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,

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

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. $\square$