HMMT 十一月 2023 · GEN 赛 · 第 10 题

HMMT November 2023 — GEN Round — Problem 10

Let ABCD be a convex trapezoid such that ∠ ABC = ∠ BCD = 90 , AB = 3 , BC = 6 , and CD = 12 . Among all points X inside the trapezoid satisfying ∠ XBC = ∠ XDA , compute the minimum possible value of CX .

解析

Let ABCD be a convex trapezoid such that ∠ ABC = ∠ BCD = 90 , AB = 3 , BC = 6 , and CD = 12 . Among all points X inside the trapezoid satisfying ∠ XBC = ∠ XDA , compute the minimum possible value of CX . Proposed by: Pitchayut Saengrungkongka √ √ Answer: 113 − 65 . Solution: P O A B X C D Let P = AD ∩ BC . Then, the given angle condition ∠ XBC = ∠ XAD implies that ∠ XBD + ∠ XP D = ◦ 180 , so X always lies on circle ⊙ ( P BD ) , which is fixed. Thus, we see that the locus if X is the arc ̂ BD of ⊙ ( P BD ) . Let O and R be the center and the radius of ⊙ ( P BD ) . Then, by triangle inequality, we get that CX ≥ CO − OX = CO − R, and the equality occurs when X is the intersection of segment CO and ⊙ ( P BD ) , as shown in the diagram above. Hence, the maximum value is CO − R . To compute CO and R , we let T be the second intersection of ⊙ ( P BD ) and CD . We can compute BP = 2 , so by Power of Point, CT · CD = CP · CB = 48 , so CT = 4 , which means that DT = 8 . The projections of O onto CD and CB are midpoints of BP and DT . Let those midpoints be M and N , respectively. Then, we get by Pythagorean theorem that √ √ √ √ ( ) ( ) 2 2 8 2 2 2 2 2 CO = CN + ON = 4 + + 6 + = 8 + 7 = 113 2 2 √ √ √ √ ( ) 2 8 2 2 2 2 2 R = BM + M O = 1 + 4 + = 1 + 8 = 65 , 2 √ √ so the answer is 113 − 65 .