HMMT 十一月 2022 · 冲刺赛 · 第 28 题

HMMT November 2022 — Guts Round — Problem 28

[15] Let ABC be a triangle with AB = 13, BC = 14, and CA = 15. Pick points Q and R on AC and ◦ AB such that ∠ CBQ = ∠ BCR = 90 . There exist two points P ̸ = P in the plane of ABC such that 1 2 △ P QR , △ P QR , and △ ABC are similar (with vertices in order). Compute the sum of the distances 1 2 from P to BC and P to BC . 1 2

解析

[15] Let ABC be a triangle with AB = 13, BC = 14, and CA = 15. Pick points Q and R on AC and ◦ AB such that ∠ CBQ = ∠ BCR = 90 . There exist two points P ̸ = P in the plane of ABC such that 1 2 △ P QR , △ P QR , and △ ABC are similar (with vertices in order). Compute the sum of the distances 1 2 from P to BC and P to BC . 1 2 Proposed by: Ankit Bisain Answer: 48 Solution 1: Let T be the foot of the A -altitude of ABC . Recall that BT = 5 and CT = 9. ′ ′ Let T be the foot of the P -altitude of P QR . Since T is the midpoint of the possibilities for P , the answer is X ′ d ( P, BC ) = 2 d ( T , BC ) . P ′ Since T splits QR in a 5 : 9 ratio, we have 9 d ( Q, BC ) + 5 d ( R, BC ) ′ d ( T , BC ) = . 14 14 ′ By similar triangles, d ( Q, BC ) = QB = 12 · , and similar for d ( R, BC ), giving d ( T , BC ) = 24, and 9 an answer of 48. ′ Solution 2: As in the previous solution, let T be the foot from A to BC , let T be the foot from P to QR . and recall that ′ d ( P , BC ) + d ( P , BC ) = 2 d ( T , BC ) . 1 2 ′ ′ ′ Now, notice that since △ P QR ∼ △ ABC , we have QT : T R = BT : T C , so T T ∥ BQ ∥ CR , implying ′ that A ∈ T T . ′ However, we recall a well-known fact that A is the midpoint of T T (can be proven by simple similar ′ triangles). Thus, d ( T , BC ) is equal to two times the altitude from A to BC . Hence, the answer is four times the altitude from A to BC , which is 48.