HMMT 二月 2019 · 几何 · 第 7 题
HMMT February 2019 — Geometry — Problem 7
题目详情
- Let ABC be a triangle with AB = 13 , BC = 14 , CA = 15. Let H be the orthocenter of ABC . Find the radius of the circle with nonzero radius tangent to the circumcircles of AHB, BHC, CHA .
解析
- Let ABC be a triangle with AB = 13 , BC = 14 , CA = 15. Let H be the orthocenter of ABC . Find the radius of the circle with nonzero radius tangent to the circumcircles of AHB, BHC, CHA . Proposed by: Michael Ren 65 Answer: 4 Solution 1. We claim that the circle in question is the circumcircle of the anticomplementary triangle of ABC , the triangle for which ABC is the medial triangle. ′ ′ ′ ′ ′ Let A B C be the anticomplementary triangle of ABC , such that A is the midpoint of B C , B is the ′ ′ ′ ′ ′ ′ ′ midpoint of A C , and C is the midpoint of A B . Denote by ω the circumcircle of A B C . Denote by ω the circumcircle of BHC , and similarly define ω , ω . A B C ′ ◦ ′ Since ∠ BA C = ∠ BAC = 180 − ∠ BHC , we have that ω passes through A . Thus, ω can be A A ′ ′ ′ ′ ′ redefined as the circumcircle of A BC . Since triangle A B C is triangle A BC dilated by a factor of 2 ′ ′ from point A , ω is ω dilated by a factor of 2 from point A . Thus, circles ω and ω are tangent at A A ′ A . By a similar logic, ω is also tangent to ω and ω . Therefore, the circumcircle of the anticomplementary B C triangle of ABC is indeed the circle that the question is asking for. abc 65 Using the formula R = , we can find that the circumradius of triangle ABC is . The circumradius 4 A 8 65 of the anticomplementary triangle is double of that, so the answer is . 4 Solution 2. It is well-known that the circumcircle of AHB is the reflection of the circumcircle of 65 ABC over AB . In particular, the circumcircle of AHB has radius equal to the circumradius R = . 8 Similarly, the circumcircles of BHC and CHA have radii R . Since H lies on all three circles (in the 65 question), the circle centered at H with radius 2 R = is tangent to each circle at the antipode of H 4 in that circle.