HMMT 二月 2026 · 几何 · 第 2 题

HMMT February 2026 — Geometry — Problem 2

Let HORSE be a convex pentagon such that ∠ EHO = ∠ ORS = ∠ SEH = 90 and ∠ HOR = ◦ ∠ RSE = 135 . Given that HO = 20 , SE = 26 , and OS = 10 , compute the area of HORSE .

解析

Let HORSE be a convex pentagon such that ∠ EHO = ∠ ORS = ∠ SEH = 90 and ∠ HOR = ◦ ∠ RSE = 135 . Given that HO = 20 , SE = 26 , and OS = 10 , compute the area of HORSE . Proposed by: Jason Mao Answer: 191 Solution: H 20 O a X a 8 8 10 R b E 20 O 6 S Y 1 ©2026 HMMT Let O be the foot of the perpendicular from O onto SE . We have O S = 6 , so HE = OO = 8 . 1 1 1 (20+26) · 8 Hence, the area of trapezoid HOSE is = 184 . 2 Now, draw circumscribing rectangle HXY E such that R lies on XY . Note that △ OXR and △ RY S are both isosceles right triangles. Let OX = XR = a and RY = Y S = b . We have a − b = O S = 6 1 and a + b = XY = HE = 8 . √ √ Thus, we have a = 7 and b = 1 . Hence, OR = 7 2 and RS = 2 , giving [ ORS ] = 7 . Finally, [ HORSE ] = [ HOSE ] + [ ORS ] = 184 + 7 = 191 .