12.64. Let the median $B M$ of triangle $A B C$ intersect the inscribed circle at points $K$ and $L$, such that $B K=K L=L M=x$. For definiteness, let the point of tangency of the inscribed circle with side $A C$ lie on segment $M C$. Then, since under symmetry with respect to the perpendicular bisector of segment $B M$, points $B$ and $M$ swap places, and the inscribed circle maps to itself, the tangent $M C$ maps to the tangent $B C$. Therefore, $B C=M C=A C / 2$, i.e., $b=2 a$.

Since $B M^{2}=\left(2 a^{2}+2 c^{2}-b^{2}\right) / 4$ (see problem 12.11, a)), then $9 x^{2}=\left(2 a^{2}+2 c^{2}-4 a^{2}\right) / 4=\left(c^{2}-a^{2}\right) / 2$. Let point $P$ be the point of tangency of the inscribed circle with side $B C$. Then $B P=(a+c-b) / 2=(c-a) / 2$. On the other hand, by the property of tangents, $B P^{2}=B K \cdot B L$, i.e., $B P^{2}=2 x^{2}$. Therefore, $2 x^{2}=((c-a) / 2)^{2}$. Multiplying the equations $9 x^{2}=\left(c^{2}-a^{2}\right) / 2$ and $((c-a) / 2)^{2}=$ $=2 x^{2}$, we get $(c+a) /(c-a)=9 / 4$, i.e., $c: a=13: 5$. In the end, we obtain that $a: b: c=5: 10: 13$.
12.64*. Find the ratio of the sides of a triangle, one of whose medians is divided into three equal parts by the inscribed circle.