HMMT 十一月 2025 · GEN 赛 · 第 9 题
HMMT November 2025 — GEN Round — Problem 9
题目详情
- Let a , b , and c be pairwise distinct nonzero complex numbers such that 1 (10 a + b )(10 a + c ) = a + , a 1 (10 b + a )(10 b + c ) = b + , b 1 (10 c + a )(10 c + b ) = c + . c Compute abc .
解析
- Let a , b , and c be pairwise distinct nonzero complex numbers such that 1 (10 a + b )(10 a + c ) = a + , a 1 (10 b + a )(10 b + c ) = b + , b 1 (10 c + a )(10 c + b ) = c + . c Compute abc . Proposed by: Pitchayut Saengrungkongka, Qiao Zhang 1 Answer: 91 Solution 1: Consider the polynomial f ( x ) = ( x + a )( x + b )( x + c ). The given conditions can be rewritten as 1 2 f (10 a ) = (10 a + a )( a + ) = 11( a + 1) , a 1 2 f (10 b ) = (10 b + b )( b + ) = 11( b + 1) , b 2 1 f (10 c ) = (10 c + c )( c + ) = 11( c + 1) . c 2 x Thus, 10 a , 10 b , and 10 c are roots to f ( x ) − 11( + 1). Since a , b , and c are pairwise distinct, these 100 three roots are distinct, so 2 x ( x + a )( x + b )( x + c ) = ( x − 10 a )( x − 10 b )( x − 10 c ) + 11 + 1 . 100 Comparing the constant term of both sides gives 11 1 abc = − 1000 abc + 11 = ⇒ abc = = . 1001 91 1 Remark. By comparing coefficients of the equations above, we get a + b + c = and ab + bc + ca = 0. 100 1 Together with abc = , this solves ( a, b, c ) up to permutation and shows that such ( a, b, c ) exist. 91 © 2025 HMMT Solution 2: Subtracting the second equation from the first yields 1 1 2 2 100( a − b ) + 9 ac − 9 bc = a − b + − . a b As a and b are distinct, we can divide both sides by a − b to find 1 100( a + b ) + 9 c = 1 − . ab A similar argument shows 1 100( b + c ) + 9 a = 1 − . bc The difference of these two equations gives us 1 1 a − c 100( a − c ) + 9( c − a ) = − = ⇒ 91( a − c ) = . bc ab abc 1 Since a and c are distinct, we can divide both sides by a − c to get abc = . 91