PUMaC 2015 · 代数（A 组） · 第 6 题

PUMaC 2015 — Algebra (Division A) — Problem 6

[ 6 ] We define the function f ( x, y ) = x + ( y − 4) x + ( y − 4 y + 4) x + ( y − 4 y + 4 y ). Then choose any distinct a, b, c ∈ R such that the following holds: f ( a, b ) = f ( b, c ) = f ( c, a ). Over all such choices of a, b, c , what is the maximum value achieved by: 4 3 2 4 3 2 4 3 2 min( a − 4 a + 4 a , b − 4 b + 4 b , c − 4 c + 4 c )?

解析

[ 6 ] We define the function f ( x, y ) = x + ( y − 4) x + ( y − 4 y + 4) x + ( y − 4 y + 4 y ). Then choose any distinct a, b, c ∈ R such that the following holds: f ( a, b ) = f ( b, c ) = f ( c, a ). Over all such choices of a, b, c , what is the maximum value achieved by: 4 3 2 4 3 2 4 3 2 min( a − 4 a + 4 a , b − 4 b + 4 b , c − 4 c + 4 c )? 2 Solution: f ( b ) − f ( a ) f ( c ) − f ( b ) f ( a ) − f ( c ) 4 3 2 Let f ( x ) = x − 4 x +4 x then the equalities become = = . b − a c − b a − c So this implies that the points ( a, f ( a )) , ( b, f ( b )) , ( c, f ( c )) lie on the same line and the question now becomes: what is the maximum, over all such lines which intersect f ( x ) at least 3 times, of the 3rd highest intersection point? We claim that the highest intersection points occurs when the line is horizontal and tangent to f ( x ) at x = 1 , f ( x ) = 1. This is simply because this is a valid point and if we choose a, b such that f ( a ) , f ( b ) > 1, then it is clear from looking at the graph of f ( x ) that the line between ( a, f ( a )) , ( b, f ( b )) can only intersect f ( x ) two times. Therefore f ( x ) = 1 is our answer. Author: Roy Zhao