\begin{aligned} & \text { Pythagorean Theorem (direct and inverse). } \\ & \text { [ Properties of medians. Center of gravity of a triangle. }]\end{aligned}

In triangle $ABC$, the medians drawn to sides $AC$ and $BC$ intersect at a right angle. It is known that $AC=b$ and $BC=a$. Find $AB$.
The medians of a triangle are divided by the point of intersection in the ratio 2:1, counting from the vertex.

## Solution

Let $M$ and $N$ be the midpoints of sides $AC$ and $BC$ of triangle $ABC$, and $O$ be the point of intersection of medians $BM$ and $AN$.

Denote $OM=x, ON=y$. Then $BO=2OM=2x, AO=2ON=2y$.

From the right triangles $AOM$ and $BON$, we find that

$$
AM^2 = AO^2 + OM^2, \quad BN^2 = BO^2 + ON^2
$$

or

$$
\left\{\begin{array}{l}
4y^2 + x^2 = \frac{a^2}{4} \\
4x^2 + y^2 = \frac{b^2}{4}
\end{array}\right.
$$

Adding these equations term by term, we get

$$
5(x^2 + y^2) = \frac{a^2 + b^2}{4}
$$

Therefore, $x^2 + y^2 = \frac{a^2 + b^2}{20}$.

From the right triangle $AOB$, we find:

$$
AB^2 = AO^2 + BO^2 = 4y^2 + 4x^2 = 4(y^2 + x^2) = \frac{a^2 + b^2}{5}
$$

Let $M$ and $N$ be the midpoints of sides $AC$ and $BC$ of triangle $ABC$, and $O$ be the point of intersection of medians $BM$ and $AN$. Denote $OM=x, ON=y$. Then $BO=2OM=2x, AO=2ON=2y$.

From the right triangles $AOM$ and $BON$, we find that

$$
AM^2 = AO^2 + OM^2, \quad BN^2 = BO^2 + ON^2
$$

or

$$
\left\{\begin{array}{l}
4y^2 + x^2 = \frac{a^2}{4} \\
4x^2 + y^2 = \frac{b^2}{4}
\end{array}\right.
$$

Adding these equations term by term, we get

$$
5(x^2 + y^2) = \frac{a^2 + b^2}{4}
$$

Therefore, $x^2 + y^2 = \frac{a^2 + b^2}{20}$.

From the right triangle $AOB$, we find:

$$
AB^2 = AO^2 + BO^2 = 4y^2 + 4x^2 = 4(y^2 + x^2) = \frac{a^2 + b^2}{5}
$$

Let $M$ and $N$ be the midpoints of sides $AC$ and $BC$ of triangle $ABC$, and $O$ be the point of intersection of medians $BM$ and $AN$. Denote $OM=x, ON=y$. Then $BO=2OM=2x, AO=2ON=2y$.

From the right triangles $AOM$ and $BON$, we find that

$$
AM^2 = AO^2 + OM^2, \quad BN^2 = BO^2 + ON^2
$$

or

$$
\left\{\begin{array}{l}
4y^2 + x^2 = \frac{a^2}{4} \\
4x^2 + y^2 = \frac{b^2}{4}
\end{array}\right.
$$

Adding these equations term by term, we get

$$
5(x^2 + y^2) = \frac{a^2 + b^2}{4}
$$

Therefore, $x^2 + y^2 = \frac{a^2 + b^2}{20}$.

From the right triangle $AOB$, we find:

$$
AB^2 = AO^2 + BO^2 = 4y^2 + 4x^2 = 4(y^2 + x^2) = \frac{a^2 + b^2}{5}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_b9b3326e3456e69a696eg-23.jpg?height=389&width=446&top_left_y=0&top_left_x=809)

## Answer

$\sqrt{\frac{a^2 + b^2}{5}}$