Pre-IMO 2017 Mock Exam

We claim that $f(x,y)$ is a polynomial in $x$ and $y$ of degree at most $d$. It is obvious that every such polynomial satisfies the desired condition. To prove the converse, let $f$ be a function satisfying the desired condition. Pick $(d+2)$ straight lines ${\ell }_1$, ${\ell }_2$, $\dots$, ${\ell }_{d+2}$ in ${\mathbb{R}}^2$ such that no two are parallel and no three are concurrent. Let the equation of ${\ell }_i$ be $h_i(x,y)=0$, where $h_i$ is a linear polynomial. For $i<j$, let $(a_{ij},b_{ij})$ be the intersection of ${\ell }_i$ and ${\ell }_j$, and consider the polynomial $$\phi (x,y)=\sum _{1\le i<j\le d+2}f(a_{ij},b_{ij})\prod _{\begin{array}{c}k=1\\ k\ne i,j\end{array}}^{d+2}\frac{h_k(x,y)}{h_k(a_{ij},b_{ij})}.$$ It is easy to see that $\phi (a_{ij},b_{ij})=f(a_{ij},b_{ij})$ for all $i<j$, and that $\mathrm{deg}\phi \le d$. We shall show that $\phi (a,b)=f(a,b)$ for all $a,b\in \mathbb{R}$. We first make an observation: if $\ell$ is a line such that $\phi (a,b)=f(a,b)$ for at least $(d+1)$ points $(a,b)$ on $\ell$, then $\phi (a,b)=f(a,b)$ for all points $(a,b)$ on $\ell$. Indeed, pick constants $A$, $B$, $C$ and $D$ such that $t\mapsto (At+B,Ct+D)$ parametrizes the line. Then, note that $\phi (At+B,Ct+D)-f(At+B,Ct+D)$ is a polynomial of degree at most $d$, and that it vanishes at least $(d+1)$ points, so $\phi (a,b)=f(a,b)$ for all $(a,b)$ on $\ell$. Now, for a fixed ${\ell }_i$, note that $f(a_{ij},b_{ij})=\phi (a_{ij},b_{ij})$ for every $j\ne i$, so $f(a,b)=\phi (a,b)$ for all points $(a,b)$ on ${\ell }_i$ from the claim. If $(c,d)$ is a point not lying on any ${\ell }_i$, then we can construct a line $\ell$ which passes through $(c,d)$, does not pass through any $(a_{ij},b_{ij})$, and is not parallel to any ${\ell }_i$. Now $f(a,b)=\phi (a,b)$ with $(a,b)={\ell }_i\cap \ell$ for various $i$, so $f(a,b)=\phi (a,b)$ for all $(a,b)$ on $\ell$. In particular, $\phi (c,d)=f(c,d)$. This completes the proof. $\square$