HMMT 十一月 2025 · 冲刺赛 · 第 22 题
HMMT November 2025 — Guts Round — Problem 22
题目详情
- [12] Suppose that a , b , and c are pairwise distinct nonzero complex numbers such that 3 2 3 2 3 2 a − 4 a + 5 bc = b − 4 b + 5 ac = c − 4 c + 5 ab = 67 . Compute abc .
解析
- [12] Suppose that a , b , and c are pairwise distinct nonzero complex numbers such that 3 2 3 2 3 2 a − 4 a + 5 bc = b − 4 b + 5 ac = c − 4 c + 5 ab = 67 . Compute abc . Proposed by: Pitchayut Saengrungkongka Answer: 42 Solution 1: For convenience, let p = abc . Then we have 4 3 x − 4 x − 67 x + 5 p = 0 for x ∈ { a, b, c } . Thus, this quartic equation has a, b, c as roots, but the product of roots is 5 p , so 5 must be the last root. This implies that 4 3 2 3 5 − 4 · 5 − 67 · 5 + 5 p = 0 = ⇒ p = 67 + 4 · 5 − 5 = 42 . Solution 2: Taking the difference of the first two equations, 3 2 3 2 3 3 2 2 0 = ( a − 4 a + 5 bc ) − ( b − 4 b + 5 ac ) = ( a − b ) − 4( a − b ) + 5 c ( b − a ) 2 2 = ( a − b )( a + ab + b ) − 4( a − b )( a + b ) − 5 c ( a − b ) 2 2 = ( a − b )[( a + ab + b ) − 4( a + b ) − 5 c ] 2 2 = ( a + ab + b ) − 4( a + b ) − 5 c, © 2025 HMMT and similar for b, c and a, c , since a, b, c are not equal. Then, take the difference of these equations: 2 2 2 2 0 = [( a + ab + b ) − 4( a + b ) − 5 c ] − [( a + ac + c ) − 4( a + c ) − 5 b ] 2 2 = ( b − c ) + a ( b − c ) − 4( b − c ) − 5( c − b ) = ( b − c )[( b + c ) + a + 1] = b + c + a + 1 , so a + b + c = − 1. Taking the cyclic sum of the first difference equation, we get 2 2 2 2 2 2 0 = [( a + ab + b ) − 4( a + b ) − 5 c ] + [( b + bc + c ) − 4( b + c ) − 5 a ] + [( a + ac + c ) − 4( a + c ) − 5 b ] 2 2 2 = 2( a + b + c ) + ( ab + bc + ca ) − 13( a + b + c ) 2 = 2[( a + b + c ) − 2( ab + bc + ca )] + ( ab + bc + ca ) − 13( a + b + c ) = 2[1 − 2( ab + bc + ca )] + ( ab + bc + ca ) + 13 = − 3( ab + bc + ca ) + 15 , so ab + bc + ca = 5. Finally, sum the original expressions: 3 2 3 2 3 2 67 · 3 = ( a − 4 a + 5 bc ) + ( b − 4 b + 5 ac ) + ( c − 4 c + 5 ab ) 3 3 3 2 2 2 201 = a + b + c − 4( a + b + c ) + 5( ab + bc + ac ) 3 2 2 2 2 2 2 = [( a + b + c ) − 3( a b + a c + b a + b c + c b + c a ) − 6 abc ] 2 − 4(( a + b + c ) − 2( ab + bc + ca )) + 5 · 5 = [ − 1 − 3[( a + b + c )( ab + bc + ca ) − 3 abc ] − 6 abc ] − 4(1 − 2 · 5) + 25 = [ − 1 − 3[ − 5 − 3 abc ] − 6 abc ] + 36 + 25 = − 1 + 15 + 3 abc + 61 = 3 abc + 75 . Therefore, abc = 42 from the equation 201 = 3 abc + 75 above.