HMMT 二月 2022 · 团队赛 · 第 7 题
HMMT February 2022 — Team Round — Problem 7
题目详情
- [50] Find, with proof, all functions f : R \ { 0 } → R such that 2 f ( x ) − f ( y ) f ( z ) = x ( x + y + z )( f ( x ) + f ( y ) + f ( z )) for all real x, y, z such that xyz = 1.
解析
- [50] Find, with proof, all functions f : R \ { 0 } → R such that 2 f ( x ) − f ( y ) f ( z ) = x ( x + y + z )( f ( x ) + f ( y ) + f ( z )) for all real x, y, z such that xyz = 1. Proposed by: Akash Das 2 1 Answer: f ( x ) = 0 or f ( x ) = x − . x 1 2 Solution 1: The answer is either f ( x ) = 0 for all x or f ( x ) = x − for all x . These can be checked x to work. Now, I will prove that these are the only solutions. Let P ( x, y, z ) be the assertion of the problem statement. 1 2 Lemma 1. f ( x ) ∈ { 0 , x − } for all x ∈ R \ { 0 } . x 1 1 Proof. P (1 , 1 , 1) yields f (1) = 0. Then, P ( x, 1 , ) and P (1 , x, ) yield x x 1 1 2 f ( x ) = x ( x + + 1)( f ( x ) + f ( )) , x x 1 1 1 − f ( x ) f ( ) = ( x + + 1)( f ( x ) + f ( )) . x x x f ( x ) 2 1 1 Thus, we have f ( x ) = − xf ( x ) f ( ) , so we have f ( x ) = 0 or f ( ) = − . Plugging in the latter into x x x the first equation above gives us 1 f ( x ) 2 f ( x ) = x ( x + + 1)( f ( x ) − ) , x x 1 2 which gives us f ( x ) = 0 or f ( x ) = x − . This proves Lemma 1. x Lemma 2. If f ( t ) = 0 for some t ̸ = 1, then we have f ( x ) = 0 for all x . 1 1 Proof. P ( x, t, ) and P ( t, x, ) give us tx tx 1 1 2 f ( x ) = x ( x + + t )( f ( x ) + f ( )) , tx tx 1 1 1 − f ( x ) f ( ) = t ( x + + t )( f ( x ) + f ( )) . tx tx tx 1 1 t 2 Thus we have tf ( x ) = − xf ( x ) f ( ) , so f ( x ) = 0 or f ( ) = − f ( x ). Plugging in the latter into the tx tx x first equation gives us 1 tf ( x ) 2 f ( x ) = x ( x + + t )( f ( x ) − ) , tx x 1 t 2 1 2 1 which gives us either f ( x ) = 0 or f ( x ) = x ( x + t + )(1 − ) = x − − ( t − ) . Note that since tx x x t 2 1 the ladder expression doesn’t equal x − , since t ̸ = 1 , we must have that f ( x ) = 0. Thus, we have x proved lemma 2. Combining these lemmas finishes the problem. Solution 2: Suppose xyz = 1 and x + y + z ̸ = 0 and that x, y, z are not all the same. Then we have 2 f ( x ) − f ( y ) f ( z ) = x ( x + y + z )( f ( x ) + f ( y ) + f ( z )) , 2 f ( y ) − f ( z ) f ( x ) = y ( x + y + z )( f ( x ) + f ( y ) + f ( z )) , 2 f ( z ) − f ( x ) f ( y ) = z ( x + y + z )( f ( x ) + f ( y ) + f ( z )) . Squaring the first equation and subtracting the second equation times the third gives us: f ( x ) F ( x, y, z ) = 2 2 3 3 3 ( x − yz ) G ( x, y, z ) , where F ( x, y, z ) = f ( x ) + f ( y ) + f ( z ) − 3 f ( x ) f ( y ) f ( z ) and G ( x, y, z ) = ( x + y + z )( f ( x )+ f ( y )+ f ( z )). If F ( x, y, z ) = 0, it is not too hard to see that we get f ( x ) = f ( y ) = f ( z ) = 0. If 2 G 2 2 2 not, then we can let K = and we substitute ( f ( x ) , f ( y ) , f ( z )) = ( K ( x − yz ) , K ( y − xz ) , K ( z − xy )) F 2 3 3 3 3 3 3 into the first equation to get K x ( x + y + z − 3 xyz ) = Kx ( x + y + z − 3 xyz ). Thus, we have K = 0 or K = 1. Thus, we have either ( f ( x ) , f ( y ) , f ( z )) = (0 , 0 , 0) or ( f ( x ) , f ( y ) , f ( z )) = 2 2 2 ( x − yz, y − zx, z − xy ). 1 2 Thus, f (0 . 5) = 0 or f (0 . 5) = 0 . 5 − . If the former is true, then for all y and z such that yz = 2 0 . 5 and y + z ̸ = 0 . 5, we have f ( y ) = 0. However, this gives that f ( y ) = 0 for all y . Likewise, if the latter 2 1 were true, we would have f ( y ) = y − for all y , so we are done. y