PUMaC 2020 · 几何（B 组） · 第 7 题

PUMaC 2020 — Geometry (Division B) — Problem 7

Triangle ABC is so that AB = 15 , BC = 22 , and AC = 20 . Let D, E, F lie on BC, AC, and AB, respectively, so AD, BE, CF all contain a point K. Let L be the second intersection of AK 11 a 2 the circumcircles of BF K and CEK. Suppose that = , and BD = 6 . If KL = , where KD 7 b a, b are relatively prime integers, find a + b.

解析

Triangle ABC is so that AB = 15 , BC = 22 , and AC = 20 . Let D, E, F lie on BC, AC, and AB, respectively, so AD, BE, CF all contain a point K. Let L be the second intersection of AK 11 2 a the circumcircles of BF K and CEK. Suppose that = , and BD = 6 . If KL = , where KD 7 b a, b are relatively prime integers, find a + b. Proposed by: Frank Lu Answer: 497 AK DC BF First, by Menalaus’s theorem, we can compute that = 1 , which in turn implies KD CB F A BF 7 22 7 AE AF BD 8 6 3 that = = . Therefore, by Ceva’s theorem, it follows that = = = . F A 11 16 8 EC F B DC 7 16 7 From here, we see that AF = 8 , AE = 6 . In particular, notice that by power of point, since AE · AC = 120 = AB · AF, it follows that A lies on the radical axes of these circles; in particular, notice that A, K, L are collinear. 2 Now, notice that the length of AD, by Stewart’s theorem, is so that BD · DC · BC + AD · BC = 2 2 2 AC · BD + AB · CD. Plugging in the values we computed, it follows that 6 · 16 · 22+ AD · 22 = 6000 − 96 · 22 2 2 2 20 · 6 + 15 · 16 = 3600 + 2400 = 6000 . In particular, it follows that AD = = 22 √ √ 3888 1944 6 = , or that AD = 18 . In particular, this means that AK = 66 . Therefore, 22 11 11 computing the power of A again, we see that AK · AL = 120 too, meaning that it follows that √ √ 120 20 66 9 66 486 2 √ AL = = . Hence, it follows that KL = , and so that KL = = 497 . 11 11 11 66