PUMaC 2015 · 个人决赛（A 组） · 第 3 题
PUMaC 2015 — Individual Finals (Division A) — Problem 3
题目详情
- Let I be the incenter of a triangle ABC with AB = 20, BC = 15, and BI = 12. Let CI intersect the circumcircle ω of ABC at D 6 = A . Alice draws a line l through D that intersects 1 ω on the minor arc AC at X and the circumcircle ω of AIC at Y outside ω . She notices 1 2 1 that she can construct a right triangle with side lengths ID , DX , and XY . What is the length of IY ? 1
解析
- Let I be the incenter of a triangle ABC with AB = 20, BC = 15, and BI = 12. Let CI intersect the circumcircle ω of ABC at D 6 = A . Alice draws a line l through D that intersects 1 ω on the minor arc AC at X and the circumcircle ω of AIC at Y outside ω . She notices 1 2 1 that she can construct a right triangle with side lengths ID , DX , and XY . What is the length of IY ? Solution: First, notice that ∠ BID = ∠ BCI + ∠ CBI = ∠ DBA + ∠ ABI = ∠ DBI , so ID = BD = CD . Then, clearly, DX > ID , so ID cannot be the hypotenuse. Additionally, drawing an accurate diagram, it is easy to see that XY < DX for any choice of the line l (this does depend on the triangle ABC , but one can verify that XY > DX can only occur when ∠ BAC > ∠ BCA , or AB < BC , which is not the case). 2 2 2 So now, we need to find the appropriate line l such that DA · DB = DX = ID + XY . We 2 will show that DX − DA · DB = AX · BX . Let F be the intersection of AX and the tangent to ω at D . We have from similar triangles 1 2 4 ADF ∼ 4 DXF ∼ 4 BXD that: DF AF = DX AD DF · AD = DX · AF = DX · ( F X − AX ) AD AD 2 AD = · F X · DX − · DX · AX DF DF 2 = DX − BX · AX 2 DX − DA · DB = AX · BX 2 As desired. It remains to find the appropriate line l such that AX · BX = XY . Note that 1 ∠ AXB = ∠ C and since XY is the angle bisector of ∠ AXB , ∠ AXY = ∠ BXY = 180 − ∠ C . 2 AX XY Since = , the two triangles AXY, Y XB must be similar. This means that ∠ AY B = XY BX 1 ∠ AY X + ∠ XY B = ∠ AY X + ∠ XAY = ∠ C . 2 Let P be the second intersection of BY and ω . Then, since AY CI are cyclic, ∠ ACI = 2 1 ∠ AY P = ∠ AY B = ∠ C , so P must lie on the angle bixector of ∠ ACB , and this P = I as 2 the line CP intersects ω at exactly the points C and I . Thus, Y is the B -excenter of 4 ABC . 2 BY Now, we have 4 BIA ∼ 4 BCY , as ∠ ABI = ∠ Y BC and ∠ BY C = ∠ BAI . Thus, = BC BA 20 · 15 ⇒ BY = = 25. Then, IY = BY − BI = 25 − 12 = 13 . BI 12 Author: Bill Huang 3