HMMT 十一月 2025 · GEN 赛 · 第 8 题
HMMT November 2025 — GEN Round — Problem 8
题目详情
- Let Γ and Γ be two circles that intersect at two points P and Q . Let ℓ and ℓ be the common 1 2 1 2 external tangents of Γ and Γ . Let Γ touch ℓ and ℓ at U and U , respectively, and let Γ touch 1 2 1 1 2 1 2 2 ℓ and ℓ at V and V , respectively. Given that P Q = 10 and the distances from P to ℓ and ℓ are 3 1 2 1 2 1 2 and 12, respectively, compute the area of the quadrilateral U U V V . 1 2 2 1
解析
- Let Γ and Γ be two circles that intersect at two points P and Q . Let ℓ and ℓ be the common 1 2 1 2 external tangents of Γ and Γ . Let Γ touch ℓ and ℓ at U and U , respectively, and let Γ touch 1 2 1 1 2 1 2 2 ℓ and ℓ at V and V , respectively. Given that P Q = 10 and the distances from P to ℓ and ℓ are 3 1 2 1 2 1 2 and 12, respectively, compute the area of the quadrilateral U U V V . 1 2 2 1 Proposed by: Aprameya Tripathy Answer: 200 Solution: U 1 X A V 1 P Q V 2 B Y U 2 © 2025 HMMT Let P Q meet ℓ and ℓ at X and Y , respectively, and let the feet from P to ℓ and ℓ be A and B , 1 2 1 2 repsectively. By symmetry, note that U U V V and U U XY are isosceles trapezoids. We will find 1 2 2 1 2 1 both the height and the midline length of U U V V to compute its area. 1 2 2 1 Since U U XY is an isosceles trapezoid, we have that ∠ P XA = ∠ P Y B . Furthermore, ∠ P AX = 2 1 ◦ 90 = ∠ P BY , so △ P XA ∼ △ P Y B . Then, since QY = P X by symmetry, P Q P Y P B 12 = − 1 = − 1 = − 1 = 3 , XP XP P A 3 10 40 P A 9 meaning XP = and XQ = XP + P Q = . Thus, sin ∠ P XA = = . Since lines P Q 3 3 XP 10 and U U are parallel, we have that ∠ P XA = ∠ U U V , so the height of trapezoid U U V V is 1 2 2 1 1 1 2 2 1 9 U V · sin ∠ P XA = U V . Now, by Power of a Point, 1 1 1 1 10 2 2 XU = XP · XQ = XV , 1 1 and 2 2 Y U = Y P · Y Q = Y V 2 2 √ 40 so X and Y are the midpoints of U V and U V , respectively, and U V = 2 XP · XQ = . 1 1 2 2 1 1 3 20 50 Therefore, XY is the midline of U U V V and has length P Q + 2 · P X = 10 + 2 · = , while the 1 2 2 1 3 3 9 height of U U V V is U V = 12. 1 2 2 1 1 1 10 50 Thus, the area of trapezoid U U V V is 12 · = 200 . 1 2 2 1 3