Lemma 0: $f(x, x) = x$ for all reals $x$. This is obvious by taking $a = b = c = x$. Lemma 1: Suppose $x < y$ are reals. Of $f(x, y)$ and $f(y, x)$, one is less than or equal to $x$, and one is greater than or equal to $y$. pf. Taking $a = b = x, c = y$, we have $x$ is the median of $f(x, x) = x$, $f(x, y)$, and $f(y, x)$. Thus, one of $f(x, y)$ and $f(y, x)$ is less than or equal to $x$, and the other is greater than or equal to $x$. By the same logic one is less than or equal to $y$ and one is greater than or equal to $y$, so Lemma 1 is proved. Lemma 2: For any real $x$, we either have $f(x, r) = x$ for all reals $r$, or $f(r, x) = x$ for all reals $r$. Pf. If there exist reals $y, z$ with $y < x < z$ so that $f(y, x) \neq x$ and $f(x, z) \neq x$, we'd have a contradiction because $f(z, y) \neq x$ by Lemma 1, yet $x$ is the median and therefore an element of $\{f(y, x), f(x, z), f(z, y)\}$. Therefore, either $f(y, x) = x$ for all $y < x$ or $f(x, z) = x$ for all $z > x$. By the same logic either $f(x, y) = x$ for all $y < x$ or $f(z, x) = x$ for all $z > x$. Thus, we have four cases: 1. $f(y, x) = x$ for all reals $y$ 2. $f(x, y) = x$ for all reals $y$ 3. $f(y, x) = f(x, y) = x$ for all $y < x$ 4. $f(y, x) = f(x, y) = x$ for all $y > x$ However, cases 3 and 4 contradict Lemma 1, so only cases 1 and 2 are possible, proving Lemma 2. Now, there can't exist $x_1 \neq x_2$ such that $f(x_1, r) = x_1$, $f(r, x_2) = x_2$ for all reals $r$, otherwise no value is possible for $f(x_1, x_2)$. Thus, the only solutions are $f(x, y) = x$ and $f(x, y) = y$, and it's easy to check that both have the property.

Your solution is incorrect. The functions $f(x,y) = \min\{x,y\}$ or $f(x,y) = \max\{x,y\}$ can easily be shown to fullfill the functional equation.

His solution is fine. Last I checked, for plugging in $a < b < c$ we don't get $b$ the median of $a,a,b$ or $b,c,c,$, which you get from having $f$ be the $\min$ or $\max$ respectively.

I have a different solution, which I'll just sketch here. Let $P(a,b,c)$ be the assertion. First we get Lemma 0 and Lemma 1 in math_explorer's solution in the same way. Let a pair $(x,y)$ with $x<y$ be called bad if $f(y,x)\ge y,f(x,y)\le x$ and let it be called good if $f(x,y)\ge y, f(y,x)\le x$. Now if we have all bad pairs, then $P(z,y,x)$ with $z>y>x$ gives $f(x,y)=x\forall x,y$. For all good pairs, $P(x,y,z)$ gives $f(x,y)=y\forall x,y$. So otherwise there is a good pair $(x,y)$ and also a bad pair. Take any constant $c<x<y$ and check $P(c,x,y)$. Obviously $f(x,y)\neq y$ and $f(y,c)\neq x$ so $f(c,x)=x$ for all $c<x$. So $(c,x)$ is good. Then we get $f(x,c)\le c$ and going back to $P(c,x,y)$ we get $(c,y)$ is also good. Check $P(x,c,y)$ to conclude that $f(y,x)=x$ for all good pairs $(x,y)$. Pick $d>x>c$ so that $(d,x)$ is bad (we can do this since $f(c,x)$ is good for $c<x$) and now $P(d,x,c)$ gives a contradiction. So the two solutions are $f(x,y)=x\forall x,y$ and $f(x,y)=y\forall x,y$.

