The Problems of Ukrainian Authors

Let's find the condition on the sides of the triangle, under which the medians are perpendicular. Without loss of generality let's consider that medians $AM$ (from the edge $A$) and $BN$ (from the edge $B$) are perpendicular. Let's use vectors: $\vec{c}=\overrightarrow{AB}$, $\vec{b}=\overrightarrow{AC}$. Then $\overrightarrow{AM}=\frac{1}{2}(\overrightarrow{AB}+\overrightarrow{AC})=\frac{1}{2}(\vec{c}+\vec{b})$, $\overrightarrow{BN}=\frac{1}{2}(\overrightarrow{BA}+\overrightarrow{BC})=\frac{1}{2}(2\overrightarrow{BA}+\overrightarrow{AC})=\frac{1}{2}(\vec{b}-2\vec{c})$. These vectors are perpendicular, so, their dot product is equal to $0$. So, $$\begin{array}{c}0=(\overrightarrow{AM},\overrightarrow{BN})=\frac{1}{4}(\vec{c}+\vec{b},\vec{b}-2\vec{c})=\frac{1}{4}(|\vec{b}{|}^2-2|\vec{c}{|}^2-(\vec{c},\vec{b})),(\ast )\\ |\overrightarrow{BC}{|}^2=(\vec{c}+\vec{b},\vec{c}+\vec{b})=|\vec{c}{|}^2+|\vec{b}{|}^2+2(\vec{c},\vec{b}).\text{ But then: }(\vec{c},\vec{b})=\frac{1}{2}(B{C}^2-|\vec{b}{|}^2-2|\vec{c}{|}^2).\end{array}$$

Let's substitute (*): $0=|\vec{b}{|}^2-2|\vec{c}{|}^2-\frac{1}{2}(B{C}^2-|\vec{b}{|}^2-|\vec{c}{|}^2)$. Let's notice, that $|\vec{c}{|}^2=A{B}^2,|\vec{b}{|}^2=A{C}^2$.

So, we can rewrite the equation like this: $5A{B}^2=B{C}^2+A{C}^2$. Let's notice, that this condition is necessary and sufficient for medians to be perpendicular. That's why the only thing that remains is to check if this equation has the solution in integers, that is true for triangle inequality. Let's find $AB$ in this form: $d^2+e^2$. Then, $$\begin{gathered} (1^2+2^2)(d^2+e^2)(d^2+e^2)=((d-2e{)}^2+(2d+e{)}^2)(d^2+e^2) \\ =(4de-d^2+e^2{)}^2+(2d^2+2de-2e^2{)}^2 \end{gathered}$$ If $e=2$ and $d=3$, the triangle with the sides $AB=13$, $BC=19$ and $AC=22$ satisfy the condition. So, we've found the triangle, 2 medians of which are perpendicular and the sides are integers.