59th Ukrainian National Mathematical Olympiad

Let triangle $△ABC$ have a right angle at $C$. Then for any triangle with even integer hypotenuse, the midpoint of hypotenuse is the point of interest (Fig. 11).

Let $BC=an$, $AC=bn$, $AB=cn$, $n\in \mathbb{N}$ and $a^2+b^2=c^2$. Let $K\in AC$, $N\in BC$, $CN=bk$, $CK=ak$, $k\in \mathbb{N}$. It suffices for lengths $AN=b\sqrt{n^2+k^2}$ and $BK=a\sqrt{n^2+k^2}$ to be integer. Thus, it suffices that $n^2+k^2=m^2$ for some $m\in \mathbb{N}$, $bk<an$ and $ak<bn$. Let $a=3$, $b=4$, $c=5$, $k=5$, $n=12$, $m=13$, hence $an=36$, $bn=48$, $ak=15$, $bk=20$. Thus, we obtain the following segments: $$AC=48,CK=15,BC=36,CN=20,AB=60.$$