HMMT 二月 2026 · 冲刺赛 · 第 16 题

HMMT February 2026 — Guts Round — Problem 16

[9] Let O and G be the circumcenter and centroid of triangle ABC , respectively, and let M be the √ √ midpoint of side BC . Given that OG = 1 , OM = 2 , and GM = 3 , compute the area of triangle ABC . ©2026 HMMT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT February 2026, February 14, 2026 — GUTS ROUND Organization Team Team ID#

解析

[9] Let O and G be the circumcenter and centroid of triangle ABC , respectively, and let M be the √ √ midpoint of side BC . Given that OG = 1 , OM = 2 , and GM = 3 , compute the area of triangle ABC . Proposed by: Jason Mao √ √ Answer: 3 30 = 270 Solution: A O G B C D M ©2026 HMMT Observe that ∠ M OG is right, so lines OG and BC are parallel. Let D be the foot of the altitude from √ A to BC so since AM = 3 · GM , we have AD = 3 · OM = 3 2 . We also note M D = 3 · OG = 3 . This means letting r be the radius of the circumcircle of △ ABC , we see 2 2 2 2 2 OM + BM = r = ( AD − OM ) + ( M D ) = 17 . √ √ Solving for BM gets BM = 15 so BC = 2 15 . This means that our area is ( ) ( ) √ √ √ 1 · 3 2 · 2 15 = 3 15 . 2