2. In an isosceles triangle $ABC, \overline{AC}=\overline{BC}=1$. For which value of $\gamma=\measuredangle ACB$, does the expression $g=\frac{\overline{AB}^{2}+2}{P_{\triangle ABC}}$ reach its minimum value.
Solution. From the cosine theorem, we have $\overline{A B}^{2}=2-2 \cos \gamma$, and for the area of the triangle, we have

$$
P_{\triangle A B C}=\frac{1}{2} \overline{A C} \cdot \overline{B C} \cdot \sin \gamma=\frac{\sin \gamma}{2}
$$

Then we get

$$
g(\gamma)=\frac{2(4-2 \cos \gamma)}{\sin \gamma}=\frac{4(2-\cos \gamma)}{\sin \gamma}
$$

We will introduce the substitution $x=\operatorname{tg} \frac{\gamma}{2}$, for which $\sin \gamma=\frac{2 x}{1+x^{2}}, \cos \gamma=\frac{1-x^{2}}{1+x^{2}}$, and we get $g=g(x)=\frac{2\left(1+3 x^{2}\right)}{x}$. Given that $\gamma \in(0, \pi)$, i.e., $\frac{\gamma}{2} \in\left(0, \frac{\pi}{2}\right)$, it is clear that $x>0$ and consequently $g>0$.

We will determine the minimum value that the function $g(x)$, $x \in(0, \infty)$, can achieve. Let $g_{\min }=y$. We will determine for which value of $x$ this is achieved. This means we solve the equation $y=g(x)=\frac{2\left(1+3 x^{2}\right)}{x}$, or the equation $6 x^{2}-y x+2=0$. The last equation must have real roots, so the discriminant must be non-negative, i.e., $D \geq 0$. Then

$$
y^{2}-48 \geq 0 \Leftrightarrow y \leq-4 \sqrt{3} \quad \text { or } y \geq 4 \sqrt{3} \text {. }
$$

Since $g>0$, we only consider the case $y \geq 4 \sqrt{3}$. Thus, the minimum value that the function can achieve is $y=4 \sqrt{3}$, and this is achieved when $6 x^{2}-4 x \sqrt{3}+2=0$. From the last equation, we find $x=\frac{\sqrt{3}}{3}$, and the corresponding angle of the triangle is $\gamma=\frac{\pi}{3}$. Since the triangle is isosceles, we conclude that it is also equilateral.