HMMT 二月 2025 · 几何 · 第 7 题
HMMT February 2025 — Geometry — Problem 7
题目详情
- Point P is inside triangle △ ABC such that ∠ ABP = ∠ ACP . Given that AB = 6, AC = 8, BC = 7, [ BP C ] BP 1 and = , compute . P C 2 [ ABC ] (Here, [ XY Z ] denotes the area of △ XY Z ).
解析
- Point P is inside triangle △ ABC such that ∠ ABP = ∠ ACP . Given that AB = 6, AC = 8, BC = 7, [ BP C ] BP 1 and = , compute . P C 2 [ ABC ] (Here, [ XY Z ] denotes the area of △ XY Z ). Proposed by: Pitchayut Saengrungkongka 7 Answer: 18 Solution 1: A P ′ C D B D Let the internal and external bisectors of ∠ BAC meet BC at D and E . Similarly, let the internal and ′ ′ ′ external bisectors of ∠ BP C meet BC at D and E . The angle condition implies that AD ∥ P D and ′ ′ ′ DE AE ∥ P E . Thus, triangles ADE and P D E are homothetic. Hence, the requested ratio is . ′ ′ D E Repeated applications of the angle bisector theorem yield 6 BD = 7 · = 3 , 6+8 6 BE = 7 · = 21 , 8 − 6 ′ 1 7 BD = 7 · = , 1+2 3 ′ 1 BE = 7 · = 7 , 2 − 1 28 / 3 ′ ′ 7 so DE = 24 and D E = 28 / 3. Hence, the answer is = . 24 18 Solution 2: A E F P C B Let D = AP ∩ BC , E = BP ∩ AC and F = CP ∩ AB . Then, BCEF is cyclic, so by Power of a Point, AE AB 3 = = . Let AE = 3 x and AF = 4 x . Then, AF AC 4 BF BP 6 − 4 x 1 △ P F B ∼ △ P EC = ⇒ = = ⇒ = . CE CP 8 − 3 x 2 4 16 14 12 40 Solving for x gives x = , so we get that AF = , F B = , AE = , and EC = . By Ceva’s 5 5 5 5 5 theorem on △ ABC and point P , we have BD AE BF AE BF 3 1 3 = · = · = · = . DC EC F A AF CE 4 2 8 Finally, by Menelaus’s theorem on △ ADC and line BE , we get that AP CB AE 11 12 11 = · = · = , P D BD EC 3 28 7 [ BP C ] 7 which implies that = . [ ABC ] 18 Solution 3: We use barycentric coordinates with respect to △ ABC . From ∠ ABP = ∠ ACP , we get [ ABP ] AB · BP 3 = = , [ ACP ] AC · CP 8 so P has coordinate ( − : 8 : 3). Let Q be the isogonal conjugate of P , so Q lies on perpendicular bisector 2 2 of BC . By Steiner ratio theorem, Q has coordinate − : 8 / 8 : 6 / 3 = ( − : 8 : 12) = ( − : 2 : 3). Let the coordinate be ( t : 2 : 3) for some real number t . Then, recall the equation for the perpendicular bisector (from Corollary 6 of https://web.evanchen.cc/handouts/bary/bary-full.pdf ): 2 2 2 a ( z − y ) + x ( c − b ) = 0 2 2 2 = ⇒ 7 (3 − 2) + t (6 − 8 ) = 0 , 2 7 (3 − 2) 7 so t = = . Hence, point Q has coordinates (7 : 8 : 12), so point P has coordinates 2 2 8 − 6 4 7 2 2 2 (7 / 7 : 8 / 8 : 6 / 12) = (7 : 8 : 3). Therefore, [ BP C ] / [ ABC ] = . 18