Problem 1177

Minthogy

$$a+b: a-b=\operatorname{tg} \frac{\alpha+\beta}{2}: \operatorname{tg} \frac{\alpha-\beta}{2}$$

and

$$\frac{\alpha+\beta}{2}=90^{\circ}-\frac{\gamma}{2}, \frac{\alpha-\beta}{2}=90^{\circ}-\left(\beta+\frac{\gamma}{2}\right)$$

therefore

$$a+b: a-b=\operatorname{ctg} \frac{\gamma}{2}: \operatorname{ctg}\left(\beta+\frac{\gamma}{2}\right)$$

or

$$\begin{gathered} \frac{a+b}{\operatorname{tg}\left(\beta+\frac{\gamma}{2}\right)}=\frac{a-b}{\operatorname{tg} \frac{\gamma}{2}} \\ (a+b) \operatorname{tg} \frac{\gamma}{2}=(a-b) \operatorname{tg}\left(\beta+\frac{\gamma}{2}\right) \\ (a+b) \operatorname{tg} \frac{\gamma}{2}=(a-b) \frac{\operatorname{tg} \beta+\operatorname{tg} \frac{\gamma}{2}}{1-\operatorname{tg} \beta \operatorname{tg} \frac{\gamma}{2}} \end{gathered}$$

which equation can also be written as:

$$(a+b) \operatorname{tg} \beta \operatorname{tg}^{2} \frac{\gamma}{2}+(a-b) \operatorname{tg} \beta=(a+b) \operatorname{tg} \frac{\gamma}{2}-(a+b) \operatorname{tg} \frac{\gamma}{2}$$

from which

$$\operatorname{tg} \beta=\frac{2 b \operatorname{tg} \frac{\gamma}{2}}{(a+b) \operatorname{tg}^{2} \frac{\gamma}{2}+(a-b)}$$

(Rezső Földes, Budapest.)

The problem was also solved by: N. Ehrenfeld, K. Epstein, I. Erdélyi, A. Ertler, P. Esztó, M. Fekete, H. Fodor, E. Freund, I. Fuchs, P. Füstös, J. Gúman, P. Heimlich, E. Kirchknopf, E. Kiss, M. Lusztig, P. Sárközy, Gy. Schuster, Gy. Schwarz, I. Szécsi, O. Szilas, D. Szőke, B. Tóth, A. Vilcsek, V. Wáhl.