HMMT 二月 2010 · GEN1 赛 · 第 9 题
HMMT February 2010 — GEN1 Round — Problem 9
题目详情
- [ 7 ] Three unit circles ω , ω , and ω in the plane have the property that each circle passes through the 1 2 3 centers of the other two. A square S surrounds the three circles in such a way that each of its four sides is tangent to at least one of ω , ω and ω . Find the side length of the square S . 1 2 3
解析
- [ 7 ] Three unit circles ω , ω , and ω in the plane have the property that each circle passes through the 1 2 3 centers of the other two. A square S surrounds the three circles in such a way that each of its four sides is tangent to at least one of ω , ω and ω . Find the side length of the square S . 1 2 3 √ √ 6+ 2+8 Answer: 4 B C ω 2 Q ω 3 ω R 1 F P E A M N D By the Pigeonhole Principle, two of the sides must be tangent to the same circle, say ω . Since S 1 surrounds the circles, these two sides must be adjacent, so we can let A denote the common vertex of the two sides tangent to ω . Let B , C , and D be the other vertices of S in clockwise order, and 1 let P , Q , and R be the centers of ω , ω , and ω respectively, and suppose WLOG that they are also 1 2 3 in clockwise order. Then AC passes through the center of ω , and by symmetry (since AB = AD ) it 1 General Test, Part 1 must also pass through the other intersection point of ω and ω . That is, AC is the radical axis of ω 2 3 2 and ω . 3 Now, let M and N be the feet of the perpendiculars from P and R , respectively, to side AD . Let E and F be the feet of the perpendiculars from P to AB and from R to DC , respectively. Then P EAM and N RF D are rectangles, and P E and RF are radii of ω and ω respectively. Thus AM = EP = 1 1 2 and N D = RF = 1. Finally, we have ◦ M N = P R · cos(180 − ∠ EP R ) ◦ = cos(180 − EP Q − RP Q ) ◦ ◦ ◦ = − cos ((270 − 60 ) / 2 + 60 ) ◦ = − cos(165 ) ◦ = cos(15 ) √ √ 6 + 2 = . 4 √ √ √ √ 6+ 2 6+ 2+8 Thus AD = AM + M N + N D = 1 + + 1 = as claimed. 4 4