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HMMT 十一月 2020 · 团队赛 · 第 6 题

HMMT November 2020 — Team Round — Problem 6

专题
Discrete Math / 离散数学
难度
L3
来源
HMMT

题目详情

  1. [40] Regular hexagon P P P P P P has side length 2. For 1 ≤ i ≤ 6, let C be a unit circle centered 1 2 3 4 5 6 i at P and ` be one of the internal common tangents of C and C , where C = C and C = C . i i i i +2 7 1 8 2 Assume that the lines { ` , ` , ` , ` , ` , ` } bound a regular hexagon. The area of this hexagon can be 1 2 3 4 5 6 √ a expressed as , where a and b are relatively prime positive integers. Compute 100 a + b . b
解析
  1. [40] Regular hexagon P P P P P P has side length 2. For 1 ≤ i ≤ 6, let C be a unit circle centered 1 2 3 4 5 6 i at P and be one of the internal common tangents of C and C , where C = C and C = C . i i i i +2 7 1 8 2 Assume that the lines { ` , ` , ` , ` , ` , ` } bound a regular hexagon. The area of this hexagon can be 1 2 3 4 5 6 √ a expressed as , where a and b are relatively prime positive integers. Compute 100 a + b . b Proposed by: Daniel Zhu Answer: 1603 Solution: The only way for the lines to bound a regular hexagon H is if they are rotationally i symmetric around the center O of the original hexagon. (A quick way to see this is to note that the ◦ angle between the two internal common tangents of C and C cannot be a multiple of 60 .) Thus i i +2 all we need to do is to compute h , the distance from the center O to the sides of H , because then we 2 √ can compute the side length of H as h and thus its area as 3 √ ( ) 2 √ 3 2 h 2 √ 6 = 2 3 h . 4 3 P 2 T 3 C 1 C 3 M P P 1 3 Q T 1 O Without loss of generality, let’s only consider . Let M be the midpoint of P P and let T and T be 1 1 3 1 3 the tangency points between and C and C , respectively. Without loss of generality, assume T is 1 1 3 1 closer to O than T . Finally, let Q be the projection of O onto ` , so that h = OQ . 3 1 ◦ Now, note that ∠ OM Q = 90 − ∠ T M P = ∠ M P T , so 4 OM Q ∼ 4 M P T . Therefore, since 1 1 1 1 1 1 OM = OP / 2 = 1, we find 2 √ √ 2 2 OQ T M M P − P T 2 1 1 1 1 h = OQ = = = = , OM M P M P 3 1 1 √ √ √ 2 since M P = P P / 2 = 3. Thus the final area is 2 3 = 16 / 3. 1 1 3 3