That's No Sun

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Section I. Beginning Step

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Since both square and unit circle have the same areas, $\begin{array}{rl} A_{\text{square}} &= A_{\text{circle}}\\ s^2 &= \pi\\ s &= \sqrt{\pi} \end{array}$ where $s$ denotes the length of the square.

Figure 1. The combined shape is symmetric.

Since the centroids of both shapes lie on the same point, we can show that (1) 4 white regions of the square extend outside of the circle, and that (2) the combined shape is symmetric with respect to the 4 lines. Splitting the square into 4 congruent squares, we can see that the diagonal length of the subsquare is $\dfrac{\sqrt{2\pi}}{2} > 1$ . In this case, since these square share the common point, the square is not fully bounded by the circle.

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Section II. Area Computation

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Figure 2. Close-up diagram

From the second diagram, observe that

• $\angle \delta\alpha\beta = \angle \varepsilon\alpha\gamma = \angle \mu\alpha\varepsilon$ . Thus, the lines intersecting the sector must have equal lengths.
• In one of the the 4 subsquares, $\angle \gamma\alpha\delta = \dfrac{\pi}{2} - 2\angle \delta\alpha\beta$

where the angle unit is in radians. The next steps are to determine and use the areas of the colored regions for the area difference.

Section II.1. Area of Concave Shape

Since $\Delta \delta\alpha\beta$ is the right triangle, Pythagorean Theorem shows that $\begin{array}{rl} |\delta\beta| &= \sqrt{|\delta\alpha|^2 - |\alpha\beta|^2}\\ &= \sqrt{1 - \dfrac{\pi}{4}}\\ &= \sqrt{\dfrac{4 - \pi}{4}}\\ &= \dfrac{1}{2}\sqrt{4 - \pi} \end{array}$ Since there are two congruent triangles in the subsquare with these common side lengths, $\begin{array}{rl} A_{\text{triangles}_1} &= 2 \cdot \dfrac{1}{2}\left(\dfrac{\sqrt{\pi}}{2}\right)\left(\dfrac{\sqrt{4 - \pi}}{2}\right)\\ &= \dfrac{1}{4}\sqrt{\pi(4 - \pi)} \end{array}$

which is equivalent to determining the area of the rectangle of two combined triangles. For the arc sector, by trigonometry, since $\begin{array}{rl} \cos\left(\angle \delta\alpha\beta\right) &= \dfrac{\sqrt{\pi}}{2}\\ \angle \delta\alpha\beta &= \arccos\left(\dfrac{\sqrt{\pi}}{2}\right) \end{array}$ and the subsquare is symmetric with respect to its diagonal, we have $\angle \gamma\alpha\delta = \dfrac{\pi}{2} - 2\angle \delta\alpha\beta = \dfrac{\pi}{2} - 2\arccos\left(\dfrac{\sqrt{\pi}}{2}\right)$ So the area of the sector is $A_{\text{sect}_1} = \dfrac{\angle \gamma\alpha\delta}{2}$ which I chose to evaluate later. Therefore, the area of the white region is $\begin{array}{rl} A_{\text{net}_1} &= A_{\text{subsquare}} - A_{\text{triangles}_1} - A_{\text{sect}_1}\\ &= \dfrac{\pi}{4} - \dfrac{1}{4}\sqrt{\pi(4-\pi)} - \dfrac{\angle \gamma\alpha\delta}{2} \end{array}$

Section II.2. Area of Convex Shape

Focus on arc sector $\mu\gamma\alpha$ . Because it also shares the common triangles with the subsquare, $A_{\text{triangles}_2} = \dfrac{1}{4}\sqrt{\pi(4 - \pi)}$ Then, since $\angle \mu\alpha\gamma = 2\angle \delta\alpha\beta$ , $A_{\text{sect}_2} = \angle \delta\alpha\beta$ Therefore, $\begin{array}{rl} A_{\text{net}_2} &= A_{\text{sect}_2} - A_{\text{triangles}_2}\\ &= \angle \delta\alpha\beta - \dfrac{1}{4}\sqrt{\pi(4 - \pi)} \end{array}$

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Section III. Final Step

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So $\begin{array}{rl} A_{\text{white}} &= 4A_{\text{net}_1} + 4A_{\text{net}_2}\\ &= 8\arccos\left(\dfrac{\sqrt{\pi}}{2}\right) - 2\sqrt{\pi(4 - \pi)} \end{array}$ Since $\arccos\left(\dfrac{\sqrt{\pi}}{2}\right) = \arctan\left( \dfrac{\sqrt{4 - \pi}}{\sqrt{\pi}}\right)$ and $2\arctan(\alpha) = \arctan\left(\dfrac{2\alpha}{1 - \alpha^2}\right)$ where $\alpha = \dfrac{\sqrt{4 - \pi}}{\sqrt{\pi}}$ , the resulting area is $\begin{array}{rl} 8\arctan\left(\dfrac{\sqrt{4 - \pi}}{\sqrt{\pi}}\right) - 2\sqrt{\pi(4 - \pi)} &= 4\left( 2\arctan\left(\dfrac{\sqrt{4 - \pi}}{\sqrt{\pi}}\right)\right) - 2\sqrt{\pi(4 - \pi)}\\ &= 4\arctan\left(\dfrac{\sqrt{\pi(4 - \pi)}}{\pi - 2}\right) - 2\sqrt{\pi(4 - \pi)} \end{array}$ where $A + B + C + D + E= 4 + 4 + 2 + 2 + 4 = \boxed{16}$