In triangle $ABC$, we draw the circle that is tangent to the sides. What are the lengths of the sides and the angles of the triangle if the sides of the triangle determined by the points of tangency are: $25 \mathrm{~cm}, 29 \mathrm{~cm}, 36 \mathrm{~cm}$?
Let $A B C$ be a triangle with sides $a, b$, and $c$, and let the points of tangency of the inscribed circle be $E, D$, and $F$. Furthermore, let the radius of the inscribed circle be denoted by $r$, and the sides of the triangle $D E F$ by $a_{1}$, $b_{1}$, and $c_{1}$. Then, as is known,

$$
r=\frac{a_{1} b_{1} c_{1}}{4 \sqrt{s\left(s-a_{1}\right)\left(s-b_{1}\right)\left(s-c_{1}\right)}}=\frac{145}{8}
$$

Furthermore, from the triangle $D O F$,

$$
\sin \frac{180^{\circ}-a}{2}=\cos \frac{\alpha}{2}=\frac{a_{1}}{2 r}=\frac{100}{145}
$$

Similarly,

$$
\cos \frac{\beta}{2}=\frac{b_{1}}{2 r}=\frac{116}{145}
$$

and

$$
\cos \frac{\gamma}{2}=\frac{c_{1}}{2 r}=\frac{144}{145}
$$

Performing the calculations,

$$
\alpha=92^{\circ} 47^{\prime} 40^{\prime \prime}, \beta=73^{\circ} 44^{\prime} 23^{\prime \prime}, \gamma=13^{\circ} 27^{\prime} 57^{\prime \prime}
$$

(Mathematical Circle of Solymár.)

The sides are calculated as follows:

$$
\begin{gathered}
a=r\left(\operatorname{ctg} \frac{\beta}{2}+\operatorname{ctg} \frac{\gamma}{2}\right)=r \frac{\cos \frac{\alpha}{2}}{\sin \frac{\beta}{2} \sin \frac{\gamma}{2}} \\
b=r \frac{\cos \frac{\beta}{2}}{\sin \frac{\alpha}{2} \sin \frac{\gamma}{2}}, \quad c=r \frac{\cos \frac{\gamma}{2}}{\sin \frac{\alpha}{2} \sin \frac{\beta}{2}}
\end{gathered}
$$

Performing the calculations,

$$
a=177.7 \mathrm{~cm}, \quad b=170.8 \mathrm{~cm}, \quad c=41.4 \mathrm{~cm}
$$

(József Jánosy, Esztergom.)