB = AC = AP . To see why, simply note that since B and C are inverses with respect to the Apollonius Circle, we get that ⊙ ( A, AB ) and the Apollonius Circle are orthogonal. This gives the claim. Finding answer is easy. Let M be the midpoint of BC , and let T be the center of that Apollonius Circle. We easily compute AD = 7, so we have two cases. − − → BP • If P lies on AD , then DP = 1. Since △ DP T ∼ △ ADM , we get that T D = 7. Thus, = P C q q BT 4 1 √ = = . T C 12 3 • Now, note that the two ratios must have product BD/DC = 3 / 5 by the Ratio lemma. So the √ 3 3 other ratio must be . 5 n o √ 1 3 3 √ Therefore, the solution set is , 1 , . 5 3