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?

  • 1
    $\begingroup$ A baby version: do we know whether such point is at most unique, provided that $A>0$ is sufficiently small? $\endgroup$ Commented Oct 8, 2015 at 14:54
  • $\begingroup$ There should in fact exist no such point for very small triangles except for a vertex sitting on the intersection of two sides with integral length. $\endgroup$ Commented Oct 8, 2015 at 15:26

1 Answer 1


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...


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.