A circle inscribed in a triangle with a radius of $3 \sqrt{2}$ passes through the points that trisect one of the triangle's medians. What are the lengths of the triangle's sides?
Let's use the notations in the figure. Since points $Q$ and $P$ divide segment $CF$ into three equal parts, the power of points $C$ and $F$ with respect to the inscribed circle is equal. Therefore, the tangent segments $CM$ and $FN$ are equal, denoted as $x$ in our figure. With $AN = AM = y$, we have $AF = FB = x + y$, and thus, by the equality of the tangent segments from $B$, $BL = 2x + y$. With $CQ = QP = PF = z$, by the power of a point theorem applied to the tangent and secant segments from $C$, we have $x^2 = z \cdot 2z$, from which $z = \frac{x}{\sqrt{2}}$ and $CF = \frac{3x}{\sqrt{2}}$.

We will solve the problem by using the cosine rule and area formulas to calculate $x$ and $y$. Applying the cosine rule to triangles $AFC$ and $BFC$:

\[
\begin{gathered}
(x+y)^{2}=\frac{9 x^{2}}{2}+(x+y)^{2}-2 \cdot \frac{3 x}{\sqrt{2}}(x+y) \cos \alpha \\
(3 x+y)^{2}=\frac{9 x^{2}}{2}+(x+y)^{2}-2 \cdot \frac{3 x}{\sqrt{2}}(x+y) \cos \left(180^{\circ}-\alpha\right)
\end{gathered}
\]

Adding the two equations:

\[
(x+y)^{2}+(3 x+y)^{2}=9 x^{2}+2(x+y)^{2}, \quad \text { from which } \quad x^{2}=4 x y
\]

and since $x \neq 0, x=4 y$. Then the sides of the triangle are $AC=5 y, BC=13 y, AB=10 y$, and the semiperimeter $s=14 y$. Writing the area of triangle $ABC$ in two ways:

\[
3 \cdot \sqrt{2} \cdot 14 y=\sqrt{14 y \cdot y \cdot 9 y \cdot 4 y}, \quad \text { from which } \quad y=\sqrt{7}
\]

The sides are $AC=5 \sqrt{7}, BC=13 \sqrt{7}, AB=10 \sqrt{7}$.

Based on the work of Ildikó Bujdosó (Bp., Veres Péter Gimn., IV. o.t.) and Margit Nagy (Vörösmarty M. Gimn., III. o.t.)

Note. From the provided solution, it only follows that if there exists a triangle satisfying the problem's requirements, then its sides can only be $5 \sqrt{7}, 13 \sqrt{7}, 10 \sqrt{7}$. However, a simple calculation shows that the incircle of this triangle indeed divides the median $CF$ into three equal parts.

![](https://cdn.mathpix.com/cropped/2024_05_02_a016b57743695d354ff4g-1.jpg?height=450&width=1000&top_left_y=1363&top_left_x=557)