$$
\begin{aligned}
\operatorname{tg} \frac{\gamma}{2} & =\operatorname{tg}\left(90^{\circ}-\frac{\alpha+\beta}{2}\right)=\operatorname{ctg}\left(\frac{\alpha+\beta}{2}\right)= \\
& =\frac{1}{\operatorname{tg}\left(\frac{\alpha}{2}+\frac{\beta}{2}\right)}=\frac{1-\operatorname{tg} \frac{\alpha}{2} \operatorname{tg} \frac{\beta}{2}}{\operatorname{tg} \frac{\alpha}{2}+\operatorname{tg} \frac{\beta}{2}}
\end{aligned}
$$

Substituting the given values for $\operatorname{tg} \frac{\alpha}{2}$ and $\operatorname{tg} \frac{\beta}{2}$, we get

$$
\operatorname{tg} \frac{\gamma}{2}=\frac{2}{5}
$$

If the radius of the inscribed circle of the triangle is $r$, then

$$
a=\left(\frac{r}{\operatorname{tg} \frac{\beta}{2}}+\frac{r}{\operatorname{tg} \frac{\gamma}{2}}\right)=\frac{87}{20} r
$$

Similarly,

$$
b=\frac{37}{10} r \text { and } c=\frac{61}{20} r
$$

Thus,

$$
a+c=\frac{74}{10} r=2 \cdot \frac{37}{10} r=2 b
$$

(Julius Póka)

The problem was also solved by: F. Aczél, E. Baranyó, I. Bartók, B. Bayer, G. Bogdán, K. Buxbaum, B. Enyedi, A. Goldstein, G. Hirschfeld, L. Izsáky, F. Kertész, E. Kiss, D. König, L. Lázár, G. Messik, P. Mixich, P. Pilczer, I. Radó, J. Sasvári, A. Schlesinger, I. Schmidl, K. Sólymos, D. Steiner, Gy. Sümegi, H. Szmodics, L. Tóbiás, S. Wohlstein.
If the angles of a triangle are $\alpha, \beta$, and $\gamma$; furthermore,

$$
\operatorname{tg} \frac{\alpha}{2}=\frac{5}{6} \quad \text { and } \quad \operatorname{tg} \frac{\beta}{2}=\frac{30}{27}
$$

then

$$
a+c=2 b
$$

What is $\operatorname{tg} \frac{\gamma}{2}$?