SAUDI ARABIAN MATHEMATICAL COMPETITIONS

Consider the right triangle $ABC$ with incircle ($I$) and $D,E$ are tangent points of $(I)$ on sides $AB,AC$. It is easy to see that $ADIE$ is a square of side $n$.



Denote $BD=x$, $CE=y$ then we can see that from a pair $(x,y)$ with $x>n,y>n$, we can construct a right triangle satisfying the given condition. Thus, to count the number of triangles, we will count the number of pairs $(x,y)$. Notice that

$$[ABC]=\frac{1}{2}AB\cdot AC=\frac{(x+n)(y+n)}{2}$$

and

$$[ABC]=\frac{1}{2}n(AB+BC+CA)=\frac{1}{2}n(x+n+y+n+x+y)=n(x+y+n).$$

Hence,

$$(x+n)(y+n)=2n(x+y+n)\iff n^2=xy-(nx+ny).$$

This implies that $(x-n)(y-n)=2n^2$.

Since $2n^2=2p_1^2{p}_2^2\cdots p_{2017}^2$, then $2n^2$ has $2\cdot 3^{2017}$ positive divisors.

These divisors can be partitioned into $3^{2017}$ pairs of the form $\left(d,\frac{2n^2}{d}\right)$, and from each pair, we can find one pair $(x,y)$ which implies that there are exactly $3^{2017}$ triangles satisfying the given conditions. $\square$