Let's determine the base $a$ of the isosceles triangle if its area is $t$ and the angle at the vertex is $\alpha$.

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I. solution: The area formulas provide a relationship between $t$, $\alpha$, and $a$. For example, starting from the formula

$$
t=\frac{a m_{a}}{2}
$$

and substituting $m_{a}=\frac{a}{2 \operatorname{tg} \frac{\alpha}{2}}$,

$$
t=\frac{a^{2}}{4 \operatorname{tg} \frac{\alpha}{2}}
$$

from which

$$
a=2 \sqrt{t \cdot \operatorname{tg} \frac{\alpha}{2}}
$$

László Kárpáti (Kecskemét, Piarist g. III. o. t.)

II. solution: Denoting the leg of the isosceles triangle by $b$ and using the formula

$$
t=\frac{1}{2} b^{2} \sin \alpha
$$

after substituting $b=\frac{a}{2 \sin \frac{\alpha}{2}}$,

$$
t=\frac{1}{2} \cdot \frac{a^{2} \sin \alpha}{4 \sin ^{2} \frac{\alpha}{2}}=\frac{4^{2} \cdot 2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}{8 \sin ^{2} \frac{\alpha}{2}}=\frac{a^{2} \cos \frac{\alpha}{2}}{4 \sin \frac{\alpha}{2}}=\frac{a^{2}}{4 \operatorname{tg} \frac{\alpha}{2}}
$$

which agrees with the result obtained in the first solution.

András Bagi (Bp. IX., Apáczai Csere János g. III. o. t.)

III. solution: We can also start from the formula

$$
t=\frac{a^{2} \sin \beta \sin \gamma}{2 \sin \alpha}
$$

In this case, $\beta=\gamma=90-\frac{\alpha}{2}$, so

$$
t=\frac{a^{2} \cos ^{2} \frac{\alpha}{2}}{4 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}=\frac{a^{2}}{4 \operatorname{tg} \frac{\alpha}{2}}
$$

Emil Klafszky (Győr, Révai g. IV. o. t.)

IV. solution: Even Heron's formula leads to the goal:

$$
1=\sqrt{\frac{2 b+a}{2} \cdot \frac{2 b-a}{2} \cdot \frac{a^{2}}{4}}=\frac{a}{4} \sqrt{4 b^{2}-a^{2}}=\frac{a}{4} \sqrt{\frac{4 a^{2}}{4 \sin ^{2} \frac{\alpha}{2}}-a^{2}}=\frac{a^{2}}{4} \sqrt{\frac{1-\sin ^{2} \frac{\alpha}{2}}{\sin ^{2} \frac{\alpha}{2}}}=\frac{a^{2}}{4} \cdot \frac{\cos \frac{\alpha}{2}}{\sin \frac{\alpha}{2}}=\frac{a^{2}}{4 \operatorname{tg} \frac{\alpha}{2}}
$$

Zoltán Papp (Sárospatak, Rákóczi g. III. o. t.)