4. Determine the triangle with the smallest perimeter among all those that have the inscribed circle with the same radius and the same angle value.
## SOLUTION:

Let $ABC$ be all triangles with this property; that is, the common angle with the same value is $A$ and the common inscribed radius is $r$.

Then the perimeter

$$
\begin{aligned}
& 2 p=a+b+c= \\
& r\left(\cot \frac{B}{2}+\cot \frac{C}{2}\right)+r\left(\cot \frac{C}{2}+\cot \frac{A}{2}\right)+r\left(\cot \frac{A}{2}+\cot \frac{B}{2}\right)=2 r\left(\cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2}\right)
\end{aligned}
$$

Since $A$ and $r$ are fixed, the perimeter will be minimized when the following expression is minimized:

$\cot \frac{B}{2}+\cot \frac{C}{2}=\frac{\sin((B+C) / 2)}{\sin B / 2 \sin C / 2}=\frac{\cos A / 2}{\sin B / 2 \sin C / 2}$, since $A+B+C=180^{\circ}$.

And then $\sin B / 2 \sin C / 2=\frac{1}{2}\left(\cos \frac{B-C}{2}-\cos \frac{B+C}{2}\right)=\frac{1}{2}\left(\cos \frac{B-C}{2}-\cos \frac{A}{2}\right)$ must be maximized, which leads to $\cos \frac{B-C}{2}$ being 1 and thus $B=C$.

Therefore, the triangles with the minimum perimeter, with the angle $A$ fixed and the inscribed radius fixed, are the isosceles triangles with $B=C=\frac{180^{\circ}-A}{2}$.