Local Mathematical Competitions

The triple $(f(x_1,x_2),A,B)$ does exist, so the answer is yes. Let $A=O=(0,0)$, $B=(x_0,y_0)=(2009+\frac{2}{3},\frac{1}{3})$. The idea is to search for a polynomial $f$ such that $f(x,y)=0$ is the equation of an ellipse centered at $B$, passing through $O$ and with tangent line $y=0$ at $O$. In fact, if $f$ is chosen like this, the ellipse $f(x,y)=0$ is completely contained in the region $\{(x,y);0\le y\le \frac{2}{3}\}$, with $O$ the only integer point on the ellipse or in its interior; clearly, the absolute minimum of $f(x,y)$ is attained at $B$ and $f(x,y)$ is positive at all integer points other than $O$. Therefore, we consider polynomials of the type $$f(X,Y)=9M(X-x_0{)}^2+9N(X-x_0)(Y-y_0)+9P(Y-y_0{)}^2-Q$$ where $M,N,P,Q$ are integers with $M,P,Q,4MP-N^2>0$. The condition that the ellipse $f(x,y)=0$ passes through $O$, with tangent line $y=0$ at $O$, is expressed by $$\{\begin{array}{l}6029^{2}M+6029N+P-Q=0\\ 2\cdot 6029M+N=0.\end{array}$$

It is then sufficient to choose $M=1$, $N=-2\cdot 6029$, $P$ any integer greater than $6029^2$ and $Q=P-6029^2$.

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Alternative solution.

(Alternative Solution. D. Schwarz) Given any integer point $A(a_1,a_2)$, there exist infinitely many points $B(b_1,b_2)$ with $b_1,b_2\in \mathbb{Q}\setminus \mathbb{Z}$, and such that $|a_1-b_1|+|a_2-b_2|=2010$, for example $b_1=a_1+\alpha +r$, $b_2=a_2+\beta +(1-r)$, with $\alpha ,\beta \in {\mathbb{Z}}_{+}$, $r\in \mathbb{Q}\cap (0,1)$, and $\alpha +\beta =2009$. We now will consider polynomials of the type $$f(X,Y)=N\left((X-a_1{)}^2+(Y-a_2{)}^2+\epsilon \right)\left((X-b_1{)}^2+(Y-b_2{)}^2\right)$$ where $\epsilon \in {\mathbb{Q}}_{+}^{\ast }$ and $N\in {\mathbb{Z}}_{+}^{\ast }$ large enough for $f(X,Y)$ to have integer coefficients. One then has $f(b_1,b_2)=0$, while $f(x,y)>0$ for all points $(x,y)$ in the plane, other than $B$. One also then has $f(a_1,a_2)=N\epsilon ((a_1-b_1{)}^2+(a_2-b_2{)}^2)$, while one has, for all integer points $(n_1,n_2)$ in the plane, other than $A$, $$\begin{gathered} \min \{(n_1-a_1{)}^2+(n_2-a_2{)}^2+\epsilon \}=1+\epsilon \text{and} \\ \min \{(n_1-b_1{)}^2+(n_2-b_2{)}^2\}=m, \end{gathered}$$ for some $m\in \mathbb{Q}\cap (0,\frac{1}{2}]$ (for example $m=\frac{1}{2}$ when $r=\frac{1}{2}$), therefore $F(n_1,n_2)>N(1+\epsilon )m$. In order to have $f(n_1,n_2)>f(a_1,a_2)$ it is thus enough that $N(1+\epsilon )m\ge N\epsilon ((a_1-b_1{)}^2+(a_2-b_2{)}^2)$, therefore let us take $$\epsilon =m/((a_1-b_1{)}^2+(a_2-b_2{)}^2-m)\in {\mathbb{Q}}_{+}^{\ast },$$ and then choose some appropriate $N$.