The following solution is joint with Andrew He. We prove the following main claim, from which repeated applications can deduce the problem. Claim: Let $a < b < c$ be arbitrary. On $\{a,b,c\}^2$, $f$ takes one of the following two forms, where the column indicates the $x$-value and the row indicates the $y$-value. \[ \begin{array}{c|rrr} f & a & b & c \\ \hline a & a & b & \ge c \\ b & \le a & b & \ge c \\ c & \le a & b & c \end{array} \qquad \text{or} \qquad \begin{array}{c|rrr} f & a & b & c \\ \hline a & a & \le a & \le a \\ b & b & b & b \\ c & \ge c & \ge c & c \end{array} \]Proof. First, we of course have $f(x,x) = x$ for all $x$. Now: By considering the assertion for $(a,a,c)$ and $(a,c,c)$ we see that one of $f(a,c)$ and $f(c,a)$ is $\ge c$ and the other is $\le a$. Hence, by considering $(a,b,c)$ we find that one of $f(a,b)$ and $f(b,c)$ must be $b$, and similarly for $f(b,a)$ and $f(c,b)$. Now, WLOG $f(b,a) = b$; we prove we get the first case. By considering $(a,a,b)$ we deduce $f(a,b) \le a$, so $f(b,c) = b$ and then $f(c,b) \ge c$. Finally, considering $(c,b,a)$ once again in conjunction with the first bullet, we arrive at the conclusion that $f(a,c) \le a$; similarly $f(c,a) \ge c$. \qedhere $\blacksquare$ From this it's easy to obtain that $f(x,y) \equiv x$ or $f(x,y) \equiv y$ are the only solutions.

Clearly, $f(x,x)=x$ by setting $x=a=b=c$. Now, we claim that for all $a$ and $b$, one of $f(a,b)$ and $f(b,a)$ has to be either $a$ or $b$. WLOG $a<b$. By considering $f(a,a),f(a,b),f(b,a)$, we have that either $f(a,b)<a<f(b,a)$ or $f(b,a)<a<f(a,b)$. Similarly, either $f(a,b)<b<f(b,a)$ or $f(b,a)<b<f(a,b)$. Suppose that $f(a,b)<a<b<f(b,a)$ (the other case follows similarly). Consider a $c>b$. We have the median of $f(a,b), f(b,c), f(c,a)$ should be $b$. Since $f(a,b)<b$, either $f(b,c)=b$ or $f(c,a)=b$. If $f(c,a)=b$, then we have that the median of $f(c,c),f(c,a),f(a,c)=c,b,f(a,c)$ is $c$, so $f(a,c)\geq c$. But then, the median of $f(a,c),f(c,b),f(b,a)$ must be $b$, but $f(a,c),f(b,a)>b$, a contradiction. If $f(b,c)=b$, then we have that the median of $f(c,c),f(c,b),f(b,c)=c,f(c,b),b$ is $c$, so $f(c,b)\geq c$. But then, the median of $f(a,c),f(c,b),f(b,a)$ must be $b$, but $(c,b),f(b,a)>b$, a contradiction. Hence, one of $f(a,b)$ and $f(b,a)$ must be $a$ or $b$. WLOG, suppose that $f(a,b)=a$. We have that the median of $f(a,b),f(b,c),f(c,a)$ is $b$, so either $f(b,c)=b$ or $f(c,a)=b$. If $f(c,a)=b$, then by the same argument as above, $f(a,c)\geq c$. Considering $f(a,c),f(c,b),f(b,a)$, we have that $f(b,a)=b$. If $f(b,c)=b$, then by the same argument as above, $f(c,b)\geq c$. Considering $f(a,c),f(c,b),f(b,a)$, we have that $f(b,a)=b$. Hence, we must have that $f(a,b)=a$ and $f(b,a)=b$ or $f(a,b)=b$ and $f(b,a)=a$ for all $a$ and $b$. To finish, note that if $f(a,b)=a$, then $b$ must appear in $f(a,b),f(b,c),f(c,a)$, so $f(b,c)=b$. This leads to $f(b,a)=b$ and $f(c,b)=c$. By considering $f(b,a),f(c,b),f(a,c)$, we must have $f(a,c)\leq b$, from which it follows that $f(a,c)=a$ and $f(c,a)=c$. Now, for any $x$ and $y$, we have that this applied to $a,b,x$ gives $f(x,a)=x$ and $f(a,x)=a$, and this applied to $a,b,y$ gives $f(y,a)=y$ and $f(a,y)=a$. Applying it to $a,x,y$ gives $f(x,y)=x$ and $f(y,x)=y$. Considering the other similar case as well, we have that our set of solutions is $f(x,y)=x$ and $f(y,x)=y$.

