HMMT 十一月 2020 · THM 赛 · 第 4 题

HMMT November 2020 — THM Round — Problem 4

Points G and N are chosen on the interiors of sides ED and DO of unit square DOM E , so that pentagon GN OM E √ a − b c has only two distinct side lengths. The sum of all possible areas of quadrilateral N OM E can be expressed as , d where a, b, c, d are positive integers such that gcd( a, b, d ) = 1 and c is square-free (i.e. no perfect square greater than 1 divides c ). Compute 1000 a + 100 b + 10 c + d .

解析

Points G and N are chosen on the interiors of sides ED and DO of unit square DOM E , so that pentagon GN OM E has only two distinct side lengths. The sum of all possible areas of quadrilateral √ a − b c N OM E can be expressed as , where a, b, c, d are positive integers such that gcd( a, b, d ) = 1 and d c is square-free (i.e. no perfect square greater than 1 divides c ). Compute 1000 a + 100 b + 10 c + d . Proposed by: Andrew Lin Answer: 10324 Solution: E M x G x D O N Since M O = M E = 1, but ON and GE are both less than 1, we must have either ON = N G = GE = x (call this case 1) or ON = GE = x, N G = 1 (call this case 2). 1+ x Either way, the area of N OM E (a trapezoid) is , and triangle N GT is a 45-45-90 triangle. In case 2 ( ) √ √ √ 2 3 − 2 1, we have 1 = ON + N T = x 1 + , so x = 2 − 2 and the area of the trapezoid is . In case 2 2 √ √ √ 2 4 − 2 2 − 2 2, we have 1 = ON + N T = x + , which yields an area of as x = . The sum of these two 2 4 2 √ 10 − 3 2 answers is . 4