Determine all functions $f: \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ such that \[ f(x,y) + f(y,z) + f(z,x) = \max \{ x,y,z \} - \min \{ x,y,z \} \]for every $x,y,z \in \mathbb{R}$ and there exists some real $a$ such that $f(x,a) = f(a,x) $ for every $x \in \mathbb{R}$.
Problem
Source: INAMO 2015 Shortlist A4
Tags: functions, algebra, functional equation, function
bruckner
30.12.2018 13:08
Not as ugly as some FEs out there!
Let $P(x,y,z)$ be the assertion. Note that $f(a,b)=\left| \frac{a-b}2 \right|$ is a solution. We shall show that it is unique.
$P(x,x,x):f(x,x)=0$
$P(x,x,a):f(x,a)=f(a,x)=\frac12 \left| a-x \right|=\frac12 (\max \{ a,x \} -\min \{ a,x \} )=\left| \frac{x-a}2 \right|$
$P(a,y,z):\left| \frac{y-a}2 \right|+\left| \frac{z-a}2 \right|+f(y,z)=\max \{ a,y,z \} - \min \{ a,y,z \}$
Case 1: $\bold{z=y,y=a \text{ or } z=a}$, already analised, so WLOG $z>y$
Case 2: $\bold{z>y>a}$, then $f(y,z)=\frac{z-y}2=\left| \frac{y-z}2 \right|$
Case 3: $\bold{z>a>y}$, then we have the same as above.
Case 4: $\bold{a>z>y}$, and again the same.
Thus, the only solution is $\boxed{f(x,y)=\left| \frac{x-y}2 \right|}$
GorgonMathDota
30.12.2018 13:25
bruckner wrote: Not as ugly as some FEs out there!
Let $P(x,y,z)$ be the assertion. Note that $f(a,b)=\left| \frac{a-b}2 \right|$ is a solution. We shall show that it is unique.
$P(x,x,x):f(x,x)=0$
$P(x,x,a):f(x,a)=f(a,x)=\frac12 \left| a-x \right|=\frac12 (\max \{ a,x \} -\min \{ a,x \} )=\left| \frac{x-a}2 \right|$
$P(a,y,z):\left| \frac{y-a}2 \right|+\left| \frac{z-a}2 \right|+f(y,z)=\max \{ a,y,z \} - \min \{ a,y,z \}$
Case 1: $\bold{z=y,y=a \text{ or } z=a}$, already analised, so WLOG $z>y$
Case 2: $\bold{z>y>a}$, then $f(y,z)=\frac{z-y}2=\left| \frac{y-z}2 \right|$
Case 3: $\bold{z>a>y}$, then we have the same as above.
Case 4: $\bold{a>z>y}$, and again the same.
Thus, the only solution is $\boxed{f(x,y)=\left| \frac{x-y}2 \right|}$
Thanks a lot!