PUMaC 2022 · 几何（A 组） · 第 6 题
PUMaC 2022 — Geometry (Division A) — Problem 6
题目详情
- Triangle △ ABC has sidelengths AB = 10 , AC = 14 , and, BC = 16 . Circle ω is tangent to 1 − − → − → − − → − → rays AB, AC and passes through B . Circle ω is tangent to rays AB, AC and passes through 2 C . Let ω , ω intersect at points X, Y . The square of the perimeter of triangle △ AXY is equal 1 2 √ a + b c to , where a, b, c, and, d are positive integers such that a and d are relatively prime, and d c is not divisible by the square of any prime. Find a + b + c + d .
解析
- Triangle △ ABC has sidelengths AB = 10 , AC = 14 , and, BC = 16 . Circle ω is tangent to 1 − − → − → − − → − → rays AB, AC and passes through B . Circle ω is tangent to rays AB, AC and passes through 2 C . Let ω , ω intersect at points X, Y . The square of the perimeter of triangle △ AXY is equal 1 2 √ a + b c to , where a, b, c, and, d are positive integers such that a and d are relatively prime, and d c is not divisible by the square of any prime. Find a + b + c + d . Proposed by Frank Lu Answer: 6272 Draw the angle bisector of BAC, which we denote as ℓ. Notice that if O is the center of ω 1 1 and O is the center of ω , then we have that O , O lie on this angle bisector. It follows that 2 2 1 2 this angle bisector must be the perpendicular bisector of XY, since XY is the radical axis of these two circles. 2 We first compute AP. To do this, consider the following operation: first, reflect the diagram √ about the angle bisector, then perform an inversion about A of radius 10 · 14 . (This latter inversion is referred to sometimes as a root bc inversion). Notice that this operation sends the circle ω to ω , and sends X to Y. Furthermore, since ℓ is the perpendicular bisector of XY, 1 2 √ 140 we have that AY = AX = , meaning that AX = 140 . To find XP, we need to now find AX AP. But this can be done by considering the triangle O XO . 1 2 We compute the side lengths of this triangle. First, we know that XO is the radius of ω . 1 1 We can then compute XO by considering the incircle: if the incircle has radius r, and l is the 1 l AB AB length of the tangent from A to the incircle, we know that = = . But if s is the semi- r BO XO 1 1 q ( s − AB )( s − BC )( s − AC ) perimeter of ABC, we know that l = s − BC = 4 , and r is equal to = s q √ √ √ 10 · 6 · 4 = 2 3 . Therefore, we see that XO = 5 3 . Similarly, XO = 7 3 . Finally, we can 1 2 20 √ √ then compute that O O = AO − AO , which using the Pythagorean theorem is 7 7 − 5 7 = 1 2 2 1 √ 2 7 . √ 2 2 Therefore, we compute that P O − P O = 72 . Since this is larger than 2 7 , we see that our 2 1 √ √ 36 7 triangle is obtuse. Thus, we have that O P + O P = , and OP − OP = 2 7 , which 1 2 2 1 7 √ √ 11 7 24 7 gives us that O P = , and so therefore AP = AO + O P = . 1 1 1 7 7 q 576 404 101 2 From here, we compute that XP = 140 − = , or that XP = 2 . Therefore, the 7 7 7 q √ √ √ 101 1616+ 3920 √ perimeter of our triangle is equal to 4 + 4 35 = . Therefore the square of 7 7 √ 5536+224 505 the perimeter is , so our answer is a + b + c + d = 5536 + 224 + 505 + 7 = 6272. 7