v_Enhance wrote: Solution with Andrew He: We prove the following main claim, from which repeated applications can deduce the problem. Claim: Let $a < b < c$ be arbitrary. On $\{a,b,c\}^2$, $f$ takes one of the following two forms: \[ \begin{array}{c|rrr} f & a & b & c \\ \hline a & \le a & b & c \\ b & \le a & b & \ge c \\ c & a & b & \ge c \end{array} \qquad \text{or} \qquad \begin{array}{c|rrr} f & a & b & c \\ \hline a & \ge c & \ge c & c \\ b & b & b & b \\ c & a & \le a & \le a. \end{array} \] Proof: First, we of course have $f(x,x) = x$ for all $x$. By considering the assertion for $(a,a,c)$ and $(a,c,c)$ we see that one of $f(a,c)$ and $f(c,a)$ is $\ge c$ and the other is $\le a$ ($\star$). Now, by considering $(a,b,c)$ we find that one of $f(a,b)$ and $f(b,c)$ must be $b$, and similarly for $f(b,a)$ and $f(c,b)$. Now, WLOG $f(b,a) = b$; we prove we get the first case. By considering $(a,a,b)$ we deduce that $f(a,b) \le a$, so $f(b,c) = b$ and then $f(c,b) \ge c$. Finally, considering $(c,b,a)$ once again in conjunction with ($\star$), we arrive at the conclusion that $f(a,c) \le a$; similarly $f(c,a) \ge c$. $\blacksquare$ From this it's easy to obtain that $f(x,y) \equiv x$ or $f(x,y) \equiv y$ are the only solutions. Shouldn't it be the following? \[ \begin{array}{c|rrr} f & a & b & c \\ \hline a & a & b & \ge c \\ b & \le a & b & \ge c \\ c & \le a & b & c \end{array} \qquad \text{or} \qquad \begin{array}{c|rrr} f & a & b & c \\ \hline a & c & \ge c & \ge c \\ b & b & b & b \\ c & \le a & \le a & a. \end{array} \]

Yes, I fixed it. Thanks.

Let $m(x, y, z)$ represent the median of $x, y,$ and $z$. So, $m(f(a, b), f(b, c), f(c, a))=m(a, b, c)$. We claim that the only solutions are $f(x, y)\equiv x$ and $f(x, y)\equiv y$. In the first case, the condition holds, since $$m(f(a, b), f(b, c), f(c, a))=m(a, b, c)$$The condition holds in the second case for similar reasons. Note $$m(f(a, a), f(a, a), f(a, a))=m(a, a, a)$$$$f(a, a)=a$$Lemma: Let $b\in \mathbb{R}$. One of the following is true: $$f(x, b)=b\quad \forall x\in \mathbb{R}$$$$f(b, x)=b\quad \forall x\in \mathbb{R}$$Proof: We first claim that one of the following is true: $$f(a, b)=b\quad \forall a<b$$$$f(b, c)=b\quad \forall c>b$$Assume towards a contradiction that neither is true. Then, there exists $a, c\in \mathbb{R}$ with $a<b<c$ such that $f(a, b)\neq b$ and $f(b, c)\neq b$. Because $$m(f(a, b), f(b, c), f(c, a))=m(a, b, c)=b,$$we know $f(c, a)=b$. So, $$a=m(a, a, c)=m(f(a, a), f(a, c), f(c, a))=m(a, b, f(c, a))$$$$\implies f(c, a)\leq a$$However, by the same logic, $f(c, a)\geq c$, a contradiction. So, one of the following is true: $$(1)\quad f(a, b)=b\quad \forall a<b$$$$(2)\quad f(b, c)=b\quad \forall c>b$$By the same logic, one of the following is also true: $$(3)\quad f(c, b)=b\quad \forall c>b$$$$(4)\quad f(b, a)=b\quad \forall a<b$$However, if $(1)$ and $(4)$ are both true, then for all $a<b$, $a=m(a, a, b)=m(f(a, a), f(a, b), f(b, a))=m(a, b, b)=b$, a contradiction. Similarly, $(2)$ and $(3)$ also cannot both be true. Therefore, either both $(1)$ and $(3)$ are true or both $(2)$ and $(4)$ are true, which, when combined with the fact that $f(b, b)=b$, completes the proof. Rest of proof: Let $a, b \in \mathbb{R}$ with $a\neq b$. By the lemma, either $$f(a, x)=a\quad \forall x\in \mathbb{R}$$or $$f(x, a)=a\quad \forall x\in \mathbb{R}$$ Assume that $f(a, x)=a\quad\forall x\in\mathbb{R}$. So, $f(a, b)=a$. Clearly it is not true that $f(x, b)=b\quad\forall x\in \mathbb{R}$, since $x=a$ violates this. So, by the lemma, $f(b, x)=b\quad\forall x\in\mathbb{R}$. So, $f(x, y)=x\quad\forall x, y\in\mathbb{R}$. In the other case, by similar logic, $f(x, y)=y\quad\forall x, y\in\mathbb{R}$.

