HMMT 十一月 2025 · 冲刺赛 · 第 33 题
HMMT November 2025 — Guts Round — Problem 33
题目详情
- [17] Four points A , B , C , and D lie on a circle with radius 2 such that CD = 3, CA = CB , and DA − DB = 1. Compute the maximum possible value of AB . © 2025 HMMT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT November 2025, November 08, 2025 — GUTS ROUND Organization Team Team ID# 8 2 2
解析
- [17] Four points A , B , C , and D lie on a circle with radius 2 such that CD = 3, CA = CB , and
DA − DB = 1. Compute the maximum possible value of AB .
Proposed by: Rohan Bodke
√
6 3
Answer:
7
Solution: We have two cases on whether C is the midpoint of arc ADB or not.
Case 1. If C is the midpoint of arc ADB , then DA < DB , and so ABDC is a cyclic quadrilateral with vertices in that order. B A D C © 2025 HMMT Applying Ptolemy’s on ABCD , AB · CD + AC · BD = AD · BC = ⇒ 3 AB = CA ( DA − DB ) = CA = CB. Thus, △ ABC is similar to one with side lengths 1, 3, and 3, and has circumradius 2. The area of a q √ 2 1 1 35 2 triangle with side lengths 1, 3, and 3 is 3 − = , so its circumradius is 2 2 4 1 · 3 · 3 9 √ = √ . 35 35 4 · 4 √ 2 35 Thus, in order for △ ABC to have a circumradius of 2, we must have AB = . 9
Case 2. If C is not the midpoint of arc ADB , the direct application of Ptolemy’s theorem on ABCD is not useful. ′ C D A B C ′ The idea is to construct the antipode C of C . We have p p √ ′ ′ 2 2 2 2 C D = ( CC ) − CD = 4 − 3 = 7 , ′ so we can apply a similar argument as before. Since DA < DB , we have that AC DB is cyclic in this ′ order. By Ptolemy’s theorem on AC DB , we obtain √ ′ ′ ′ ′ ′ ′ AC · DB + AB · C D = AD · BC = ⇒ 7 AB = BC ( DA − DB ) = BC = AC . √ √ ′ Thus, △ ABC is similar to one with side lengths 1, 7, and 7 and has circumradius 2. The area of q √ √ √ 2 1 1 3 3 a 1- 7- 7 triangle is 7 − = , so its circumradius is 2 2 4 √ √ √ 1 · 7 · 7 7 3 √ = . 3 3 9 4 · 4 √ ′ 6 3 Thus, in order for △ ABC to have a circumradius of 2, we must have AB = . 7 √ √ 6 3 2 35 Conclusion. The two possible values of AB are and . However, we note that 7 9 r r √ √ √ 6 3 108 140 2 35 = > 2 > = , 7 49 81 9 √ 6 3 so is the maximum value. 7 © 2025 HMMT 8 2 2