HMMT 二月 2026 · 几何 · 第 5 题
HMMT February 2026 — Geometry — Problem 5
题目详情
- In the diagram below, three circles of radius 2 are internally tangent to a circle Ω centered at O of radius 11 , and three chords of Ω are each tangent to two of the three circles. Given that O lies inside √ the triangle formed by the three chords and two of the chords have length 4 30 , compute the length of the third chord. O
解析
- In the diagram below, three circles of radius 2 are internally tangent to a circle Ω centered at O of radius 11 , and three chords of Ω are each tangent to two of the three circles. Given that O lies inside √ the triangle formed by the three chords and two of the chords have length 4 30 , compute the length of the third chord. O Proposed by: Jason Mao √ Answer: 8 6 Solution: Let A , B , and C be the centers of the three smaller circles, let the three chords be ℓ , ℓ , 1 2 and ℓ , and let X , Y , and Z be the midpoints of sides BC , AC , and AB , respectively, as shown below. 3 ©2026 HMMT A ℓ ℓ 2 1 √ √ 6 2 6 2 9 Z O Y 3 3 √ √ 6 2 9 9 6 2 ℓ 3 B C X Since ℓ is tangent to circles A and B of radius 2 , it follows that ℓ ∥ AB , and that the distance 1 1 between ℓ and AB is exactly 2 . The same holds for ℓ and AC , as well as for ℓ and BC . 1 2 3 We may now compute OY = OZ to be 2 more than the distance from O to ℓ , which is: 1 √ ( ) 2 √ 1 2 OY = OZ = 2 + 11 − · 4 30 = 3 . 2 Furthermore, since circle A of radius 2 is internally tangent to circle O of radius 11 , it follows that OA = 11 − 2 = 9 . Similarly, OB = OC = 9 . Hence, O is the circumcenter of isosceles △ ABC . Therefore, △ AZO and △ AY O are right, so the Pythagorean Theorem yields: √ √ 2 2 AZ = BZ = AY = CY = 9 − 3 = 6 2 . We now compute OX by noting that △ AZO ∼ △ AXB , so: √ AZ AX 6 2 OX + 9 √ = = ⇒ = = ⇒ OX = 7 . AO AB 9 12 2 √ √ 2 2 Thus, the distance from O to ℓ is OX − 2 = 5 , so the length of ℓ is 2 · 11 − 5 = 8 6 . 3 3