by setting $a = b= c$ its easy to see $f(x, x) = x$. Lemma 1: For any $x > y$, one of $f(x, y), f(y, x)$ is at least $x$, and the other is at most $y$ Proof: Consider $a = b = x < c$. Then the median of $x, f(x, c), f(c, x)$ is $x$. This means one of $f(x, c), f(c, x) \geq x$ and the other is at most $x$. We can similarly do $a < x = b = c$ to show that for any $f(x, y), f(y, x)$, one is at most $x$ and the other is at least $x$. Similarly one is at most $y$ and the other is at least $y$. WLOG $x > y$, then if any of them where between $x$ and $y$, we would arrive at a contradiction. Thus one is at most $y$ and the other is at least $x$. Lemma 2: For all $a < b < c$, either $f(a, b) = b, f(c, b) =b$ or $f(b, a) = b, f(b, c) = b$. Proof: One of $f(a, b), f(b, c), f(c, a)$ is $b$ and similarly one of $f(b, a), f(c, b), f(a, c)$ is $b$. By lemma $1$, neither $f(a, c)$ or $f(c, a)$ can be $b$, and exactly one of $f(a, b), f(b, a)$ can be $b$, similarly exactly one of $f(b, c), f(c, b)$ can be $b$. Thus, we must have either $f(b, a) = f(b, c) =b$ or $f(a, b) = f(c, b) = b$. if there existed $c_0$ so $f(c_0, b) = b$, then by Lemma 2 $f(a, b) =b$ for all $a < b$. If there existed $c_1$ so $f(b, c_1) = b$, then by Lemma 2 $f(b, a) = b$ for all $a<b$. Thus either $f(x, b) = b$ for all $x$ or $f(b, x) = b$ for all $x$. Say there was some $a$ such that $f(a, x) = a$, for all $x$ and some $b \neq a$ so $f(x, b) = b$. Then $f(a, b) = a = b$ a contradiction. Thus either $f(x, y) \equiv x$ or $f(x, y)\equiv y$.

mathmax12 wrote:

hey @above so you’re proving something where $a>b$ and you assume $a=b$? this can’t be right..

