What are the conditions for two medians of a triangle to be perpendicular to each other?
Let the medians from vertices $A$ and $B$ of triangle $ABC$ be perpendicular to each other. Thus, from their common point, the centroid $S$, the side $AB$ is seen at a right angle, meaning $S$ lies on the circumference of the Thales circle with diameter $AB$, so the distance from $S$ to the midpoint $C_1$ of side $AB = c$ is equal to half the side: $SC_1 = AB / 2$. And since $S$ trisects the median $C_1C$, it follows that $C_1C = 3AB / 2$, meaning $C$ lies on the circle centered at $C_1$ with radius $s_c = 3c / 2$. - Our reasoning can be reversed, so the condition is not only necessary but also sufficient.

![](https://cdn.mathpix.com/cropped/2024_05_02_6b5d67b1113868420ed6g-1.jpg?height=628&width=672&top_left_y=388&top_left_x=716)

The median $CC_1 = s_c$ in any triangle can be simply expressed in terms of the three sides, so the above condition can be transformed into a metric condition involving the sides of the triangle. Let the reflection of $C$ over $C_1$ be $C^*$, then in the parallelogram $CACA^*B$ the relationship between the diagonals and sides is:

$$
\begin{aligned}
CC^{*2} + AB^2 & = AC^2 + CB^2 + BC^{*2} + C^*A^2 \\
4s_c^2 + c^2 & = 2a^2 + 2b^2
\end{aligned}
$$

In our case, the condition specializes to $2s_c = 3c$:

$$
5c^2 = a^2 + b^2
$$

and this is also necessary and sufficient for the medians corresponding to sides $a$ and $b$ to be perpendicular in a triangle with sides $a$, $b$, and $c$.

Remark. From the fact that in the examined triangles $C$ lies on the circle centered at $C_1$ with radius $3c / 2 = 3 \cdot AC_1 = 3 \cdot BC_1$, it follows that the other two sides are in magnitude between $c$ and $2c$: $c < a, b < 2c$, because for the points $B'$ and $A'$ on the line $AB$ closest to $A$ and farthest from $A$ respectively, we have $B'A = AB = BA' = c$, $AA' = 2c$, and if $C$ coincided with $B'$ or $A'$, a proper triangle would not be formed.

However, we must also specify that $c$ is the shortest side of the triangle. If, for example, $a = \vartheta c$ where $0 < \vartheta \leq 1$, then from (1):

$$
b^2 = (5 - \vartheta^2)c^2 \geq 4c^2, \quad b \geq 2c
$$

and in this case, a triangle with sides $a$, $b$, and $c$ cannot be constructed because the triangle inequality is not satisfied, as $b \geq a + c$.[^0]

[^0]: The auxiliary theorem used is obtained by applying the Pythagorean theorem to the right triangles $CC^*C_0$, $ABB_0$, $ACC_0$, and $BC^*B_0$, where $B_0$ and $C_0$ are the projections of $B$ and $C$ onto the line $AC^*$.