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

Question

I can't solve the following problem:

Prove that triangle $ABC$ with known side $AC$ and angle $B$ has maximum perimeter when $AB=BC$.

Purely geometric proof is preferred, calculus is not allowed.

I started by assuming the triangle is circumscribed by a circle uniquely determined by $AC$ and $\angle B$, and showing that triangle height perpendicular to $AC$ is maximized when it divides $AC$ in half. Which means that area is maximized when $AB=BC$. Which leads to $AB\times BC \gt AM\times MC$, where M is some other point on the circle. And there I'm stuck, not sure how to continue further.

Answer 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.

Answer 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.

Answer 3

We know $b$ and $B$, and have $$b^2=a^2+c^2-2ac\cos B=\frac{1-\cos B}{2}(a+c)^2+\frac{1+\cos B}{2}(a-c)^2$$ by the cosine rule. Since $0<B<\pi$, we have $-1<\cos B<1$, and in particular $$\frac{1-\cos B}{2},\frac{1+\cos B}{2}>0.$$ We want to maximise $a+b+c$; since $b$ is known and everything is positive, this is the same as maximising $\frac{1-\cos B}{2}(a+c)^2$, i.e. the same as minimising $b^2-\frac{1-\cos B}{2}(a+c)^2=\frac{1+\cos B}{2}(a-c)^2$, and this clearly happens (only) when $a=c$.

Answer 4

Law of cosine. If the angle is $\beta$ the length is $b,$ then $$b^2=a^2+c^2-2ac\cos \theta\tag1$$ and the perimeter $$p=a+b+c$$

You want $$0=\frac{dp}{da}=1+\frac{dc}{da}$$ or $$\frac{dc}{da}=-1$$

Differentiating (1), $$\begin{align}0&=2a+2c\frac{dc}{da}-2c\cos\beta-2a\frac{dc}{da}\cos \beta\\&=(a-c)\left(2-2\cos\beta\right) \end{align}$$

So $a=c$ is the only possible extremum for the perimeter.

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Now take the second derivative of (1):

$$\begin{align}0&=2+2\left(\frac{dc}{da}\right)^2+2c\frac{d^2c}{da^2}-4\frac{dc}{da}\cos\beta-2a\frac{d^2c}{da^2}\cos \beta \end{align}$$

When $a=c$ and $\frac{dc}{da}=-1,$ we get, when dividing by $2,$

$$\begin{align}0&=2(1+\cos\beta)+ c(-\cos \beta)\frac{d^2c}{da^2} \end{align}$$

When $0<\beta<\pi,$ this means $$0> \frac{d^2c}{da^2}= \frac{d^2p}{da^2}$$

So $p$ is maximized when $a=c.$

Answer 5

The locus of point B is a circular arc with endpoints $A, C$; radius $R=AC/(2\sin\angle B)$ and angular measure $\Phi=360°-2×\angle B$. Within this arc we have components $AB$ measuring $\theta$ and $BC$ measuring $\Phi-\theta$.

Then $AB=2R\sin[\theta/2]$ and $BC=2R\sin[(\Phi-\theta)/2]$. Adding these together and applying the trigonometric sum-product relations gives

$AB+BC=2R[\sin[\theta/2]+\sin[(\Phi-\theta)/2]=4R\sin[\Phi/4]\color{blue}{\cos[\Phi/4-\theta/2]}$

Then the blue function contains all the variation with the location of $B$, and we see this is maximized at unity when $\theta=\Phi/2=\Phi-\theta$. Thus the maximizing point occurs when $B$ splits the locus in half and $AB=BC=2R\sin[\Phi/2]$.