HMMT 二月 2021 · 团队赛 · 第 2 题
HMMT February 2021 — Team Round — Problem 2
题目详情
- [50] Let ABC be a right triangle with ∠ A = 90 . A circle ω centered on BC is tangent to AB at D and AC at E . Let F and G be the intersections of ω and BC so that F lies between B and G . If lines DG and EF intersect at X , show that AX = AD .
解析
- [50] Let ABC be a right triangle with ∠ A = 90 . A circle ω centered on BC is tangent to AB at D and AC at E . Let F and G be the intersections of ω and BC so that F lies between B and G . If lines DG and EF intersect at X , show that AX = AD . Proposed by: Krit Boonsiriseth ◦ ◦ Solution 1: In all solutions, let O be the center of ω . Then ∠ DOE = 90 , so ∠ DF E = 45 , so − − → ◦ ′ ∠ DXE = 135 . Let Γ be the circle centered at A with radius AD = AE , and let X = AX ∩ Γ. Then ′ ◦ ′ ∠ DXE = ∠ DX E = 135 , so X = X . Solution 2: Let N be the midpoint of arc F DEG . Note that DOEA is a square. Also, DXEN is ◦ a parallelogram; one way to see this is that by considering inscribed angles, ∠ N EX = ∠ N EF = 45 , ◦ ◦ ∼ ∠ N DX = ∠ N DG = 45 , and ∠ EN D = 135 . This means that 4 AXD 4 ON E , because AD = = OE , XD = N E , and ∠ ADX = ∠ OEN by considering parallel lines. So AX = ON = OE = AD . Solution 3: Let Y = DF ∩ EG . Since GX ⊥ F Y and F X ⊥ GY , X is the orthocenter of 4 F GY . Since ( AODE ) passes through the midpoint of F G , and the feet of altitudes from F and G , ( AODE ) is the nine-point circle of 4 F GY . Since AO is the diameter of ( AODE ), it follows that A is the midpoint of XY , so A is the center of ( DXEY ), so AX = AD . Solution 4: Let Z = DE ∩ F G , possibly at infinity. Then A and X are on the polar of Z with respect to ω , so AX ⊥ BC . Let J = AX ∩ BC . Then ( XJF D ) is cyclic, so ∠ ADX = ∠ DF J = ∠ DXA , so AX = AD . Solution 5: We use complex numbers, with ( F DEG ) being the unit circle, and d = − 1 , e = − i . As F G is a diameter of the unit circle, we have g = − f . We have a = − 1 − i , from either intersecting the tangents to the unit circle at D and E or noting that ADOE is a square. Now, intersecting the chords DG and EF , we obtain dg ( e + f ) − ef ( d + g ) f ( i + f ) + if (1 − f ) f − i − i − if 1 − i x = = = = − 1 − i + f · . dg − ef f + if 1 + i 1 + i So, √ ∣ ∣ ∣ ∣ 1 − i 2 ∣ ∣ √ | x − a | = f · = 1 · = 1 . ∣ ∣ 1 + i 2 So AX and AD have the same length (1 unit), as desired.