HMMT 二月 2024 · ALGNT 赛 · 第 9 题
HMMT February 2024 — ALGNT Round — Problem 9
题目详情
- Suppose a , b , and c are complex numbers satisfying 2 a = b − c, 2 b = c − a, and 2 c = a − b. Compute all possible values of a + b + c .
解析
- Suppose a , b , and c are complex numbers satisfying 2 a = b − c, 2 b = c − a, and 2 c = a − b. Compute all possible values of a + b + c . Proposed by: Rishabh Das √ Answer: 0 , ± i 6 2 2 2 Solution: Summing the equations gives a + b + c = 0 and summing a times the first equation and 3 3 3 2 2 2 2 etc. gives a + b + c = 0. Let a + b + c = k . Then a + b + c = 0 means ab + bc + ca = k / 2, and 3 3 3 3 3 3 2 2 2 3 a + b + c = 0 = ⇒ − 3 abc = a + b + c − 3 abc = ( a + b + c )( a + b + c − ab − bc − ca ) = − k / 2 , 3 so abc = k / 6. This means a , b , and c are roots of the cubic 3 2 2 3 x − kx + ( k / 2) x − ( k / 6) = 0 for some k . Next, note that ∑ 4 4 4 2 2 3 a + b + c = a ( ka − ( k / 2) a + ( k / 6)) cyc ∑ 2 2 3 2 2 3 = k ( ka − ( k / 2) a + ( k / 6)) − ( k / 2) a + ( k / 6) a cyc ∑ 2 2 3 4 = ( k / 2) a − ( k / 3) a + ( k / 6) cyc 4 4 = − k / 3 + k / 2 4 = k / 6 . After this, there are two ways to extract the values of k . • Summing squares of each equation gives ∑ 4 4 4 2 2 2 2 2 a + b + c = ( a − b ) = 2( a + b + c ) − 2( ab + bc + ca ) = − k , cyc so 4 √ k 2 = − k = ⇒ k = 0 , ± i 6 . 6 2 • Summing a times the first equation, etc. gives 6 ∑ k 4 4 4 2 2 2 2 a + b + c = a ( b − c ) = − ( a − b )( b − c )( c − a ) = − a b c = − , 36 cyc so 4 6 √ k k = − = ⇒ k = 0 , ± i 6 . 6 36 √ √ 3 2 We can achieve k = 0 with a = b = c = 0. Letting a , b , and c be the roots of x − ( i 6) x − 3 x + ( i 6) 2 2 will force one of a = b − c and all other equalities or a − c − b and all other equalities to hold, if the √ latter happens, swap b and c . Finally, for these ( a, b, c ), take ( − a, − c, − b ) to get − i 6. Thus, all of these are achievable.