HMMT 二月 2026 · 几何 · 第 6 题
HMMT February 2026 — Geometry — Problem 6
题目详情
- Let ABC be a triangle, and M be the midpoint of segment BC . Points P and Q lie on segments 1 AB and AC , respectively, so that ∠ P M B = ∠ QM C = ∠ BAC . Given that AP = 1 , AQ = 3 , and 2 BC = 8 , compute BP .
解析
- Let ABC be a triangle, and M be the midpoint of segment BC . Points P and Q lie on segments 1 AB and AC , respectively, so that ∠ P M B = ∠ QM C = ∠ BAC . Given that AP = 1 , AQ = 3 , and 2 BC = 8 , compute BP . Proposed by: Jason Mao √ Answer: 17 − 1 Solution 1: A P Q B C T M ©2026 HMMT Claim 1. BP = CQ . Proof. We have that ∠ BP M + ∠ M QC ◦ = 360 − ∠ M BP − ∠ P M B − ∠ CM Q − ∠ QCM ◦ = 360 − ∠ CBA − ∠ BAC − ∠ ACB ◦ = 180 . Thus by the Law of Sines, BM · sin ∠ P M B CM · sin ∠ M QC BP = = = CQ. sin ∠ BP M sin ∠ CM Q Let x be the length of BP , and let the circle ( AP M Q ) meets BC again at T . Then by Power of a Point, x ( x + 1) + x ( x + 3) = BT · 4 + CT · 4 = 32 . √ Thus x = 17 − 1 . Solution 2: Let T be the foot of the angle bisector of ∠ BAC . Let a , b , c be the lengths BC , AC , AB respectively, we are given a = 8 . BT c ac From the angle-bisector theorem, = . Since BT + CT = a , we get BT = . We also have that CT b b + c △ BP M ∼ △ BT A because ∠ M BP = ∠ ABT and ∠ P M B = ∠ T BA . From these two, we calculate 2 2 2 BM · BT a 2 bc + 2 c − a AP = AB − BP = AB − = c − = . BA 2( b + c ) 2( b + c ) 2 2 2 bc +2 b − a Similarly, AQ = . We can now solve for b and c . We have that 2( b + c ) 2 2 2 b − 2 c b − c = = AQ − AP = 2 . 2( b + c ) Thus, b = c + 2 . Substituting a = 8 and b = c + 2 into the equation for AP , we get 2 2 c ( c + 2) + 2 c − 64 2 = 1 = ⇒ 4 c + 4 c − 64 = 4 c + 4 2(2 c + 2) √ = ⇒ c = 17 . √ Hence BP = AB − AP = c − 1 = 17 − 1 .