I guess I should have tried to construct an example explicitly.

Now I'm wondering why Mathematica failed to find a solution. Maybe I typed in the formula incorrectly.

Thanks.

Regarding Mathematica.

Heron's formula:

\(\displaystyle A=\sqrt{\frac{(a+b+c)(a+b-c)(b+c-a)(c+a-b)}{16}}\)

Let \(\displaystyle a=b\). Then

\(\displaystyle A=\sqrt{\frac{(2a+c)(2a-c)(c)(c)}{16}}\)

\(\displaystyle =\sqrt{\frac{((2a)^2-c^2)(c^2)}{16}}\)

\(\displaystyle 16A^2=(4a^2-c^2)(c^2)\)

\(\displaystyle 16A^2-(4a^2-c^2)(c^2)=0\)

Screenshot from Mathematica:

Edit: Of course it's also possible to avoid Heron's formula by dividing the isosceles triangle into two right triangles to begin with.

Label the two congruent sides \(\displaystyle a\) and the other side \(\displaystyle b\). Treat \(\displaystyle b\) as the base and draw an altitude from the base to the opposite vertex.

\(\displaystyle \left(\frac{b}{2}\right)^2+h^2=a^2\)

and

\(\displaystyle A=\left(\frac{1}{2}\right)bh\)

\(\displaystyle h = \frac{2A}{b}\)

Substitute

\(\displaystyle \left(\frac{b}{2}\right)^2+\left(\frac{2A}{b}\right)^2=a^2\)

etc.