HMMT 十一月 2024 · THM 赛 · 第 10 题
HMMT November 2024 — THM Round — Problem 10
题目详情
- Isabella the geologist discovers a diamond deep underground using an X-ray A D machine. The diamond has the shape of a convex cyclic pentagon P ABCD with AD ∥ BC . Soon after the discovery, her X-ray breaks, and she only recovers partial information about its dimensions. She knows that AD = 70 , BC = 55 , P A : P D = 3 : 4 , and P B : P C = 5 : 6 . Compute P B . P
解析
- Isabella the geologist discovers a diamond deep underground via an X-ray machine. The diamond has the shape of a convex cyclic pentagon P ABCD with AD ∥ BC . Soon after the discovery, her X-ray breaks, and she only recovers partial information about its dimensions. She knows that AD = 70, BC = 55, P A : P D = 3 : 4, and P B : P C = 5 : 6. Compute P B . B C A D P Proposed by: Pitchayut Saengrungkongka √ Answer: 25 6. Solution 1: B C X Y A D p q r P Let X = P B ∩ AD and Y = P C ∩ AD . Let AX = p , XY = q , and Y D = r . From AB ∥ CD , we get that AB = CD , and so ∠ AP X = ∠ DP Y . Thus, we may apply Steiner ratio theorem on △ P AD and △ P XY to get that 2 2 p ( p + q ) 3 p ( q + r ) 5 = , = . 2 2 r ( q + r ) 4 r ( p + q ) 6 Multiplying these two equations gives p : r = 5 : 8, and using each individual equations gives p : q : r = 5 : 22 : 8. Thus, p = 10, q = 44, and r = 16. 2 Now, from XY ∥ BC , we have P X : XB = 4 : 1, so set P X = 4 t and XB = t . However, 4 t = √ √ √ P Y · Y C = 10 · 60 = 600. Solving this gives t = 150 = 5 6, hence P B = 5 t = 25 6 . Solution 2: B C A D P AP DP BP CP Let AB = CD = a , AC = BD = b , = = x , and = = y . Applying Ptolemy’s theorem 3 4 5 6 for the quadrilaterals ABCP , BCDP , and ABCD yields: b · 5 y = 55 · 3 x + a · 6 y (1) b · 6 y = 55 · 4 x + a · 5 y (2) 2 2 b = 55 · 70 + a (3) Equating the left-hand sides of (1) and (2) leads to 6 · (165 x + 6 ay ) = 5 · (220 x + 5 ay ) = ⇒ 110 x = 11 ay = ⇒ 10 x = ay. (4) 9 Substituting 220 x = 22 ay into (2) implies 27 ay = 6 by , or b = a . Plugging this into (3), we find 2 √ √ √ √ 2 a = 200, so a = 10 2, and therefore b = 45 2. Furthermore, x = y 2 after replacing a with 10 2 in (4). We now apply Law of Cosines for △ ABC and △ AP C : √ √ √ 2 2 2 55 + (10 2) − 2 · (10 2) · 55 cos θ = (45 2) (5) √ √ √ 2 2 2 (3 2 y ) + (6 y ) + 2 · (3 2 y ) · (6 y ) cos θ = (45 2) (6) where θ = ∠ ABC . Solving (5) yields √ √ √ 2 2 2 55 + (10 2) − (45 2) 3 2 cos θ = √ = − . 8 2 · (10 2) · 55 Plugging this into (6), we can compute: √ 2 (45 2) 4050 2 y = √ = = 150 . 2 2 27 (3 2) + 6 − 27 √ √ Therefore, BP = 5 y = 5 150 = 25 6 .