HMMT 十一月 2025 · 冲刺赛 · 第 26 题

HMMT November 2025 — Guts Round — Problem 26

[13] Rectangles ABXP and CDXQ lie inside semicircle S such that A , B , C , and D lie on the arc of S , and P and Q lie on the diameter of S . Given that BX = 7, P X = 6, and QX = 8, compute DX . 7

解析

[13] Rectangles ABXP and CDXQ lie inside semicircle S such that A , B , C , and D lie on the arc of S , and P and Q lie on the diameter of S . Given that BX = 7, P X = 6, and QX = 8, compute DX . Proposed by: Marin Hristov Hristov, Sarunyu Thongjarast √ Answer: 114 − 3 Solution: © 2025 HMMT D B C A X P O Q ′ B ′ D Let O be the circumcenter of S , which is the intersection of the perpendicular bisectors of AB and CD . The perpendicular bisector of AB is also the perpendicular bisector of P X , and the perpendicular bisector of CD is also the perpendicular bisector of QX . Therefore O is the intersection of the ◦ perpendicular bisectors of P X and QX . However, O lies on P Q ; therefore ∠ P XQ = 90 . Thus B , X , and Q are collinear, and similarly, D , X , and P are collinear. ′ ′ Let Γ be the circle that contains S . Let B as the second intersection of BX and Γ, and D be the ′ second intersection of DX and Γ. Since O lies on the perpendicular bisector of QX and BB , we have ′ ′ that BX = QB = 7, and similarly DX = P D . Let x = DX . Then power of a point on X gives ′ ′ 7(8 + 7) = BX · XB = DX · XD = x (6 + x ) . √ √ Solving for x gives x = ± 114 − 3. The positive solution is DX = 114 − 3 . 7