We show that the only solutions are $f(a,b)=a$ and $f(a,b)=b.$ These clearly work. First, denote the assertion as $P(a,b,c).$ Taking $P(a,a,a)$ gives $f(a,a)=a.$ Taking $P(a,a,b)$ with $a \ne b$ gives that the median of $a,f(a,b),f(b,a)$ is $a,$ and taking $P(a,b,b)$ with $a \ne b$ gives that the median of $b,f(a,b),f(b,a)$ is $b.$ First, notice that $f(a,b) \ne f(b,a)$ since otherwise would imply that $f(a,b)=f(b,a)$ is both medians and must equal both $a,b$ which is impossible. Then assume without loss of generality that $f(a,b)<f(b,a).$ We see that our first result gives $f(a,b) \le a$ and $f(b,a) \ge a.$ Similarly, we have $f(a,b) \le b$ and $f(b,a) \ge b.$ Therefore, we have that $f(a,b)$ cannot be between $a$ and $b$ exclusive. Call this result $(1).$ Now take $P(1,2,3).$ From $(1)$ we see that $f(3,1) \ne 2,$ so either $f(1,2)=2$ or $f(2,3)=2.$ We will only consider the first case, since the second case is symmetric. First, suppose $f(b,c)=c$ and take $P(d,c,b).$ We have $f(c,b) \ne f(b,c)=c$ and $f(b,d) \ne c$ from $(1),$ so $f(d,c)=c.$ Then take $P(a,c,d).$ We get $f(c,d) \ne f(d,c)=c$ and $f(d,a) \ne c$ from $(1),$ so $f(a,c)=c.$ Additionally, now supposing only $f(a,c)=c$ we can take $P(d,c,a).$ We get $f(c,a) \ne f(a,c)=c$ and $f(a,d) \ne c$ from $(1),$ so $f(d,c)=c.$ Then, from $P(b,c,d)$ we get $f(c,d) \ne c$ and $f(d,b) \ne c$ so $f(b,c)=c.$ Thus, we have $f(b,c)=c$ implies $f(d,c)=c$ and $f(a,c)=c,$ and $f(a,c)=c$ implies $f(b,c)=c.$ Since $a,b,c,d$ are arbitrary, these together with $f(x,x)=x$ show that $f(x_0,y_0)=y_0$ for some $x_0,y_0$ implies $f(x,y_0)=y_0$ for all $x.$ Next, suppose $f(a,b)=b.$ Take $P(c,b,a)$ to get $f(c,b)=b.$ Take $P(d,c,b)$ to get $f(d,c)=c.$ Take $P(a,c,d)$ to get $f(a,c)=c.$ Then suppose only $f(a,c)=c.$ Take $P(d,c,a)$ to get $f(d,c)=c.$ Take $P(b,c,d)$ to get $f(b,c)=c.$ Take $P(a,b,c)$ to get $f(a,b)=b.$ Then suppose only $f(c,d)=d.$ Take $P(b,c,d)$ to get $f(b,c)=c.$ Take $P(a,b,c)$ to get $f(a,b)=b.$ Take $P(c,b,a)$ to get $f(c,b)=b.$ Thus, we have $f(a,b)=b$ implies $f(a,c)=c,$ $f(a,c)=c$ implies $f(a,b)=b$ and $f(c,d)=d$ implies $f(c,b)=b.$ Since $a,b,c,d$ are arbitrary, these together with $f(x,x)=x$ show that $f(x_0,y_0)=y_0$ for some $x_0,y_0$ implies $f(x_0,y)=y$ for all $y.$ Combining these two results with the assumption that $f(1,2)=2$ shows that $f(x,y)=y$ for all $y.$ The other case follows similarly, so our only solutions are $f(x,y)=y$ and $f(x,y)=x.$

Denote the assertion as $P(a, b, c)$. Note that $P(a, a, a)$ implies that $f(a, a) = a$. Then, plugging in $P(a, a, b)$ and $P(a, b, b)$ implies that one of $f(a, b)$ and $f(b, a)$ is $\leq a$, and the other must be $\geq b$ if $a < b$. Then, considering the original assertion for $a < b < c$, $P(a, b, c)$ and $P(c, b, a)$ implies that either $f(a, b) = b$ or $f(b, c) = b$ and either $f(b, a) = b$ or $f(c, b) = b$. But then, note \[ f(a, b) = b \implies f(b, a) \leq a \implies f(c, b) = b \implies f(b, c) \geq c \implies f(c, a) \leq b \implies f(a, c) \geq c \]We get something similar if $f(b, a) = b$. Now, assume by contradiction that for some $b_0$, $f(x, b_0) = b_0$ for all $x$, but there exists some $a$ and $b$ such that $f(a, b) = a$. Then considering the different orderings of $a$, $b$ and $b_0$, we find all of them give a contradiction, and we extract the answers of $f(a, b) = b$ or $f(b, a) = a$. $\blacksquare$

