PUMaC 2022 · 几何（B 组） · 第 8 题
PUMaC 2022 — Geometry (Division B) — Problem 8
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 . (Write answers on next page.) 1 Name: Team: Write answers in table below: Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 2
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. 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 3 √ √ 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 2 576 404 101 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 4