In the isosceles triangle $ABC$, the midpoint of the base $AB$ is $F$, and the orthocenter is $M$. We know that the centroid of the triangle lies on the incircle, and that $FM = \sqrt{6}$. What are the lengths of the sides of the triangle?
Solution. The sides of the triangle are $AB=c, AC=BC=a$, the center of the inscribed circle is $O$, and its radius is $r$. The center of the circle lies on the altitude $FC$, and since the centroid lies on the circle, $FS=2r$ and $FC=m=6r$. Let's write the area of the triangle in two ways:

$$
T=\frac{cm}{2}=\frac{r(2a+c)}{2}
$$

Rearrange the equation: $c \cdot 6r = r(2a + c)$, and from this

$$
a=\frac{5}{2}c
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_eb529608178805f76b55g-1.jpg?height=854&width=379&top_left_y=587&top_left_x=862)

We haven't used the given $\sqrt{6}$ value yet. Draw perpendiculars from the center $O$ of the circle and from point $A$ to side $BC$, with feet at $P$ and $T$. The line $AT$ clearly passes through point $M$ because it is the altitude to side $BC$. From the right triangle $OPC$:

$$
CP=\sqrt{(5r)^2 - r^2} = 2r\sqrt{6}
$$

Triangle $AFM$ is similar to triangle $COP$ (they have two equal angles), so the ratios of their corresponding sides are equal:

$$
\frac{\sqrt{6}}{\frac{c}{2}} = \frac{r}{2r\sqrt{6}}
$$

From this, $c=24$, and from (1) $a=60$.