A triangle's inscribed circle divides one of the medians into three equal parts. In what ratio does the other two medians divide into parts?

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The inscribed circle of a triangle divides one of the medians into three equal parts. In what ratio do the other two medians divide into parts?
Let the triangle in question be $ABC$, with the median $AD$ dividing it into equal parts. We choose one-third of the length of segment $AD$ as the unit of length.

![](https://cdn.mathpix.com/cropped/2024_05_02_393fc2d31abec4711cb5g-1.jpg?height=458&width=1182&top_left_y=259&top_left_x=447)

Thus, the incircle $k$ of the triangle passes through the endpoints of the unit-length segment $GS$ located at the midpoint of segment $AD$, where $S$ is the centroid of the triangle. Let $B$ be the vertex among $B$ and $C$ that lies on the side containing the center of circle $k$ relative to line $AD$, and let $AB = x$. Furthermore, let the points of tangency on sides $BC$, $CA$, and $AB$ be denoted by $K$, $L$, and $M$, respectively. According to the theorem of secants and tangents,

$$
AM = DK = \sqrt{2}
$$

Triangle $ABD$ is isosceles, and by the cosine rule,

$$
x^2 = 9 + x^2 - 6x \cos \delta
$$

where $\delta = \angle ADB$. In triangle $ACD$, $AC = AL + LC = AM + CK$, so

$$
(x + 2\sqrt{2})^2 = 9 + x^2 + 6x \cos \delta
$$

Adding these two equations gives $x = \frac{5\sqrt{2}}{4}$, and from the first equation,

$$
x \cos \delta = \frac{3}{2}
$$

Using the cosine rule again in triangles $BSD$ and $CSD$,

$$
\begin{array}{ll}
BS^2 = 1 + x^2 - 2x \cos \delta = \frac{9}{8}, & BS = 3 \cdot \frac{\sqrt{2}}{4} \\
CS^2 = 1 + x^2 + 2x \cos \delta = \frac{57}{8}, & CS = \frac{\sqrt{114}}{4}
\end{array}
$$

Let the midpoints of sides $AB$ and $AC$ be $F$ and $E$, respectively, and the intersection points of the incircle $k$ with the medians $BE$ and $CF$ (other than $S$) be $H$ and $J$. $H$ lies on segment $BS$, and $J$ lies on segment $SF$, because the quadrilateral $ABKG$ contains the arc $KG$ of the circle that includes point $M$, and thus $SB$ and $SF$ intersect this arc.

Applying the secant and tangent theorem to secants $BS$ and $CJ$,

$$
\begin{gathered}
BH \cdot BS = BK^2 = (BD - DK)^2 = (x - \sqrt{2})^2 \\
BH = \frac{(x - \sqrt{2})^2}{BS} = \frac{\sqrt{2}}{12}
\end{gathered}
$$

and

$$
\begin{gathered}
CS \cdot CJ = CK^2 = (CD + DK)^2 = (x + \sqrt{2})^2 \\
CJ = \frac{(x + \sqrt{2})^2}{CS} = \frac{27 \sqrt{114}}{76}
\end{gathered}
$$

Thus, the desired ratios are

$$
\begin{gathered}
BH : HS : SE = BH : (BS - BH) : \frac{1}{2} BS = \\
\frac{\sqrt{2}}{12} : \left(\frac{3}{4} \sqrt{2} - \frac{\sqrt{2}}{12}\right) : \frac{3 \sqrt{2}}{8} = 2 : 16 : 9 \\
CS : SJ : JF = CS : (CJ - CS) : \left(\frac{3}{2} \cdot CS - CJ\right) = \\
\frac{\sqrt{114}}{4} : \left(\frac{27 \sqrt{114}}{76} - \frac{\sqrt{114}}{4}\right) : \left(\frac{3 \sqrt{114}}{8} - \frac{27 \sqrt{114}}{76}\right) = 38 : 16 : 3.
\end{gathered}
$$