Bounding the number of points at integral distance from vertices of a triangle Can the number of points at integral distance to all three points of a non-degenerate triangle of area $A$ be bounded by $1+cA$ for some suitable constant $c$? 
Remark: Since it is easy to bound this number by $4(D+1)^2$ where $D$ is the diameter of the triangle, a sequence giving rise to counterexamples must consist of very thin triangles.
Addendum: Ilya Bogdanov's example shows that the original statement is too optimistic. There are two ways (both suggested in Ilya's answer) to strengthen it: Replace $1$ by a constant or ask the question asymptotically (for $A$ large enough). Both strengthenings are interesting but I prefer the first one:
Are there constants $b\geq 2$ and $c$ such that the number of points at integral distance to a triangle of area $A$ is bounded by the affine function $b+cA$ of the area?
 A: Even a baby version turns out to be wrong: there exist triangles with arbitrarily small area such that there are two points at integral distances to all three vertices; this already shows that there is no such $c$. 
Let $R$ be a large integer, and take a triangle $PAQ$ with $PA=R$, $PQ=R+1$, $AQ=2$. Choose a point $B$ on the side $PQ$ with $QB=1$. Finally, reflect $A$ with respect to $PQ$ to get a point $C$.
Both $P$ and $Q$ have integer distances to $A$, $B$, and $C$. Moreover, $P$ is the circumcenter of $\triangle ABC$, and $R$ is its circumradius. We are left to show that the area of $\triangle ABC$ can be arbitrarily small.
By Stewart's theorem, we have
$$
  AB^2=\frac{AQ^2\cdot BP+AP^2\cdot BQ}{PQ}-PB\cdot QB
  =\frac{4R+R^2}{R+1}-R=\frac{3R}{R+1}<3.
$$
Thus the area of $\triangle ABC$ is 
$$
  \frac{AB\cdot AC\cdot BC}{4R}<\frac{3\cdot 3\cdot 6}{4R}\to0
$$
as $R\to\infty$.
Perhaps, it is more interesting to ask whether such $c$ exists for all sufficiently large $A$, or to replace $1$ in `$1+cA$' by some other constant...
