Quadrilaterals in a Semicircle

Figure 1. Full circle of a regular polygon Consider the full circle with the inscribed polygon of $6$ edges. For the polygon to have maximized area, it must be a cyclic regular polygon. In this case, the inscribed polygon is a regular hexagon.

Note: Another way to visualize this is to draw the chords of the same length. The alternative mathematical proof is here .

So by slicing the circle in half as shown in the diagram, we obtain the semicircle, now containing inscribed polygon of $4$ edges (because the slice creates an extra edge). Here is the new diagram: Figure 2. Sliced circle, now the semicircle Thus, since there are three equivalent isosceles triangles of $60^{\circ}$ angle and two edges of length $1$ , by the following formula, $A = \dfrac{1}{2}bc\sin(\alpha)$ we have $\begin{array}{rl} \text{maximized area of the inscribed polygon} &= 3 \cdot \text{area of one isosceles triangle}\\ &= \dfrac{3}{2}\sin\left(60^{\circ}\right)\\ &= \dfrac{3\sqrt{3}}{4}\approx \boxed{1.299038} \end{array}$

---

In case you want to take the challenges even further, see comment under my solution!

With no prior intuitive grasp of the problem, I did mine as a maximisation using this setup:

$|AC| = |CB|$ , so $\triangle APC$ , $\triangle PQC$ and $\triangle CQB$ are isosceles, since $C$ is the centre of $AB\\$

Using the sine form for the area of a triangle ( $A = \frac{1}{2}ab\sin C$ ), we come up with the function for the area of the quadrilateral:

$f(x,y) = z = \frac{1}{2}(1)(1)\sin x + \frac{1}{2}(1)(1)\sin y + \frac{1}{2}(1)(1)\sin (\pi - x - y)$

$z = \frac{1}{2}(\sin x + \sin y + \sin (\pi - x - y)\\$

Observing that $\sin (\pi - A) \equiv sin(A)$

$z = \frac{1}{2}(\sin x + \sin y + \sin (x+y))\\$

Taking partial derivatives and equating to zero for maximisation

$\frac{\delta z}{\delta x} = \cos x + \cos (x+y) = 0 \quad (1)$

$\frac{\delta z}{\delta y} = \cos y + \cos (x+y) = 0 \quad (2)\\$

Subtracting equation (2) from equation (1)

$\cos x - \cos y = 0 \\$

For $x,y \in (0,\pi)$ , we conclude that $x = y$ , since $f(x) = \cos x$ is one-to-one in the interval

Now the maximisation problem is simplified. We can now write our area function in the form $z = g(x)$

$z = \frac{1}{2}(2 \sin x + \sin 2x)\\$

Taking the derivative and maximising by equating to zero

$\frac{dz}{dx} = \cos x + \cos 2x = 0\\$

Observing the double-angle identity $\cos 2x \equiv 2 \cos ^{2}x - 1$

$\cos x + 2 \cos ^{2}x - 1 = 0$

$(2 \cos x - 1)(\cos x + 1) = 0$

$\cos x = \frac{1}{2}, -1 \Rightarrow x = \frac{\pi}{3}, \pi \\$

Since $\pi$ is not in the domain for $x$ , we conclude that $x = \frac{\pi}{3}$

Substituting back into our area function, we find that

$z = \frac{1}{2}(2 \sin \frac{\pi}{3} + \sin \frac{2\pi}{3}) = \frac{1}{2}(2(\frac{\sqrt{3}}{2}) + \frac{\sqrt{3}}{2}) = \frac{3 \sqrt{3}}{4} = \boxed{1.299...}$