Maximizing the area enclosed by a curve of fixed length that passes through two fixed points: do the two arcs have equal length?

Question

A closed curve of length $L>2$ passes through two fixed points that are $1$ unit apart. Suppose the enclosed area is maximized. The isoperimetric problem implies that each of the two sections of the curve (between the two fixed points) is a circular arc.

Then I wondered, do the two arcs have equal length?

I know that if $L>\pi$, then the answer is no. This is because (according to the isoperimetric problem) the curve is a circle; and since the two fixed points are not endpoints of a diameter of the circle, the two arcs have unequal length.

My question is:

If $2<L<\pi$, do the two arcs have equal length?

I did some numerical investigation using desmos, and it seems that, unlike the first case, the answer is yes, but I do not know how to prove it. I would prefer an intuitive proof, but an algebraic proof would be welcome also.

My attempt:

Suppose the two arcs have angles $2\theta_1, 2\theta_2$, and radii $r_1,r_2$. The sum of the two arc lengths is $L$, so $2r_1 \theta_1+2r_2 \theta_2=L$. The endpoints of each arc are $1$ unit apart, so $2r_1\sin{\theta_1}=2r_2\sin{\theta_2}=1$. So we have

$$\frac{\theta_1}{\sin{\theta_1}}+\frac{\theta_2}{\sin{\theta_2}}=L \text{ where } 2<L<\pi$$

The enclosed area is

$$A=\frac{2\theta_1-\sin{2\theta_1}}{8\sin^2{\theta_1}}+\frac{2\theta_2-\sin{2\theta_2}}{8\sin^2{\theta_2}}$$

We want to prove that, when $A$ is maximized, $\theta_1=\theta_2$. But I do not know how to do so.

Answer 1

(Self-answering; after further investigation, I have found a proof.)

Consider isosceles triangle $ABC$ inscribed in a circle, where $AB=AC=1$ and the total length of minor arcs $\overset{\huge\frown}{AB}$ and $\overset{\huge\frown}{AC}$ is $L$ (so $2<L<\pi$).

Assume, for the sake of contradiction, that minor arcs $\overset{\huge\frown}{AB}$ and $\overset{\huge\frown}{AC}$ can be redrawn so that their total length is still $L$ but their individual lengths are no longer equal, and the total area of segments $AB$ and $AC$ is larger than before. This contradicts the isoperimetric problem. Therefore, the total area of segments $AB$ and $AC$ is maximized when $\overset{\huge\frown}{AB}$ and $\overset{\huge\frown}{AC}$ have the same length.

In the OP, we also have two circular segments in which the distance between vertices is $1$, and the total length of the arcs is $L$, where $2<L<\pi$. By the same principle, their total area is maximized when their individual arc lengths are equal.

(I got the main idea of this proof from here.)

Answer 2

As I understood the problem the solution found by Calculus of Variations for the Dido's problem ( max area for given enclosing arc length) is one single full circle passing through A and B valid for a single Circle.

$$\int y ~ dx - \lambda \int \sqrt{1+y^{'2}} dx $$

Lagrangian

$$y-\lambda\sqrt{1+y^{'2}}$$ Apply Euler-Lagrange equation

$$ y-\lambda \left( \sqrt{1+y^{'2}}-\frac{y^{'2}}{\sqrt{1+y^{'2}}}\right)= c $$

If $\phi$ is tangent inclination to x-axis, the ode of the displaced circle defines a single Lagrange multiplier $ \lambda$ interpreted as the Dido circle radius and an arbitrary constant circle center displacement $c$.

$$\cos \phi= \frac{y-c}{\lambda}$$

The same circle solution includes both cases.

Let $L_1$ be given fixed major arc length and $L_2$ given fixed minor arc length. We have

$$ c=0, L_1=L_2= \pi/2$$

$$ c>0, L_1>\pi/2,~ L_2< \pi/2$$

$$ c<0, L_1<\pi/2,~ L_2> \pi/2$$

There are no two radii or no two Lagrange multipliers in this optimization problem.