HMMT 十一月 2024 · 冲刺赛 · 第 1 题
HMMT November 2024 — Guts Round — Problem 1
题目详情
- [5] A circle of area 1 is cut by two distinct chords. Compute the maximum possible area of the smallest resulting piece.
解析
- We characterize a − b that works. 3 3 3 Claim 1. Let a and b be positive integers. Then, gcd( a − b , ( a − b ) ) is squarefree if and only if gcd( a, b ) = 1, a − b is squarefree, and a − b is not divisible by 3. 3 Proof. If gcd( a, b ) = d > 1, then g is divisible by d , hence not squarefree. Thus, we now restrict our attention to the case gcd( a, b ) = 1. In that case, we factor out a − b from the gcd and simplify it as follows: 3 3 3 2 2 2 gcd( a − b , ( a − b ) ) = ( a − b ) gcd( a + ab + b , ( a − b ) ) 2 2 = ( a − b ) gcd ( a − b ) + 3 ab, ( a − b ) 2 = ( a − b ) gcd(( a − b ) , 3 ab ) . 2 Moreover, since gcd( a, b ) = 1, we have that gcd( a − b, a ) = gcd( a − b, b ) = 1, and so gcd(( a − b ) , ab ) = 1, 2 which implies that gcd(( a − b ) , 3 ab ) is either 1 or 3. Thus, each of the following is equivalent to the next. 3 3 3 • gcd( a − b , ( a − b ) ) is squarefree. 2 • ( a − b ) gcd(( a − b ) , 3 ab ) is squarefree. 2 • a − b is squarefree and at least one of a − b and gcd(( a − b ) , 3 ab ) is not divisible by 3. • a − b is squarefree and a − b is not divisible by 3. The claim implies that c = a − b is possible only if c is squarefree and not divisible by 3. For any such c , we may construct ( a, b ) that works by taking a = c + 1 and b = 1. Hence, the answer is simply the number of squarefree integers up to 50 that are not divisible by 3. There are 16 multiples of 3 from 1 to 50. Among the remaining numbers, 4, 8, 16, 20, 28, 32, 40, 44, 25, 50, and 49 are not squarefree. This leaves 50 − 16 − 11 = 23 possible values of a − b . d