HMMT 十一月 2024 · 冲刺赛 · 第 13 题

HMMT November 2024 — Guts Round — Problem 13

[9] Let f and g be two quadratic polynomials with real coefficients such that the equation f ( g ( x )) = 0 has four distinct real solutions: 112 , 131 , 146 , and a . Compute the sum of all possible values of a .

解析

[9] Let f and g be two quadratic polynomials with real coefficients such that the equation f ( g ( x )) = 0 has four distinct real solutions: 112, 131, 146, and a . Compute the sum of all possible values of a . Proposed by: Derek Liu Answer: 389 Solution: The key observation is the following. Claim 1. If a , b , c , d are roots of f ( g ( x )), then one can permute them so that a + b = c + d . Proof. Let v be the point for which g ( v ) is the local minimum or maximum. Note that if g ( x ) = g ( y ), then x and y are symmetric around v , or x + y = 2 v . Moreover, if a , b , c , d roots of f ( g ( x )), then we can permute them so that g ( a ) and g ( b ) are equal to one root of f and g ( c ) and g ( d ) are equal to another root of f . This means that a + b = c + d = 2 v . In the case of our problem, if three roots are r , s , and t , then the fourth can be r + s − t , r + t − s , or s + t − r , with sum r + s + t . Using the given values, we get that the answer is 112 + 131 + 146 = 389 . 2 Observe that these are all possible; indeed, if we let g ( x ) = x − ( r + s ) x , then g ( r ) = g ( s ) and g ( t ) = g ( r + s − t ). Now let f ( x ) = ( x − g ( r ))( x − g ( t )); then f ( g ( x )) has roots r , s , t , and r + s − t . The other two values are similarly achievable.