Prove that a triangle with a given base and angle must be isosceles to have maximum perimeter

Solution 1:

The locus of $B$ given $AC$ and the angle is a circular arc.

The locus of $B$ given $AC$ and the perimeter is an ellipse with $A$ and $C$ as the foci and $B$ on the edge.

If $B$ is not at the farthest point of the circle, then the ellipse crosses through the inside of the circle. This means there's another triangle $ADC$ where $D$ is on the ellipse inside the circle so it has the same perimeter as $ABC$, and there's a third triangle $AEC$ with $E$ on the circle, so it has the correct angle, and which contains $ADC$, so it's larger and has a higher perimeter, so given a $B$ that isn't the farthest point on the circle, there's another point $E$ that gives a triangle with a higher perimeter.

But if $B$ is at the farthest point of the circle, then the ellipse does not cross through the inside of the circle, so the rest of the construction does not work. This farthest point puts $B$ equidistant from $A$ and $C$, so it is an isosceles triangle.

Solution 2:

Let triangle have side $a,b,c$ and let the angle opposite to it be $A,B,C$ then by law of sines, the perimeter is given as:

$$ p = \kappa( \sin A + \sin B + \sin C)$$

We are also given that the $\kappa$ is fixed since $ \frac{b}{\sin B} = \kappa$, by am gm:

$$ \frac{\sin A + \sin C}{2} \geq \sqrt{\sin A \sin C} = \sqrt{ \frac{\cos(A-C)-\cos(A+C) }{2}} = \sqrt{\frac{\cos(A-C)+ \cos(B)}{2}}$$

Now $\cos(A-C)$ is maximized when $A=C$ , reapplying law of sine, we find the required QED

Note: $b$ is the fixed side and $B$ is the fixed angle.