HMMT 十一月 2021 · THM 赛 · 第 10 题

HMMT November 2021 — THM Round — Problem 10

Let n be the answer to this problem. Suppose square ABCD has side-length 3. Then, congruent n non-overlapping squares EHGF and IHJK of side-length are drawn such that A, C, and H are 6 collinear, E lies on BC and I lies on CD . Given that AJG is an equilateral triangle, then the area of √ AJG is a + b c , where a, b, c are positive integers and c is not divisible by the square of any prime. Find a + b + c .

解析

Let n be the answer to this problem. Suppose square ABCD has side-length 3. Then, congruent n non-overlapping squares EHGF and IHJK of side-length are drawn such that A, C, and H are 6 collinear, E lies on BC and I lies on CD . Given that AJG is an equilateral triangle, then the area of √ AJG is a + b c , where a, b, c are positive integers and c is not divisible by the square of any prime. Find a + b + c . Proposed by: Akash Das Answer: 48 Solution: F A B E I C G D H M K J The fact that EHGF and IHJK have side length n/ 6 ends up being irrelevant. Since A and H are both equidistant from G and J , we conclude that the line ACHM is the perpen- dicular bisector of GJ . ′ ′ Now, define the point C so that the spiral similarity centered at J sends M and H to C and I , ′ ′ ′ respectively. Since 4 JM C ∼ 4 JHI , JM ⊥ M C , so C is on line AM . Moreover, since the spiral ◦ ′ ◦ ′ similarity rotates by ∠ HJI = 45 , we conclude that IC is at a 45 angle to HM , implying that C is ′ ◦ on line CD . Therefore C = C , implying that ∠ M JC = ∠ HJI = 45 . As a result, J lies on line BC . ◦ ◦ To finish, simply note that ∠ BAJ = 75 , so by AJ = AB/ cos 75 . So √ √ √ √ √ 3 9 3 1 9 3 2 9 3 2 [ AJG ] = AJ = = = √ = 18 3 + 27 . 2 ◦ ◦ 4 4 cos 75 4 1 + cos 150 2 − 3