HMMT 二月 2016 · 几何 · 第 10 题
HMMT February 2016 — Geometry — Problem 10
题目详情
- The incircle of a triangle ABC is tangent to BC at D . Let H and Γ denote the orthocenter and circumcircle of 4 ABC . The B -mixtilinear incircle , centered at O , is tangent to lines BA and BC B and internally tangent to Γ. The C -mixtilinear incircle , centered at O , is defined similarly. Suppose C √ that DH ⊥ O O , AB = 3 and AC = 2. Find BC . B C
解析
- The incircle of a triangle ABC is tangent to BC at D . Let H and Γ denote the orthocenter and circumcircle of 4 ABC . The B -mixtilinear incircle , centered at O , is tangent to lines BA and BC B and internally tangent to Γ. The C -mixtilinear incircle , centered at O , is defined similarly. Suppose C √ that DH ⊥ O O , AB = 3 and AC = 2. Find BC . B C Proposed by: Evan Chen √ √ 1 Answer: (7 + 2 13) 3 Let the B -mixtilinear incircle ω touch Γ at T , BA at B and BC at B . Define T ∈ Γ, C ∈ CB , B B 1 2 C 1 C ∈ CA , and ω similarly. Call I the incenter of triangle ABC , and γ the incircle. 2 C We first identify two points on the radical axis of the B and C mixtilinear incircles: • The midpoint M of arc BC of the circumcircle of ABC . This follows from the fact that M , B , 1 T are collinear with B 2 2 M B = M C = M B · M T 1 B and similarly for C . • The midpoint N of ID . To see this, first recall that I is the midpoint of segments B B and 1 2 C C . From this, we can see that the radical axis of ω and γ contains N (since it is the line 1 2 B through the midpoints of the common external tangents of ω , γ ). A similar argument for C B shows that the midpoint of ID is actually the radical center of the ω , ω , γ . B C Now consider a homothety with ratio 2 at I . It sends line M N to the line through D and the A -excenter I (since M is the midpoint of II , by “Fact 5”). Since DH was supposed to be parallel to line M N , A A it follows that line DH passes through I ; however a homothety at D implies that this occurs only if A H is the midpoint of the A -altitude. √ Let a = BC , b = CA = 2 and c = AB = 3. So, we have to just find the value of a such that the orthocenter of ABC lies on the midpoint of the A -altitude. This is a direct computation with the Law of Cosines, but a more elegant solution is possible using the fact that H has barycentric coordinates 1 2 2 2 ( S S : S S : S S ), where S = ( b + c − a ) and so on. Indeed, as H is on the A -midline we B C C A A B A 2 deduce directly that 1 1 2 2 2 2 2 S S = S ( S + S ) = a S = ⇒ ( a − 1)( a + 1) = a (7 − a ) . B C A B C A 4 2 2 Solving as a quadratic in a and taking the square roots gives √ √ 1 4 2 3 a − 14 a − 1 = 0 = ⇒ a = (7 + 2 13) 3 as desired. 1