Mekkora a $T$ területű és $a$ alapú egyenlőszárú háromszögbe írt egyenlőoldalú háromszög $t$ területe?

What is the area $t$ of an equilateral triangle inscribed in an isosceles triangle with base $a$ and area $T$?
Given that the equilateral triangle $A_{1} B_{1} C_{1}$ is inscribed in the isosceles triangle $A B C$, and the side $A_{1} C_{1}$ of the equilateral triangle forms an angle $\varepsilon$ with the side $B C(a)$, then

$$
\begin{gathered}
a=B A_{1}+A_{1} C=\frac{a_{1} \sin (\varepsilon+\beta)}{\sin \beta}+\frac{a_{1} \sin \left(60^{\circ}+\varepsilon-\beta\right)}{\sin \beta}= \\
=\frac{a_{1}}{\sin \beta}\left[\sin (\varepsilon+\beta)+\sin \left(60^{\circ}+\varepsilon-\beta\right)\right]= \\
=\frac{2 a_{1}}{\sin \beta} \cdot \sin \left(\varepsilon+30^{\circ}\right) \cdot \cos \left(\beta-30^{\circ}\right)= \\
=2 a_{1} \sin \left(\varepsilon+30^{\circ}\right) \frac{\cos \beta \cdot \cos 30^{\circ}+\sin \beta \cdot \sin 30^{\circ}}{\sin \beta}= \\
=a_{1} \sin \left(\varepsilon+30^{\circ}\right)(\sqrt{3} \operatorname{ctg} \beta+1)
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_6335e166aef8bbe44197g-1.jpg?height=396&width=458&top_left_y=732&top_left_x=823)

And since

$$
T=\frac{a^{2}}{4} \operatorname{tg} \beta
$$

from this equation:

$$
\operatorname{ctg} \beta=\frac{a^{2}}{4 T}
$$

therefore:

$$
a_{1}=\frac{4 a T}{\left(a^{2} \sqrt{3}+4 T\right) \cdot \sin \left(\varepsilon+30^{\circ}\right)}
$$

and thus:

$$
t=\frac{a_{1}^{2}}{4} \sqrt{3}=\left[\frac{2 a T}{\left(a^{2} \sqrt{3}+4 T\right) \sin \left(\varepsilon+30^{\circ}\right)}\right]^{2} \cdot \sqrt{3}
$$

If $\varepsilon=60^{\circ}$, then the minimum value of $t$ is

$$
t=\left[\frac{2 a T}{4 T+a^{2} \sqrt{3}}\right]^{2} \cdot \sqrt{3}
$$

(István Pivnyik, Nyíregyháza.)

The other solvers only solved the problem for the special case where $\varepsilon=60^{\circ}$.