HMMT 二月 2025 · 几何 · 第 9 题
HMMT February 2025 — Geometry — Problem 9
题目详情
- Let ABCD be a rectangle with BC = 24. Point X lies inside the rectangle such that ∠ AXB = 90 . Given that triangles △ AXD and △ BXC are both acute and have circumradii 13 and 15, respectively, compute AB .
解析
- Let ABCD be a rectangle with BC = 24. Point X lies inside the rectangle such that ∠ AXB = 90 . Given that triangles △ AXD and △ BXC are both acute and have circumradii 13 and 15, respectively, compute AB . Proposed by: Pitchayut Saengrungkongka √ Answer: 14 + 4 37. Solution 1: Let M be the midpoint of AB . Let O and O be the circumcenters of △ AXD and 1 2 △ BXC , respectively. Since O M is the perpendicular bisector of AX and O M is the perpendicular 1 2 ◦ bisector of BX , we get that ∠ O M O = 90 . 1 2 Let P and P be the projections of O and O onto segment AB , respectively, and let AB = 2 x . By 1 2 1 2 p √ 2 2 2 2 the Pythagorean theorem, P A = O A − O P = 15 − 12 = 5, so M P = M A − P A = x − 5. 1 1 1 1 1 1 √ 2 2 Likewise, M P = M B − 15 − 12 = x − 9. Since △ M P O ∼ △ O P M , we know 2 1 1 2 2 2 ( x − 5)( x − 9) = M P · M P = O P · P O = 12 . 1 2 2 2 1 1 √ √ Solving this, we get x = 7 + 2 37, which implies that AB = 2 x = 14 + 4 37 . (The condition that √ △ AXD and △ BXC are acute rules out 14 − 4 37.) P 1 P M 2 A B O O 1 2 X D C Solution 2: A B X D C P Q Let P be the antipode of A in ⊙ ( AXD ) and Q be the antipode of B in ⊙ ( BXC ). From ∠ P DA = ◦ ◦ ∠ QCB = 90 , we get that P and Q lie on CD . Moreover, from ∠ P XA = 90 , we get that P ∈ BX , and similarly Q ∈ AX . √ 2 2 Being a diameter, AP = 2 · 13 = 26, so by the Pythagorean theorem, DP = 26 − 24 = 10. Similarly, √ 2 2 BQ = 30 and CQ = 30 − 24 = 18. Letting AB = x , we get P Q = x − 28. Quadrilateral ABQP has 2 2 2 2 2 2 2 2 perpendicular diagonals, so AB + P Q = AP + BQ , which means that x + ( x − 28) = 26 + 30 . √ Solving this quadratic gives x = 14 + 4 37 . (The condition that △ AXD and △ BXC are acute rules √ out 14 − 4 37.)