PUMaC 2023 · 几何（B 组） · 第 7 题
PUMaC 2023 — Geometry (Division B) — Problem 7
题目详情
- Let △ ABC have AB = 15, AC = 20, and BC = 21 . Suppose ω is a circle passing through A that is tangent to segment BC. Let point D ̸ = A be the second intersection of AB with ω, and let point E ̸ = A be the second intersection of AC with ω. Suppose DE is parallel to BC. a If DE = , where a, b are relatively prime positive integers, find a + b. b √ √ ′ ′ ′ ′ ′
解析
- Let △ ABC have AB = 15, AC = 20, and BC = 21 . Suppose ω is a circle passing through A that is tangent to segment BC. Let point D ̸ = A be the second intersection of AB with ω, and let point E ̸ = A be the second intersection of AC with ω. Suppose DE is parallel to BC. a If DE = , where a, b are relatively prime positive integers, find a + b. b Proposed by Frank Lu Answer: 361 First, since DE is parallel to BC, we have that triangles ADE, ABC are similar. Further- more, we have a homothety that sends triangle ADE to ABC. Notice that the image of this homothety also sends ω to the circumcircle of ABC. We thus need to determine the ratio of this homothety. To do this, let X be the tangency point of ω to BC. Draw line AX, and let M be the second AX DE intersection of line AX with the circumcircle. Then, we know from homothety that = ; AM BC 2 we just need to compute AX, AM. We furthermore note that ω is tangent to the circumcircle by homothety. From this configuration, we thus find that M is the midpoint of the minor arc BC, meaning that AX is the angle bisector of angle ∠ BAC. First, to compute AX, we can BX 3 employ Stewart’s theorem. We know from the angle bisector theorem that = , meaning CX 4 2 that BX = 9 and CX = 12 . Therefore, we have by Stewart’s theorem that AX · 21+9 · 12 · 21 = 2 2 2 2 2 2 2 15 · 12 + 20 · 9 , or that 21 AX = 5 · 3 · (3 · 4 + 4 · 3) − 9 · 12 · 21 = 5 · 3 · 4 · 21 − 9 · 12 · 21 , √ 2 or that AX = 300 − 108 = 192 , so AX = 8 3 . From here, we compute AM. The method that we use to compute this is Ptolemy’s theorem and Law of Cosines chasing. First, consider BM, CM. Note that ∠ BM C = 180 − ∠ BAC, and so cos ∠ BAC = − cos ∠ BM C. But now by Law of Cosines, we know that cos ∠ BAC = 2 2 2 15 +20 − 21 23 2 2 2 196 = . Therefore, we have that BC = BM (2 − 2 cos ∠ BM C ) = BM , meaning 2 · 15 · 20 75 75 √ 15 3 that BM = CM = . Finally, by Ptolemy’s Theorem, we have that AM · BC = BM ( AB + 2 √ √ 15 3 25 3 35 AC ) , or that AM = = . 21 2 2 AX 336 It follows that DE = BC = , so our answer is 361 . AM 25 √ √ ′ ′ ′ ′ ′