PUMaC 2021 · 团队赛 · 第 11 题
PUMaC 2021 — Team Round — Problem 11
题目详情
- ABC is a triangle where AB = 10, BC = 14, and AC = 16 . Let DEF be the triangle with smallest area so that DE is parallel to AB, EF is parallel to BC, DF is parallel to AC, and the circumcircle of ABC is DEF ’s inscribed circle. Line DA meets the circumcircle of ABC 2 again at a point X. Find AX .
解析
- ABC is a triangle where AB = 10, BC = 14, and AC = 16 . Let DEF be the triangle so that DE is parallel to AB, EF is parallel to BC, DF is parallel to AC, the circumcircle of ABC is DEF ’s inscribed circle, and D, A are on the same side of BC. Line EB meets the circumcircle 2 of ABC again at a point X. Find BX . Proposed by: Frank Lu Answer: 196 First, observe that by our parallel lines, we have that DEF is homothetic to triangle ABC. Let P be the center of this homothety. In addition, let I be the incenter of ABC and O be the circumcenter of ABC. Notice in particular that the homothety that sends triangle ABC to triangle DEF also sends the incenter of ABC to the circumcenter of ABC. Notice, however, that the semiperimeter of our triangle is equal to 20 , yielding us an area for the triangle as √ √ √ 20 · 10 · 6 · 4 = 40 3 . Therefore, we find that the inradius of triangle ABC is equal to 2 3 √ 10 · 14 · 16 14 3 √ and the circumradius of triangle ABC is equal to = . In particular, we see that 3 4 · 40 3 7 the ratio of our homothety is equal to , with the order of points being P, I, O by the condition 3 that A, D are on the same side of BC. ′ ′′ From here, if I is the point at which the incircle of ABC is tangent to AB, and I the point ′ 10+14 − 16 where the circumcircle of ABC is tangent to DE, we see that, first, that BI = = 4 , 2 ′′ 28 and so therefore that EI = . 3 Now, by our homothety, notice that E, B, P are collinear, with EP/BP = 7 / 3 . We now wish to find what BP is equal to. To find this, drop the perpendiculars from P and O onto AB, ′ ′ ′ ′ meeting AB at points P and O , respectively. We already know that BO = 5 and BI = 4 , 4 3 ′ ′ ′ ′ meaning that by our homothety we have that BP is so that BI = BP + BO , or that 7 7 √ 13 14 3 ′ BP = . Similarly, we see that, since the circumradius of ABC is equal to , we have 4 3 q q √ √ 196 121 11 3 ′ ′ ′ that OO = − 25 = = and II = 2 3 , meaning that again we have that P P 3 3 3 √ 4 3 3 3 ′ ′ ′ ′ is so that II = P P + OO , or that P P = . This means that BP, by the Pythagorean 7 7 4 √ 1 7 14 theorem, is equal to 169 + 27 = , and so hence that BE, by homothety, is equal to . 4 2 3 784 ′′ 2 However, by power of a point, we also know that BE · EX = ( EI ) = , meaning that 9 784 3 56 56 14 2 EX = = , and so hence that BX = − = 14 , and so BX = 196 . 9 14 3 3 3