PUMaC 2020 · 几何（A 组） · 第 8 题
PUMaC 2020 — Geometry (Division A) — Problem 8
题目详情
- A A A A is a cyclic quadrilateral inscribed in circle Ω , with side lengths A A = 28 , A A = 1 2 3 4 1 2 2 3 √ √ 12 3 , A A = 28 3 , and A A = 8 . Let X be the intersection of A A , A A . Now, for 3 4 4 1 1 3 2 4 i = 1 , 2 , 3 , 4 , let ω be the circle tangent to segments A X, A X, and Ω , where we take i i i +1 indices cyclically (mod 4) . Furthermore, for each i, say ω is tangent to A A at X , A A at i 1 3 i 2 4 Y , and Ω at T . Let P be the intersection of T X and T X , and P the intersection of T X i i 1 1 1 2 2 3 3 3 and T X . Let P be the intersection of T Y and T Y , and P the intersection of T Y and 4 4 2 2 2 3 3 4 1 1 T Y . Find the area of quadrilateral P P P P . 4 4 1 2 3 4 1
解析
- A A A A is a cyclic quadrilateral inscribed in circle Ω , with side lengths A A = 28 , A A = 1 2 3 4 1 2 2 3 √ √ 12 3 , A A = 28 3 , and A A = 8 . Let X be the intersection of A A , A A . Now, for 3 4 4 1 1 3 2 4 i = 1 , 2 , 3 , 4 , let ω be the circle tangent to segments A X, A X, and Ω , where we take i i i +1 indices cyclically (mod 4) . Furthermore, for each i, say ω is tangent to A A at X , A A at i 1 3 i 2 4 Y , and Ω at T . Let P be the intersection of T X and T X , and P the intersection of T X i i 1 1 1 2 2 3 3 3 3 and T X . Let P be the intersection of T Y and T Y , and P the intersection of T Y and 4 4 2 2 2 3 3 4 1 1 T Y . Find the area of quadrilateral P P P P . 4 4 1 2 3 4 Proposed by: Frank Lu Answer: 784 First, we claim that the P all lie on the circle. To show this, we first claim that P , P are i 1 3 midpoints of opposite arcs for A A . To see this, we first notice that P is the midpoint of 1 3 1 the arc A A opposite of T . To see this, notice that the midpoint of this arc lies on T X ; 1 3 1 1 1 we can see this by taking a homothety centered at T , which takes ω to Ω . This is a well- 1 1 known result of a circle inscribed in a segment. But this holds for T X as well, meaning that 2 2 this point is P . A similar result holds for P , P , P ; in particular, these all lie on the circle. 1 2 3 4 Notice furthermore that P P and P P are diameters, meaning that in particular we see that 1 3 2 4 P P P P this is a rectangle. 1 2 3 4 From here, we need to find the side lengths of the rectangle, which requires us to first find the circumradius of this quadrilateral. To see this, we first find the diagonal length A A , which 1 3 2 we set as c, and the angle ∠ A A A , which we set as α. Then, we know that c = 784+432 − 2 · 1 2 3 √ √ 2 28 · 12 3 cos α. But by cyclic quadrilaterals, we also know that c = 2352+64+2 · 28 3 · 8 cos α. √ √ − 5 3 Equating these two, we find that 2 · 28 3 · 20 = − 2416 + 1216 = 1200 , or that cos α = . 14 √ √ 11 2 From here, we see that sin α = . Therefore, we see that A A = 784 + 432 + 48 3 · 5 3 = 1 3 14 44 1216 + 720 = 1936 , and so thus A A = 44 . It thus follows that the circumradius is R = = 22 1 3 14 28 . Finally, we need to find the angle θ between A A and A A , as then the angle between P P 1 3 2 4 1 3 and P P is equal to θ as well. We now seek to find this angle. 2 4 To find this angle, notice that the area of this quadrilateral is going to be the product of the 1 diagonals times the sine of the angle between these diagonals times . But by Ptolemy’s, 2 √ √ √ √ the product of this is 28 · 28 3 + 8 · 12 3 = (784 + 96) 3 = 880 3 . As for the area of the quadrilateral, we employ the fact that we know the side lengths and split this up into four √ √ 1 3 3 3 1 triangles: we see that the corresponding inscribed angles to them have sines , , , and , 2 14 2 7 √ √ √ 3 39 3 3 respectively. Therefore, we see that the sines of the central angles are equal to , , , 2 98 2 √ 8 3 and . Finally, notice that the sum of the cosines of the three smaller angles yields us the 49 larger angle, meaning that the center doesn’t lie in this quadrilateral. To verify this: we see √ √ √ 3 13 1 3 3 5 3 11 that the sum of the first two angles has cosine of − = , with sine , and adding 2 14 2 14 14 14 √ √ 5 3 4 3 11 1 1 this to the next smaller angle yields us a cosine of − = . 14 7 14 7 2 √ √ √ √ √ √ 1 2 3 39 3 3 8 3 Our area of the quadrilateral A A A A is thus 28 ( + − + ) = 156 3+64 3 = 1 2 3 4 2 2 98 2 49 √ 1 220 3 . But this means that the angle between the two diagonals has a sine of . But this means 2 1 2 1 our rectangle has area of 56 = 784 . 2 2 4