Let $P(x,y,z)$ be the given assertion. First $P(x, x, x)$ yields $f(x,x) = x.$ Now consider $P(x, x, y), P(y, y, x)$ for $x < y.$ The medians are $x, y$ respectively. Noting $f(x,x) = x, f(y, y) = y$ either, \begin{align*} f(x, y) \le x &< y \le f(y, x) \\ f(y, x) \le x &< y \le f(x,y). \end{align*} Assume the former. For $P(x, y, z)$ where $x < y < z,$ we have $f(x, y) \le x < y.$ Note $f(z, x) \ne y$ since $x, z$ can vary. Hence $f(y, z) = y.$ Using $P(z, y, x)$ and $f(z, y) > y$, we receive $f(y, x) = y$ and $f(x, z) \le x < y.$ Synthesizing, \begin{align*} f(x,y) \le x < y = f(y, z) < z \le f(z,x) \\ f(x,z) \le x < y = f(y, x) < z \le f(z, y). \end{align*}Thus $f(y, y) = f(y, x) = f(y,z) = y$ and it is clear that the function $f(x,y) = x.$ Using the latter case from above, we can follow a similar derivation and receive, \begin{align*} f(y,x) \le x < y = f(z, y) < z \le f(x,z) \\ f(z,x) \le x < y = f(x, y) < z \le f(y, z). \end{align*}So $f(y, y) = f(z, y) = f(x,y) = y$ means that the function $f(x,y) = y.$

Denote the assertion as $A(a,b,c)$. $A(x,x,x)$ tells us $f(x,x)=x$. $A(x,x,y)$, $A(x,y,y)$ tells us one of $f(x,y)$, $f(y,x)$ is at least $y$ and the other is at most $x$, where $x \leq y$ WLOG. In particular, neither value is between $x$ and $y$. Fix $x$ and $y$, and choose a real value $t$ with $x < t < y$. Considering $A(x,t,y)$, we note we must have either $f(t,y)=t$ or $f(x,t)=t$. In our first case, we must have $f(y,t) \ge y$. Now looking at $A(y,t,x)$, we must have $f(t,x)=t$. We can continue just for verification to conclude \[f(t,x) = f(t,t) = f(t,y) = t.\] Varying $x$, $y$ tells us we have $f(t,u)=t$ for all $t$, $u$ in this case. Analogously, the second case tells us $f(t,u)=u$. Hence each value of $f(x,y)$ is either the first or second parameter. To avert this pointwise trap, notice it only takes one pair to force all the others to follow suit, so our only solutions are $\boxed{f(x,y) = x, \quad f(x,y) = y}$, which can be tested to work. $\blacksquare$

The answer is $f(x,y) = x$ or $f(x,y) = y$. Let $P(a,b,c)$ be the assertion. Plugging in $P(x,x,x)$, it follows that $f(x,x) = x$ for all reals $x$. Plugging in $P(x,x,y)$, it follows that the median of $x, f(x,y), f(y,x)$ is $x$; plugging in $P(x,y,y)$, it follows that the median of $y, f(x,y), f(y,x)$ is $y$. Therefore, $f(x,y)$ and $f(y,x)$ must fall on opposite sides of the range $(x,y)$ and in particular, do not lie inside of this range. Now we plug in $P(x,y,z)$, where WLOG $x < y < z$. Since $y$ lies in the range $(x,z)$, it follows that $f(z,x) \neq y$. So, at least one of $f(x,y)$ or $f(y,z)$ is equal to $y$. Since we have complete control over $x$ and $z$ (so long as $x < y < z$), it follows that at least one of the following is true: $f(x,y) = y$ for all $x < y$, or $f(y,z) = y$ for all $z > y$. In the former case, it follows that $f(y,x)$ must be no greater than $x$, and hence strictly less than $y$. So, if we plug in $P(z,y,x)$, since $f(y,x)$ and $f(x,z)$ are both not equal to $y$, it follows that $f(z,y) = y$. So, $f(a,y) = y$ for all $a$. The latter case goes similarly. So we know that $f(a,b)$ equals either $a$ or $b$ for all $a$ and $b$; furthermore, we know that if $f(a,b) = a$ for one value of $b$, then the equality holds for all values of $b$ (similar statement if $f(a,b) = b$). In general, $f(x,y) =x \implies f(y,x) = y$ which allows us to pivot between all real numbers, handling the pointwise trap.

Clearly we have that $f(a, a)=a$. We also have that for any $x$, $f(x, y) \geq x \leq f(y, x)$ or $f(y, x)\leq f(x, y)$. Let $x, y, z$ be values such that $x<y<z$. Clearly we can't have $f(x, z)=y$. Thus we have that either $f(x, y)=y$ or $f(y, z)=y$ suppose that $f(x, y)=y$, thus we get that $f(y, x) \neq y$, so $f(z, y)=y$. Clearly by repeating this over all combinations we can get that $f(a, b)=a$ for all $a, b$ or $f(a, b)=b$ for all $a, b$.