PUMaC 2015 · 几何（B 组） · 第 8 题
PUMaC 2015 — Geometry (Division B) — Problem 8
题目详情
- [ 8 ] Triangle ABC is inscribed in a unit circle ω . Let H be the intersection of all altitudes of the vertices and let D be the foot of the perpendicular from A to BC . Let △ XY Z be the triangle formed by drawing the tangents to ω at A, B, C . If AH = HD and the side lengths of p △ XY Z form an arithmetic sequence, the area of △ ABC can be expressed in the form for q positive coprime integers p, q . What is p + q ? 2
解析
- [ 8 ] Triangle ABC is inscribed in a unit circle ω . Let H be the intersection of all altitudes of the vertices and let D be the foot of the perpendicular from A to BC . Let △ XY Z be the triangle formed by drawing the tangents to ω at A, B, C . If AH = HD and the side lengths of p △ XY Z form an arithmetic sequence, the area of △ ABC can be expressed in the form for q positive coprime integers p, q . What is p + q ? Answer: 11 Solution: Let O be the center of ω and extend AD to P and AO to Q . Let K be the intersection of AO and BC . Then, we have ∠ HBC = ∠ CAD = ∠ CBP , so HD = DP and since P Q ∥ BC , AK AD 1 = = 2 and OK = . KQ DP 3 WLOG X is opposite A , Y opposite B , and Z opposite C . Let M be the midpoint of Y Z and N the midpoint of minor arc Y Z of ( XY Z ) . We claim that X, K, M are collinear. Let T be the intersection of BC and a line l through X parallel to Y Z . Now, from the law of sines on triangle pairs △ XT C, △ XT B and △ OKC, △ OKB , we have: sin C sin ∠ COK X ◦ sin 90 + T C sin Z KC 2 sin ∠ OKC = = = = sin X + Z sin ∠ BOK T B sin Y KB X ◦ sin ∠ OKB sin 90 − 2 4 ′ So ( B, C ; K, T ) = − 1, and since XT ∥ Y Z , if M is the intersection of XK and Y Z , then ′ ′ Y M = 1 (since XT intersects Y Z at infinity) and M = M . Thus, X, K, M are collinear, as ′ ZM desired. Now, since AO is the angle bisector of ∠ Y XZ , ∠ M Y N = ∠ CXO , and since OC ⊥ XY and N M ⊥ Y Z , we have △ M Y N ∼ △ CXO . Furthermore, N M ∥ OK since N M, OK ⊥ Y Z , so △ N XM ∼ △ OXK . N Y XN Let α = . Angle chasing, we have ∠ N Y O = ∠ N OY , so N O = N Y and = 1 + α . XO XO 1 1 + α 1 1 + α 1 Then, 3 ⋅ OK = OC = N M ⋅ = OK ⋅ ⋅ , so = 3 and α = . This means that α 1 α α 2 Y Z = 2 ⋅ Y M = 2 ⋅ α ⋅ XC = XC , so XY + XZ = 3 Y Z . Then, since the side lengths of △ XY Z form an arithmetic sequence, it must be that WLOG Y Z = 3 k , XZ = 4 k , XY = 5 k for some k . 2 1 The area of △ XY Z is then [ XY Z ] = ⋅ 3 k ⋅ 4 k = 6 k , as XY Z is a right triangle. The inradius 2 r of △ XY Z then satisfies: 1 2 6 k = [ XY Z ] = ( 3 k + 4 k + 5 k ) r = 6 k ⋅ r ⇔ r = k 2 Since r = 1, we must have k = 1. Then, it is easy to calculate the areas of △ BOC, △ COA, △ AOB using similar triangles. We have: 2 OB 1 3 2 1 [ BOC ] = ⋅ [ OBXC ] = ⋅ ( 1 ⋅ 3 ) = [ COA ] = [ AOB ] = 3 3 OX 10 5 2 1 + 3 1 3 2 12 6 Then, summing, [ ABC ] = + + = = , so p = 6 , q = 5 and p + q = 11 . 2 10 5 10 5 5