HMMT 二月 2026 · 团队赛 · 第 9 题
HMMT February 2026 — Team Round — Problem 9
题目详情
- [55] Let ABC be an acute scalene triangle. Let D be the foot of the altitude from A to BC , and let M be the midpoint of BC . There exists a unique point P strictly inside triangle ABC such that ◦ ◦ ∠ DP M = 90 and P B/P C = AB/AC . Prove that ∠ BP C = 180 − | ∠ ABC − ∠ ACB | .
解析
- [55] Let ABC be an acute scalene triangle. Let D be the foot of the altitude from A to BC , and let M be the midpoint of BC . There exists a unique point P strictly inside triangle ABC such that ◦ ◦ ∠ DP M = 90 and P B/P C = AB/AC . Prove that ∠ BP C = 180 − | ∠ ABC − ∠ ACB | . Proposed by: Aprameya Tripathy Answer: Solution 1: A P C B D M ′ A ′ Without loss of generality, assume that AB < AC . Let A be the reflection of A across BC . We claim ′ that P be the center of spiral similarity that sends AA to BC . We show that this P satisfied the ′ required condition. First, note that since △ P AB ∼ △ P A C , we get that P B AB AB = = , ′ P C A C AC verifying the Apollonius circle condition. ◦ ′ Next, we show that ∠ DP M = 90 . To do this, note that the spiral similarity send midpoint D of AA ′ ◦ to midpoint M of BC . Since AA ⊥ BC , this spiral similarity is a 90 -rotation, so combining with the ◦ previous sentence gives ∠ DP M = 90 as desired. ′ ◦ To show the conclusion, we note that since △ P AB ∼ △ P A C , we get that ∠ BP C is equal to 180 ′ ′ minus the angle between AB and A C . The angle between AB and A C is ∠ B − ∠ C , so we have that ◦ ∠ BP C = 180 − ∠ B + ∠ C , as desired. Solution 2: ©2026 HMMT A P C B D M ′ A X The following is a variant of the first solution. BT BA ′ Let A be the reflection of A across BC . Recall that the locus of point T such that = is the T C BC Apollonius circle Ω . The key claim is the following. ′ Claim 1. If lines AB and A C meet at X , then X lies on Ω . We present two proofs of the claim. BA CA BA BX Proof 1. By angle bisector theorem on △ ACX , we have that = , which implies that = , BX CX CA CX so X lies on Ω . BD BA Proof 2. Let the angle bisector of ∠ BAC meets BC at D . By angle bisector theorem, = , so DC AC D lies on Ω . We now do angle chasing: ′ ∠ ADA = 2 ∠ ADB ( ) ∠ A ◦ = 2 180 − − ∠ B 2 ◦ = 360 − ∠ A − 2 ∠ B ◦ = 180 − ∠ B + ∠ C ′ ◦ ∠ AXA = 180 − ∠ XBC − ∠ XCB = ∠ B − ∠ C, ′ ′ ◦ ′ so ∠ ADA + ∠ AXA = 180 , implying that X lies on ⊙ ( ADA ) = Ω . To finish, redefine point P as the second intersection of ⊙ ( BXC ) and Ω . Since ∠ BXC = ∠ B − ∠ C , ◦ ′ it suffices to show that ∠ DP M = 90 . To do this, note that since P is the intersection of ⊙ ( AXA ) ′ ′ and ⊙ ( BXC ) , we get that P is the center of spiral similarity that sends AA to BC . Since AA ⊥ BC , ◦ this spiral similarity must be rotation by 90 . Moreover, this spiral similarity sends D to M . Hence, ◦ ∠ DP M = 90 . ◦ Solution 3: Again, we assume AB < AC . The crux of this solution is to claim that ∠ AP B = 90 . ©2026 HMMT This claim can be obtained by working backward from the conclusion (as shown in after the proof of the claim). Redefine point P to be the intersection