A friend is a children's librarian and she has a copy of The Miscalculations of Lightning Girl, a YA book about a girl who is struck by lightning and gains mathematical prowess. Her teacher gives her the following problem (Putnam 2000, A3):
The octagon $P_1P_2P_3P_4P_5P_6P_7P_8$ is inscribed in a circle, with the vertices around the circumference in the given order. Given that the polygon $P_1P_3P_5P_7$ is a square of area $5$ and the polygon $P_2P_4P_6P_8$ is a rectangle of area $4$, find the maximum possible area of the octagon.
Here was my solution. Start by drawing a picture:
Then the area of the octagon is the area of the square, plus the areas of the four triangles with base on the square and apex on even vertices of the octagon:
If the area of the square is $5$, its sides have length $\sqrt{5}$. Then the diagonal of the square is the diameter of the circle, which is $\sqrt{2}\cdot\sqrt{5}=\sqrt{10}$. (This is what she noticed immediately, in the book.) Let's highlight this diagonal against the rectangle:
Now time for equations and notation. Let $\alpha$ between 0 and $\pi/2$ be the arc of $P_1P_2$; in other words, $\alpha$ is the angle with vertex at the center $O$ of the circle and rays $\overrightarrow{OP_1}$, $\overrightarrow{OP_2}$. Using SOH CAH TOA, the sides of the rectangle have lengths $\sqrt{10}\cos(\alpha)$ and $\sqrt{10}\sin(\alpha)$. Since we know the area of the rectangle, we can solve for $\alpha$: $$ \left(\sqrt{10}\cos(\alpha)\right)\left(\sqrt{10}\sin(\alpha)\right)=4;\qquad 10\cos(\alpha)\sin(\alpha)=4 $$By double-angle $$ 2\cos(\alpha)\sin(\alpha) = \frac{4}{5}\qquad \sin(2\alpha) = \frac{4}{5}\qquad \alpha = \frac{1}{2}\arcsin\left(\frac{4}{5}\right) $$This tells us where the even vertices are: they have angles $\alpha$, $\pi-\alpha$, $\pi+\alpha$, $-\alpha$. Using some trig identities, we can clean this up a bit: $$ \alpha = \frac{1}{2}\arcsin\left(\frac{4}{5}\right);\qquad \alpha = \arcsin\left(\frac{1}{\sqrt{5}}\right) = \arccos\left(\frac{2}{\sqrt{5}}\right) $$Further, by symmetry, each of the four green triangles have equal area. So it suffices to find the area of $P_1P_2P_3$ and multiply that by 4. There are several ways to do this (Heron's formula, $\frac{1}{2}bh$, cross products, etc.) so we'll skip the details, but the end result is: $$ A_{\text{one green triangle}} = \frac{5}{4}\left(\cos(\alpha)+\sin(\alpha)-1\right) $$Using the cleaned-up versions of $\alpha$, we have $$ A_{\text{one green triangle}} =\frac{5}{4}\left(\frac{2}{\sqrt{5}}+\frac{1}{\sqrt{5}}-1\right) = \frac{1}{4}\left(3\sqrt{5}-5\right) $$Multiplying by $4$ gives $$ A_{\text{green triangles}} =\left(3\sqrt{5}-5\right) $$So, in summary we have $$ A_{\text{octagon}}=A_{\text{square}}+A_{\text{green triangles}} =5+ \left(3\sqrt{5}-5\right) = 3\sqrt{5} $$
I'm looking for two things: an assessment of if this proof could plausibly be given by a bright middle-school student (that is, could Lightning Girl have come up with this by herself), as well as any alternate approaches to the problem.