[ Extreme properties. Problems on maximum and minimum.]

Among all triangles $ABC$ with a given angle $C$ and side $AB$, find the triangle with the maximum possible perimeter.
Construct point $A_{1}$, symmetric to vertex $A$ with respect to the bisector of the external angle $C$ of triangle $A B C$.

## Solution

Let $A_{1}$ be the point symmetric to vertex $A$ with respect to the bisector of the external angle $C$. Then

$$
B A_{1}=A_{1} C+B C=A C+B C
$$

Since $A C B$ is the external angle of the isosceles triangle $A C A_{1}$, we have $\angle A A_{1} B=\frac{1}{2} \angle A C B$. Therefore, the geometric locus of points $A_{1}$ is a fixed arc of a circle passing through points $A$ and $B$, located on the same side of the line $A B$ as point $C$, and the angular measure of this arc is equal to the angular measure of the given angle $A C B$.

The segment $B A_{1}$ is maximal if $B A_{1}$ is the diameter of the circle. Then point $C$ is the center of this circle.

Thus, $C A=C B$.

![](https://cdn.mathpix.com/cropped/2024_05_06_e1837d72990d2467a29dg-15.jpg?height=443&width=481&top_left_y=585&top_left_x=797)

## Answer

Isosceles triangle.