PUMaC 2020 · 几何（A 组） · 第 7 题
PUMaC 2020 — Geometry (Division A) — Problem 7
题目详情
- Let ABC be a triangle with sides AB = 34 , BC = 15 , AC = 35 and let Γ be the circle of smallest possible radius passing through A tangent to BC . Let the second intersections of Γ and sides AB, AC be the points X, Y . Let the ray XY intersect the circumcircle of the triangle p ABC at Z . If AZ = for relatively prime integers p and q, find p + q. q
解析
- Let ABC be a triangle with sides AB = 34 , BC = 15 , AC = 35 and let Γ be the circle of smallest possible radius passing through A tangent to BC . Let the second intersections of Γ and sides AB, AC be the points X, Y . Let the ray XY intersect the circumcircle of the triangle p ABC at Z . If AZ = for relatively prime integers p and q, find p + q. q Proposed by: Aleksa Milojevic Answer: 173 There are two solutions: the first one is elementary, but a bit trickier to find, while the second one uses inversion. First solution: Let D be the foot of the perpendicular from A to BC (or alternatively the tangency point of Γ and BC ). First, we will extend XY to the other intersection with the circumcircle of ABC and denote that intersection by W . Then, we will note that the triangle AXY is similar to ACB , by a simple angle chasing argument. Finally, if we denote the tangent to the circumcircle at the point A by t , this means that XY || t . As ZW is the chord of the circumcircle and it is parallel to t , we conclude AZ = AW . Now, we aim to prove AZ = AD . To do this, note the angle equality ∠ ZCA = ∠ ZW A = 2 ∠ AW Z , meaning that the triangles AZY and ACZ are similar. This implies AZ = AY · AC . Similarly, as ∠ ADC = ∠ DY A , we see that triangles AY D and ADC are also similar, meaning 2 AD = AY · AC , which sums up to give AD = AZ . From here, we compute AD using Heron’s formula: the semiperimeter of the triangle is equal to 42 , meaning that the area is √ 2 · 252 168 42 · 7 · 8 · 27 = 7 · 9 · 4 = 252 . Therefore, we have that AZ = AD = = , yielding us 15 5 with 168 + 5 = 173 . Second solution. As in the first solution, we aim to show AD = AZ directly. Consider the √ √ inversion with the center at A and radius AY · AC = AX · AB . It swaps pairs of points ( X, B ) and ( Y, C ). Thus, the circle ABC is sent to the line XY . Note that Z , defined as the intersection of XY and the circumcircle must therefore be sent to the intersection of the circumcircle and XY , i.e. to Z . Thus, Z stays fixed under this inversion. Similarly, D is the intersection of the circumcircle of AXY and the line BC . As these two objects are swapped under this inversion, by an argument similar to the one for Z , we see that D stays constant √ under this inversion. But, this means that AD = AY · AC = AZ , which completes the solution.