PUMaC 2022 · 几何（A 组） · 第 7 题
PUMaC 2022 — Geometry (Division A) — Problem 7
题目详情
- Let △ ABC be a triangle with BC = 7, CA = 6, and, AB = 5. Let I be the incenter of △ ABC . Let the incircle of △ ABC touch sides BC, CA, and AB at points D, E, and F . Let the circumcircle of △ AEF meet the circumcircle of △ ABC for a second time at point X ̸ = A . m Let P denote the intersection of XI and EF . If the product XP · IP can be written as for n relatively prime positive integers m, n , find m + n .
解析
- Let △ ABC be a triangle with BC = 7, CA = 6, and, AB = 5. Let I be the incenter of △ ABC . Let the incircle of △ ABC touch sides BC, CA, and AB at points D, E, and F . Let the circumcircle of △ AEF meet the circumcircle of △ ABC for a second time at point X ̸ = A . m Let P denote the intersection of XI and EF . If the product XP · IP can be written as for n relatively prime positive integers m, n , find m + n . Proposed by Sunay Joshi Answer: 629 We begin by performing a general calculation. Consider a generic triangle △ ABC . Let D denote the foot of the altitude from A to BC . Let O denote the circumcenter of △ ABC and let M denote the midpoint of BC . We will compute the lengths BD , DE , and DO . It is easy to 2 2 2 2 2 2 a + c − b a + b − c see that BD = c cos B = and DE = b cos C = by the law of cosines. Next, 2 a 2 a 2 2 2 the Pythagorean Theorem applied to △ OM D yields OD = OM + BM , or equivalently 2 2 2 2 2 2 2 a 2 2 a 2 a a a + c − b b − c OD = ( − c cos B ) +( R − ( ) ). By the law of cosines, − c cos B = − = . 2 2 2 2 a 2 a 2 a q 2 2 b − c a 2 2 2 Therefore OD = ( ) + ( R − ( ) ). 2 a 2 Returning to the problem, the key fact is that P is the foot of the altitude from D to EF . This can be seen by inverting about the incircle of △ ABC . Next, by power of a point, the desired product is XP · IP = P E · P F . Note that I is the circumcenter of △ DEF . It therefore suffices to evaluate the distances computed in the first paragraph of this solution, for the triangle △ DEF . Let x, y, z denote the lengths EF, F D, DE and r denote the inradius of △ ABC . Note q q 2 2 a − ( b − c ) A A 1 − cos A that x = 2( s − a ) sin , and similarly for y, z . Note that sin = = , and 2 2 2 4 bc B C similarly for sin , sin . 2 2 We now compute. Note that a = 7, b = 6, c = 5, so that s = 9, s − a = 2, s − b = 3, s − c = 4. √ √ √ The area of △ ABC is rs = 9 · 2 · 3 · 4, so the inradius is r = 6 6 / 9 = 2 6 / 3. Next, note 3 q q q q q q 2 2 2 2 2 2 A 7 − 1 2 B 6 − 2 8 C 5 − 1 1 that sin = = , sin = = , sin = = . Therefore 2 4 · 6 · 5 5 2 4 · 7 · 5 35 2 4 · 6 · 7 7 q q q √ 2 8 1 224 288 320 8 10 2 2 2 x = 2 · 2 · , y = 2 · 3 · , z = 2 · 4 · , and so x = , y = , z = . 2 x = . 5 35 7 35 35 35 5 2 2 2 256 / 35 x + z − y √ Plugging the above into our formulae, we see that P E = = and P F = 2 x 8 10 / 5 q q q 2 2 2 2 2 192 / 35 32 / 35 x + y − z y − z x 8 56 4 16 2 2 2 2 2 √ √ √ = . Also, IP = ( ) + ( r − ( ) ) = ( ) + ( − ) = ( ) + ( ) = 2 x 2 x 2 3 35 15 8 10 / 5 8 10 / 5 7 10 q √ 16 16 2 3030
- = . Therefore the desired product is 490 15 105 256 / 35 192 / 35 384 √ √ XP · IP = P E · P F = · = 245 8 10 / 5 8 10 / 5 and our final answer is 384 + 245 = 629.