Construct a triangle given two sides such that the medians of these sides are perpendicular to each other.

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Let the desired triangle be constructed. $A M=M C, B K=K C, A K M B$. From the right triangles $A M O$ and $OKB$, we write: $O M^{2}+O A^{2}=A M^{2}, O K^{2}+O B^{2}=K B^{2}, A C=2 b, A M=b, B C=2 a, K B=a, A K=3 x, A O=2 x, M B=3 y, O B=2 y$, $O M=y, O K=x$ (see figure). Substituting the introduced values into the given equations

$$
\left\{\begin{array}{l}
y^{2}+(2 x)^{2}=b^{2} \\
x^{2}+(2 y)^{2}=a^{2}
\end{array}\right.
$$

Solving this system, we get $x=\sqrt{4 b^{2}-a^{2}} / \sqrt{15}, y=\sqrt{4 a^{2}-b^{2}} / \sqrt{15}$. It is sufficient to construct one of these segments to construct the triangle given two sides and the median. The segments $\sqrt{4 b^{2}-a^{2}}$ and $\sqrt{15} a$ can be constructed using the Pythagorean theorem. Now, $x=\left(\sqrt{4 a^{2}-b^{2}} \cdot a\right) / \sqrt{15} a$, it is clear that it can be constructed.

![](https://cdn.mathpix.com/cropped/2024_05_06_55a07599adfd5314dc75g-23.jpg?height=654&width=529&top_left_y=885&top_left_x=776)

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