HMMT 二月 2026 · 几何 · 第 10 题
HMMT February 2026 — Geometry — Problem 10
题目详情
- Let ABC be a triangle with centroid G and circumcenter O . Suppose that the orthocenter of triangle AGO lies on line BC . Given that AB = 11 and AC = 13 , compute BC . ©2026 HMMT
解析
- Let ABC be a triangle with centroid G and circumcenter O . Suppose that the orthocenter of triangle AGO lies on line BC . Given that AB = 11 and AC = 13 , compute BC . Proposed by: Aprameya Tripathy √ √ 580 1740 Answer: = 3 3 Solution 1: ©2026 HMMT A O G P B C M H ′ A Let H be the orthocenter of △ AGO and P be the foot of altitude from O to AG . Let M be the midpoint of BC . Notice that 2 • Since △ P OM ∼ △ P M H , we get that P M = P O · P H . • Since △ P OG ∼ △ P AH , we get that P O · P H = P G · P A . 2 Thus, we get that P M = P G · P A . Note that 3 M G = AM implies that 3( P M + P G ) = P A + P M and 2 P M = P A − 3 P G . Substituting 2 2 2 2 back to get that 4 P G · P A = 4 P M = ( P A − 3 P G ) and 0 = P A − 10 P G · P A + 9 P G . Hence, ( ) ( ) 2 P A P A 0 = − 10 + 9 . Solving this quadratic equation, we get that P A = P G or P A = 9 P G . P G P G 2 Clearly, P A ̸ = P G , so P A = 9 P G . From P M = P G · P A , we have AP = 3 P M too. ′ Let AP = 3 ℓ and P M = ℓ . Let AM intersect the circumcircle again at A . Because OP is perpendicular ′ ′ ′ ′ to AA , P is the midpoint of AA , we have AA = 6 ℓ , so M A = 6 ℓ − 4 ℓ = 2 ℓ . By power of point, we have √ 2 ′ 2 2 M B = M B · M C = M A · M A = (4 ℓ )(2 ℓ ) = 8 ℓ = ⇒ M B = 8 ℓ. Thus, the length of median formula gives 2 2 2 2 2 2 2 AB + AC = 2( AM + M B ) = 2(16 ℓ + 8 ℓ ) = 48 ℓ , 2 2 2 AB + AC which implies that ℓ = . Hence, 48 580 2 2 2 2 2 2 2 2 2 BC = 4 BM = 32 ℓ = ( AB + AC ) = (11 + 13 ) = , 3 3 3 √ 580 which implies BC = . 3 Solution 2: ©2026 HMMT P A G O X C B M H T Let ω be the circumcircle of △ ABC . Let H be the orthocenter of △ AGO , which lies on BC . Let X be the intersection of line BC and the tangent line to ω at A . Let T be the intersection of tangent lines to ω at B and C . Let P be the foot from A to OM . Observe the following: • Since both GH and XA are perpendicular to AO , we get that AX ∥ GH and XH : HM = AG : GM = 2 : 1 . • By La Hire’s theorem, X is the pole of AT with respect to ⊙ ( ABC ) , so OX ⊥ AT . • We have △ OM X ∪ H ∼ △ AP T ∪ M because OM ⊥ AP , OX ⊥ AT , and OH ⊥ AM . Hence, the first and third bullet point gives T M : M P = XH : HM = 2 : 1 . In particular, [ BT C ] = 2[ BAC ] . We now show two different ways to finish from here. • Note that T has barycentric coordinate ( − 2 , x, y ) for some real numbers x and y such that x + y = 2 2 2