HMMT 二月 2026 · ALGNT 赛 · 第 4 题

HMMT February 2026 — ALGNT Round — Problem 4

Let a , b , and c be pairwise distinct complex numbers such that 2 2 a + ab + b = 3( a + b ) , 2 2 a + ac + c = 3( a + c ) , 2 2 b + bc + c = 5( b + c ) + 1 . Compute a .

解析

Let a , b , and c be pairwise distinct complex numbers such that 2 2 a + ab + b = 3( a + b ) , 2 2 a + ac + c = 3( a + c ) , 2 2 b + bc + c = 5( b + c ) + 1 . Compute a . Proposed by: Pitchayut Saengrungkongka 7 Answer: = 3 . 5 2 Solution: We multiply the first and second equation by ( a − b ) and ( a − c ) respectively. From this, we get the equations 3 3 2 2 a − b = 3 a − 3 b , 3 3 2 2 a − c = 3 a − 3 c . 3 2 3 2 3 2 3 2 In particular, these rearrange to a − 3 a = b − 3 b and a − 3 a = c − 3 c . Therefore, there exists a complex number k , such that 3 2 3 2 3 2 a − 3 a = b − 3 b = c − 3 c = k. 3 2 Thus, a , b , and c are the three roots of the cubic x − 3 x − k = 0 . By Vieta’s formulas, we have a + b + c = 3 and ab + bc + ca = 0 . Hence, 2 2 2 2 a + b + c = ( a + b + c ) − 2( ab + bc + ca ) = 9 . Therefore, we have 2 2 2 18 = 2( a + b + c ) + ( ab + bc + ca ) 2 2 2 2 2 2 = ( a + ab + b ) + ( b + bc + c ) + ( c + ca + a ) = 3( a + b ) + 3( b + c ) + 5( b + c ) + 1 = 6( a + b + c ) + 2( b + c ) + 1 = 18 + 2( b + c ) + 1 . 1 7 Therefore, b + c = − , whence a = 3 − ( b + c ) = . 2 2 ©2026 HMMT