Dual Line Segment Area

Suppose we find angles ${\theta}_{1}$ and ${\theta}_{2}$ such that the area bounded by the perpendicular sides have the maximum area. Then we reflect about the x-axis and the line $x = \cos \theta_1 + \cos \theta_1$ to form an 8-sided polygon.

What 8-sided polygon of all equal edge length has the maximum area? The regular octagon, which, for an edge length of $1$ , has an area of $2(1+\sqrt{2})$ , and so from there we have $1.207...$ as the answer for this problem.

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Let's prove that of all 8-sided polygons of side length 1, the regular octagon maximizes the area.

Proof: Label the vertices $V_1, V_2, \ldots , V_8$ in order.
If the polygon is not convex, we can easily "stretch" it out to get a larger area. Hence, we may assume that the polygon is convex.
Consider the line $V_1 V_5$ . If the areas $V_1V_2V_3V_4V_5$ and $V_5V_6V_7V_8V_1$ are not the same, then we can just reflect one across the other to get a larger area. Hence, we may assume that the areas are the same (though we do not know if they are symmetric).
Let's focus on the $V_1V_2V_3V_4V_5$ half. For any vertex $V_2, V_3, V_4$ , if the triangle $V_1 V_i V_5$ is not a right angle at $V_i$ , then we can expand the area by making $V_1 V_i V_5$ a right triangle, while keeping everything else the same. This shows that $\angle V_ 1 V_i V_5 = 90 ^ \circ$ , and thus all of these points lie on the same (semi)circle.

Another proof: Pick any three consecutive vertices, e.g. $V_3, V_4, V_5$ , and one more elsewhere, e.g. $V_7$ . While keeping other vertices fixed relative to diagonals $V_3 V_7$ and $V_5 V_7$ , we can adjust the four vertices $V_3, V_4, V_5\, V_7$ until they form a cyclic quadrilateral with maximum area. Thus, unless any four vertices of a n-gon chosen in this manner form cyclic quadrilaterals, it will always be possible to increase the area of the n-gon with fixed side lengths. Since any three consecutive vertices define an unique circumcircle, and because the choice of the $4$ th vertex is arbitrary, all of the vertices must lie on the same common circumcircle.

Oddly enough, though, it takes more work to prove that a quadrilateral with fixed side lengths has maximum area only if it's cyclic. I'll get back to this later if there's a non-calculus proof.

Edit: See Bretschneider's formula for a non-calculus proof. The area is maximum when the quadrilateral is cyclic. However, the derivation is complicated.

The area in question can be expressed as the sum of areas of two right triangles and a rectangle.

We know that the area of a right triangle is $\dfrac12 \times \text{ base } \times \text{ height}$ , and the area of the rectangle is the product of its width and length.

Let $S_{\theta_1 \theta_2}$ denote the area in question with parameters $0 \leq \theta_1 , \theta_2 \leq \frac{\pi}{2}$ . Then,

$\begin{aligned} S_{\theta_1 , \theta_2} &=& \left ( \dfrac12 \times \sin \theta_1 \cos \theta_1 \right) + \left ( \dfrac12 \times \sin \theta_2 \cos \theta_2 \right) +( \times \sin \theta_1 \cos \theta_2 ) \\ &=& \dfrac14 \left [ \sin(2\theta_1 ) + \sin(2\theta_2) \right ] + \sin\theta_1 \cos\theta_2 \end{aligned}$

Since the area function $S_{\theta_1 ,\theta_2}$ is a continuous function on a compact space, then the global maximum of this area function is either achieved at its turning point in the interior or else on the boundary.

The boundary of $S_{\theta_1, \theta_2}$ is either $S_{0,\pi/2} = 0$ or $S_{\pi/2,0} = 1$ .

Now let's find the turning point of this area function!

We have \dfrac{\partial S}{\partial \theta_1} = \dfrac12 \cos(2\theta_1) + \dfrac12 \cos\theta_1 \cos \theta_2 = 0 \qquad \qquad \tag1 and
\dfrac{\partial S}{\partial \theta_2} = \dfrac12 \cos(2\theta_2) - \dfrac12 \sin\theta_1 \sin \theta_2 = 0\qquad \qquad \tag2

Adding these two equations gives $\dfrac12 [ \cos(2\theta_1) + \cos(2\theta_2) + \cos(\theta_1 + \theta_2) ] = 0$ .

Using the compound angle formula , $\cos(A\pm B) = \cos A \cos B \mp \sin A \sin B$ , the equation above can be simplified to 2\cos(\theta_1 + \theta_2)\cos(\theta_1 - \theta_2) + \cos(\theta_1 + \theta_2) = 0 \quad \Leftrightarrow \quad \cos(\theta_1 + \theta_2) \, [ 2 \cos(\theta_1 - \theta_2) + 1 ] = 0.

By zero product property , we have $\cos(\theta_1 + \theta_2) = 0$ and/or $\cos(\theta_1 - \theta_2) = -\dfrac12$ under the restriction $0 \leq \theta_1 + \theta_2 \leq \pi$ and $| \theta_1 - \theta_2 | \leq \dfrac\pi 2$ . But $\cos(\theta_1 - \theta_2)= -\dfrac12$ cannot hold true. Thus, $\cos(\theta_1 + \theta_2) = 0 \Rightarrow \theta_2 = \dfrac\pi2 - \theta_1$ only. Substituting this back into our area function gives

$S_{\theta_1 , \theta_2} = S_{\theta_1} = \dfrac12 \sin(2\theta_1 ) + \sin^2\theta_1 = \sin\theta_1 \cos\theta_1 + \sin^2\theta_1 .$

A simple differentiation shows that $S' = 0 \Rightarrow \theta_1 = \dfrac{3\pi}8 , \theta_2 = \dfrac{\pi}8 \Rightarrow S_{3\pi/8, \pi/8} = \dfrac12 (\sqrt2 + 1)$ .

Our answer is $\max( S_{0,\pi/2} , S_{\pi/2,0} , S_{3\pi/8, \pi/8} ) = \dfrac12 (\sqrt2 + 1) \approx \boxed{1.207}$ .

Here's a brute calculus approach. See the solution by @Michael Mendrin for a nice geometrical interpretation of the problem. There are essentially two parts to the brute calculus approach:

1) Take partial derivatives of the area expression with respect to $\theta_1$ and $\theta_2$ and set them to zero (this is standard).
2) Solve the resulting nonlinear system using multivariate Newton Raphson iteration

The derivation is given below. The iterative algorithm converges on $\theta_1 = 67.5 \, deg$ and $\theta_2 = 22.5 \, deg$ . The resulting area is approximately 1.207. This result can also be verified again using brute search, with nested loops incrementing $\theta_1$ and $\theta_2$ .