HMMT 二月 2026 · 几何 · 第 3 题

HMMT February 2026 — Geometry — Problem 3

Let ABCD be a rectangle with AB = 12 and BC = 16 . Points W , X , Y , and Z lie on sides AB , BC , CD , and DA , respectively, such that W XY Z is a rhombus with area 120 . Compute XY . ◦

解析

Let ABCD be a rectangle with AB = 12 and BC = 16 . Points W , X , Y , and Z lie on sides AB , BC , CD , and DA , respectively, such that W XY Z is a rhombus with area 120 . Compute XY . Proposed by: Pitchayut Saengrungkongka √ Answer: 5 5 Solution: N Z A D W M O Y B C X Let O be the intersection of the diagonals W Y and XZ . Since OW = OY , we see that O is equidistant from lines AB and CD . Similarly, O is equidistant from lines BC and DA . Thus, O is the center of the rectangle ABCD too. Let M and N be the midpoint of AB and AD . Since W Y is perpendicular to XZ , we get that △ OM W ∼ △ ON Z . Therefore, OW : OZ = OM : ON = 8 : 6 = 4 : 3 . Let OW = 4 x and OZ = 3 x . Then the diagonals of the rhombus have length 8 x and 6 x . Since the rhombus have area 120 , we get that √ 1 (8 x )(6 x ) = 120 = ⇒ x = 5 . 2 √ √ 2 2 Finally, by Pythagorean theorem, the side length of rhombus is (3 x ) + (4 x ) = 5 x = 5 5 . ◦