PUMaC 2021 · 团队赛 · 第 9 题
PUMaC 2021 — Team Round — Problem 9
题目详情
- Let AX be a diameter of a circle Ω with radius 10 , and suppose that C lies on Ω so that ′ AC = 16. Let D be the other point on Ω so CX = CD. From here, define D to be the ′ reflection of D across the midpoint of AC, and X to be the reflection of X across the midpoint 1 p ′ ′ of CD. If the area of triangle CD X can be written as , where p, q are relatively prime, find q p + q.
解析
- Let AX be a diameter of a circle Ω with radius 10 , and suppose that C lies on Ω so that ′ AC = 16 . Let D be the other point on Ω so CX = CD. From here, define D to be the ′ reflection of D across the midpoint of AC, and X to be the reflection of X across the midpoint p ′ ′ of CD. If the area of triangle CD X can be written as , where p, q are relatively prime, find q p + q. Proposed by: Frank Lu Answer: 1367 First, we consider the triangle ADC. It is well-known that the reflection of the orthocenter of this triangle, reflected across the midpoint of DC, is the diametrically opposite point on this ′ circle. In particular, notice that X is the orthocenter of ADC. But this means that we know ′ ′ ′ that D is the orthocenter of AX C. This implies that X D is a diameter of the circumcircle ′ ′ ′ of AX C. In particular, CD X is a right triangle. Now, observe that, since CX = CD, 3 3 we have that ∠ CAD = ∠ CAX. But sin ∠ CAX = . Therefore, we see that sin ∠ CAD = . 5 5 However, we also know that, since AX is a diameter, we have that CX = 12 , and so hence that √ CD = 12 . But notice that this is less than 10 2 , meaning that D lies on the same semicircle defined by AX. In particular, this means that ∠ CDA is obtuse. Therefore, it follows that, AC 4 3 by the Law of Sines, we have that sin ∠ CDA = sin ∠ CAD = , and so cos ∠ CDA = − . CD 5 5 ′ But it is well-known property of orthocenters that ∠ CX A = 180 − ∠ CDA. Similarly, we see 3 ′ ′ ′ that ∠ X CA = 90 − ∠ DX C = 90 − ∠ DAC. Therefore, we have that cos ∠ CX A = and 5 3 ′ ′ cos ∠ X CA = , meaning that this triangle is isosceles, with AC = AX . Hence, as AC = 16 , 5 3 96 16 ′ ′ ′ we have that CX = 216 = . But notice then that X D = = 20 , meaning that we have 4 5 5 5 ′ ′ 28 96 · 28 96 · 14 1344 that D X = , and so the area is = = , and so p + q = 1367 . 5 2 · 25 25 25