HMMT 十一月 2017 · 团队赛 · 第 4 题
HMMT November 2017 — Team Round — Problem 4
题目详情
- [ 30 ] An equiangular hexagon has side lengths 1 , 1 , a, 1 , 1 , a in that order. Given that there exists a circle that intersects the hexagon at 12 distinct points, we have M < a < N for some real numbers M N and N . Determine the minimum possible value of the ratio . M 2017
解析
- [ 30 ] An equiangular hexagon has side lengths 1 , 1 , a, 1 , 1 , a in that order. Given that there exists a circle that intersects the hexagon at 12 distinct points, we have M < a < N for some real numbers M N and N . Determine the minimum possible value of the ratio . M Proposed by: Yuan Yao √ 3 3 3+3 √ Answer: OR 2 3 − 1 √ We claim that the greatest possible value of M is 3 − 1, whereas the least possible value of N is 3. To begin, note that the condition requires the circle to intersect each side of the hexagon at two points on its interior . This implies that the center must be inside the hexagon as its projection onto all six sides must be on their interior. Suppose that the hexagon is ABCDEF , with AB = BC = DE = EF = 1, CD = F A = a , and the center O . √ √ 3 When a ≤ 3 − 1, we note that the distance from O to CD (which is ) is greater than or equal to 2 a +1 the distance from O to B or E (which is ). However, for the circle to intersect all six sides at two 2 points each, the distance from the center of the circle to CD and to F A must be strictly less than that from the center to B and to E , because otherwise any circle that intersects CD and F A at two points each must include B or E on its boundary or interior, which will not satisfy the condition. WLOG assume that the center of the circle is closer to F A than to CD , including equality (in other words, the center is on the same side of BE as F A , possibly on BE itself), then note that the parabola with foci B and E and common directrix CD intersects on point O , which means that there does not exist a point in the hexagon on the same side of BE as F A that lies on the same side of both parabola as CD . This means that the center of the circle cannot be chosen. √ √ 3 When a = 3 − 1 + for some very small real number > 0, the circle with center O and radius r = 2 intersects sides AB, BC, DE, EF at two points each and is tangent to CD and F A on their interior. ′ ′ ′ Therefore, there exists a real number > 0 such that the circle with center O and radius r = r + satisfy the requirement. 1 a When a ≥ 3, we note that the projection of BF onto BC has length | − | ≥ 1, which means that the 2 2 projection of F onto side BC is not on its interior, and the same goes for side EF onto BC . However, for a circle to intersect both BC and EF at two points, the projection of center of the circle onto the two sides must be on their interior, which cannot happen in this case. √ 3 When a = 3 − for some very small real number > 0, a circle with center O and radius r = ( a + 1) 4 intersects AF and CD at two points each and is tangent to all four other sides on their interior. ′ ′ ′ Therefore, there exists a real number > 0 such that the circle with center O and radius r = r + satisfy the requirement. √ √ N 3 3 3+3 √ With M ≤ 3 − 1 and N ≥ 3, we have ≥ = , which is our answer. M 2 3 − 1 2017