HMMT 二月 2025 · 几何 · 第 6 题

6. Trapezoid ABCD , with AB ∥ CD , has side lengths AB = 11, BC = 8, CD = 19, and DA = 4. Compute the area of the convex quadrilateral whose vertices are the circumcenters of △ ABC , △ BCD , △ CDA , and △ DAB . Proposed by: Karthik Venkata Vedula, Pitchayut Saengrungkongka √ Answer: 9 15 Solution: B A O B O C O A O D C D Let O , O , O , and O be the circumcenters of △ BCD , △ CDA , △ DAB , and △ ABC , respectively. A B C D Note that O O is the perpendicular bisector of AD . Similarly, O O ⊥ AC and O O ⊥ AB . B C B D C D

As AB ∥ CD , we have O O ⊥ CD . Then, △ O O O ∼ △ ADC , as their corresponding sides are C D B C D

perpendicular. Likewise, △ O O O ∼ △ CBA , so O O O O ∼ BADC . D A B A B C D Therefore, we only need to compute the area of ABCD and the ratio of similarity between the two trapezoids. Draw a line parallel to AD passing through B . Let this line intersect CD at X . Then, BX = AD = 4, CB = 8, and CX = CD − AB = 8. The height h of ABCD is given by √ 2 2 √ 2[ △ BXC ] BX · d ( C, BX ) 4 8 − 2 h = d ( B, XC ) = = = = 15 . XC XC 8 On the other hand, the height of O O O O is the distance between the perpendicular bisectors A B C D ′ ′ of AB and CD , which pass through the midpoints M of AB and N of CD . Let A and M be the projections of A and M onto CD , respectively. Then, the height of O O O O is given by A B C D p CD AB 19 − 11 ′ ′ ′ 2 2 M N = DN − DM = − − DA = − 4 − h = 3 . 2 2 2 3 √ Therefore, the similarity ratio between the two trapezoids is . We know that the area of ABCD is 15 2 √ √ √ √ 1 3 √ (11 + 19) 15 = 15 15, so the area of O O O O is · 15 15 = 9 15 . A B C D 2 15