of circle with diameter AB and the Apollonius circle Ω . ◦ Claim 2. With the new definition of P , we have that ∠ DP M = 90 . Proof. Let T be the center of Ω , which is well-known to be the intersection of the tangent at A of ⊙ ( ABC ) and line BC . Let M be the midpoint of AB . Notice that AP ⊥ T M , so T M ∥ BP . C C C A K M C P C T B D M ′ A Construct the parallelogram AT BK . Then since T K ∥ BP and AK ∥ BC , we get that ∠ AKT = P B AB T A T A ∠ P BC . Moreover, from △ T BA ∼ △ T AC , we get that = = = . The previous P C AC T B AK two sentences give △ KAT ∼ △ BP C . The corresponding midpoint of sides are M and M . Hence, C ◦ ∠ BP M = ∠ M AK = 180 − ∠ B . C ◦ Finally, since ABP D is cyclic, we get that ∠ BP D = ∠ BAD = 90 − ∠ B . Combining with the previous ◦ paragraph gives ∠ DP M = 90 . We now prove the conclusion. This part can be cited from IMO 1996 P2, but we reprove it from scratch √ to make the solution self-contained. Perform the inversion at A with radius AB · AC followed by ∗ reflection across angle bisector of ∠ BAC . Suppose that this inversion sends point P to P . First, the ∗ ∗ similarities △ ABP ∼ △ AP C and △ ACP ∼ △ AP B gives ∗ ∗ CP BP CP BP = = = , ∗ ∗ AP BA CA AP ∗ ∗ ∗ so BP = CP . Thus, P lies on the perpendicular bisector of BC . ©2026 HMMT A P C B ∗ P ∗ ∗ ◦ ∗ ◦ ∗ Thus, ∠ P BC = ∠ P CB . Moreover, from ∠ AP B = 90 , we have ∠ ACP = 90 . Thus, ∠ P CB = ◦ ∗ ◦ ◦ 90 − ∠ C , implying that ∠ ACP = ∠ ABP = 90 + ∠ B − ∠ C . Hence, ∠ BP C = 180 − ∠ C + ∠ B , as desired. BP BA Solution 4: Again, we assume AB < AC . We prove the converse: that if = and ∠ BP C = P C AC ◦ ◦ 180 − ∠ B + ∠ C , then ∠ DP M = 90 . √ Perform the inversion at A with radius AB · AC followed by reflection across angle bisector of ∠ BAC , ∗ ∗ and denote the images of U by U . Notice that D is an antipode of A with respect to ⊙ ( ABC ) and ∗ ∗ M is the point such that ABCM is a cyclic harmonic quadrilateral. A O B C ∗ Q ∗ D ∗ M ∗ P ∗ We now locate P . To do this, we have to translate the two condition. First, the similarities △ ABP ∼ ∗ ∗ △ AP C and △ ACP ∼ △ AP B gives ∗ ∗ CP BP CP BP = = = , ∗ ∗ AP BA CA AP ∗ ∗ ∗ so BP = CP . Thus, P lies on the perpendicular bisector of BC . ◦ Now, we translate the condition ∠ BP C = 180 − ∠ B + ∠ C . Note that ∗ ∗ ∗ ∠ BP C = ∠ BP A + ∠ AP C = ∠ ABP + ∠ ACP = ∠ BP C − ∠ A ◦ = 180 − ∠ B + ∠ C − ∠ A = 2 ∠ C, ©2026 HMMT ∗ ∗ ◦ ∗ ∗ which implies that ∠ P BC = ∠ P CB = 90 − ∠ C . In particular, P , C, D are collinear. ∗ ∗ ∗ ∗ We have to show that AD is tangent to ⊙ ( DP M ) , or equivalently, AD is tangent to ⊙ ( D P M ) . To ∗ ∗ ∗ ∗ ∗ do this, we let the perpendicular bisector of BC meet BD and point Q . Since △ D P Q ∼ △ ABC ∗ ∗ ∗ ∗ with corresponding sides perpendicular, we deduce that AD is tangent to ⊙ ( D P Q ) . Therefore, it ∗ ∗ ∗ ∗ suffices to show that M lies on this circle. To do this, let the ⊙ ( D P Q ) meet ⊙ ( ABC ) again at ∗ ∗ ∗ ∗ ∗ ∗ ∗ K . Since OD = OK , we get that OK is tangent to ⊙ ( D P Q ) . Thus, ⊙ ( D P Q K ) is a harmonic ∗ ∗ ∗ quadrilateral, so D K ⊥ AM , so K = M . Hence, we are done.