PUMaC 2023 · 几何（A 组） · 第 8 题
PUMaC 2023 — Geometry (Division A) — Problem 8
题目详情
- Similar to the last 6 problems, let △ ABC be a triangle with AB = 4 and AC = . Let ω 2 denote the A -excircle of △ ABC . Let ω touch lines AB, AC at the points D, E , respectively. Let Ω denote the circumcircle of △ ADE . Consider the line ℓ parallel to BC such that ℓ is tangent to ω at a point F and such that ℓ does not intersect Ω. Let ℓ intersect lines AB, AC at the points X, Y , respectively, with XY = 18 and AX = 16. Let the perpendicular bisector of XY meet the circumcircle of △ AXY at P, Q , where the distance from P to F is smaller than − − → m 2 the distance from Q to F . Let ray P F meet Ω for the first time at the point Z . If P Z = n for relatively prime positive integers m, n , find m + n . 1 (Write answers on next page.) Name: Team: Write answers in table below: Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 2
解析
- Similar to the last 6 problems, let △ ABC be a triangle with AB = 4 and AC = . Let ω 2 denote the A -excircle of △ ABC . Let ω touch lines AB, AC at the points D, E , respectively. Let Ω denote the circumcircle of △ ADE . Consider the line ℓ parallel to BC such that ℓ is tangent to ω at a point F and such that ℓ does not intersect Ω. Let ℓ intersect lines AB, AC at the points X, Y , respectively, with XY = 18 and AX = 16. Let the perpendicular bisector of XY meet the circumcircle of △ AXY at P, Q , where the distance from P to F is smaller than − − → m 2 the distance from Q to F . Let ray P F meet Ω for the first time at the point Z . If P Z = n for relatively prime positive integers m, n , find m + n . Proposed by Sunay Joshi Answer: 1159 Below, we let ( XY Z ) denote the circumcircle of the triangle △ XY Z . We restate the problem (changing the names of points) as follows: let △ ABC be a triangle with sidelengths a = 9, b = 7, and c = 8. Let M denote the midpoint of the minor arc BC in ( ABC ). Let D denote the point of tangency between the incircle ω of △ ABC and BC . Let 2 ⃗ M D intersect ω for the first time at X . We seek M X . Note that our problem is simply this, 2 scaled up by a factor of 2. Therefore at the end, we multiply the value of M X we obtain by a factor of 4. The first key observation is that P, X, M are collinear, where P ̸ = A is the second intersection of ω and ( ABC ). The next key observation is the spiral similarity △ AXM ∼ △ AI O , where A 4 M X OI A I denotes the center of ( AP I ). This yields = , which when rearranged implies A M A OA OI A M X = M A · . R To find M A , we may apply Ptolemy’s theorem to the cyclic quadrilateral ABM C : M A · a = b + c A b + c ( b + c ) · BM , which implies M A = M B · = 2 R sin · . a 2 a To find OI , we apply the power of a point to I and the circle ( ABC ). The power of I is A A A 1 1 given by AI · I M , or equivalently AI · ( AM − AI ). By the definition of power, this also A A 2 2 2 2 equals R − OI . A r By trigonometry, we have AI = . By the Law of Sines in ( ABC ), we have M A = A sin 2 p √ A A A 2 R sin( + B ). By the Law of Cosines, we find cos A = 2 / 7, sin = 5 / 14, cos = 3 / 14, 2 2 2 √ √ √ √ 21 5 √ √ cos B = 2 / 3, sin B = 5 / 3. Also, we find R = and r = 5. Therefore AI = = 14 2 5 5 / 14 √ 1 14 A A A and AI = . Also, by the sine addition formula, sin( + B ) = sin cos B + sin B cos = 2 2 2 2 2 q √ √ √ 5 5 14 5 5 21 5 1 √ √ . Next, AM = 2 · · = . Hence AM − AI = 2 14. Thus the power of I A 3 14 3 2 2 2 5 14 √ √ 1 2 2 equals 14 · 2 14 = 14. Equating the two expressions for power, we have OI = R − 14 = A 2 √ 2 21 161 A b + c 5 14 175 2 − 14 = . Next, M A = 2 R sin · , which reduces to , so that M A = . 20 20 2 a 2 2 21 441 2 2 √ Finally, note R = ( ) = . 20 2 5 2 161 OI 2 2 175 575 A 20 Putting everything together, we find M X = M A · = · = . Recall that we 2 441 R 2 18 20 575 1150 2 2 must scale up by a factor of 4. Therefore the true value of M X is M X = 4 · = , so 18 9 that our answer is 1150 + 9 = 1159. 5