Let's construct the triangle if $a=5 \mathrm{~cm}, \alpha=72^{\circ}$, and we know that the following relationship holds between its sides: $\left(a^{2}-c^{2}\right)^{2}=b^{2}\left(2 c^{2}-b^{2}\right)$
The given relationship can be transformed as follows:

$$
\begin{gathered}
a^{4}+b^{4}+c^{4}-2 a^{2} c^{2}-2 b^{2} c^{2}=0 \\
\left(a^{2}+b^{2}-c^{2}\right)^{2}=2 a^{2} b^{2} \\
\left(\frac{a^{2}+b^{2}-c^{2}}{2 a b}\right)^{2}=\cos ^{2} \gamma=\frac{1}{2}, \quad \cos \gamma= \pm \frac{1}{\sqrt{2}}
\end{gathered}
$$

From this, $\gamma_{1}=45^{\circ}, \gamma_{2}=135^{\circ}$. However, with $\gamma_{2}$, $\alpha+\gamma_{2}>180^{\circ}$, which is not a solution.

The construction of the angle $\alpha=72^{\circ}$ was seen in problem 1213. We obtain $\gamma$ by bisecting the right angle, thus reducing the construction of the triangle to the basic case of "1 side and 2 angles."

Éva Vincze (Budapest, Kossuth Zsuzsa g. IV. o. t.)[^0]


[^0]:    ${ }^{1}$ K. M. L. $27(1963 / 9) 18$.