HMMT 二月 2021 · 几何 · 第 10 题

HMMT February 2021 — Geometry — Problem 10

Acute triangle ABC has circumcircle Γ. Let M be the midpoint of BC . Points P and Q lie on Γ so ◦ that ∠ AP M = 90 and Q 6 = A lies on line AM . Segments P Q and BC intersect at S . Suppose that √ 7 2 BS = 1, CS = 3, P Q = 8 , and the radius of Γ is r . If the sum of all possible values of r can be 37 a expressed as for relatively prime positive integers a and b , compute 100 a + b . b

解析

Acute triangle ABC has circumcircle Γ. Let M be the midpoint of BC . Points P and Q lie on Γ so ◦ that ∠ AP M = 90 and Q 6 = A lies on line AM . Segments P Q and BC intersect at S . Suppose that √ 7 2 BS = 1, CS = 3, P Q = 8 , and the radius of Γ is r . If the sum of all possible values of r can be 37 a expressed as for relatively prime positive integers a and b , compute 100 a + b . b Proposed by: Jeffrey Lu Answer: 3703 ′ ′ Solution: Let A be the A -antipode in Γ, let O be the center of Γ, and let T := AA ∩ BC . Note that ′ A lies on line P M . The key observation is that T is the reflection of S about M ; this follows by the ′ Butterfly Theorem on chords P A and AQ . P Q P Q P M Let θ := ∠ AM P and x = OT = OS . Observe that cos θ = = = . We find the area of ′ AM AA 2 r ′ 4 AM A in two ways. First, we have ′ 2[ AM A ] = AM · M A · sin θ M B · M C = AM · · sin θ P M = 4 tan θ √ 37 1 = 8 r − . 2 448 4 r On the other hand, ′ ′ 2[ AM A ] = M T · AA · sin ∠ OT M √ 1 = 2 r 1 − . 2 x 37 4 1 Setting the two expressions equal and squaring yields − = 1 − . By Power of a Point, 2 2 28 r x 2 2 2 2 2 3 = BS · SC = r − x , so x = r − 3. Substituting and solving the resulting quadratic in r gives 16 a 37 2 2 r = and r = 7. Thus = , so 100 a + b = 3703. 3 b 3