What a Nice Looking Logo

This is an approach using calculus.

Suppose the square is $ABCD$ , the center of the inscribed circle is $O$ , circle $O$ touches $AD$ at $S$ and $CD$ at $T$ , the quarter circle intersects circle $O$ at $R$ and $Q$ . Connect $BD$ , $BD$ intersects $\stackrel\frown {AC}$ at $P$ and $\stackrel\frown {ST}$ at $X$ .

Let's first calculate the area of region bounded by $\stackrel\frown {RPQ}$ and $\stackrel\frown {RSTQ}$ . For simplicity's sake, let's call this region " crescent-shaped hotdog ".

Let $XD=k$ , then using Pythagoras' theorem in $\triangle ODT$ , $(k+1)^2=1^2+1^2 \\ \therefore k=\sqrt2-1$ Since $\triangle ODT$ and $\triangle BCD$ are similar, by similarity, $PD=2XD$ . $\therefore PX=XD=k=\sqrt2-1$

Make $BD$ the $y$ -axis and the tangent of circle $O$ at $X$ the $x$ -axis. Circle $O$ is represented by the graph $x^2+(y-1)^2=1^2$ or $y=\pm\sqrt{1-x^2}+1$ , since we only consider the lower part of the circle, we take $y=-\sqrt{1-x^2}+1$

Meanwhile, the quarter circle can be represented by the lower part of the graph $x^2+[y-2-(\sqrt2-1)]^2=2^2$ or $y=-\sqrt{4-x^2}+\sqrt2+1$ .

The $x$ -coordinate of point $Q$ can be calculated by letting $-\sqrt{1-x^2}+1=-\sqrt{4-x^2}+\sqrt2+1$ and solving for $x$ which we could get $x=\sqrt{\frac{7}{8}}$ .

Thus, the area of the crescent-shaped hotdog is $2\left[\int\limits_0^{\sqrt{\frac{7}{8}}}(-\sqrt{4-x^2}+\sqrt2+1)dx-\int\limits_0^{\sqrt{\frac{7}{8}}}(-\sqrt{1-x^2}+1)dx \right]\approx 0.58552$

Finally, denote $S_\text{a shape}$ as the area of a shape, the area of the shaded region is $\begin{aligned} S_{\text{quarter circle}}-S_{\odot O}+2S_{\text{crescent-shaped hotdog}}&=\frac{\pi(2)^2}{4}-\pi+2\times0.58552 \\ &\approx1.171 \end{aligned}$

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The fig. to the right, and note are addition by Challenge Master Calvin Lin. I thank him for the improvement.

Referring to the image, we have 6 unknowns. So we will need 6 linearly independent equations. If we naively list them out what we can see, we only have 5 linearly independent equations:

• $A + C = \frac{ 4 - \pi } { 4}$
• $B = \frac{ 4 - \pi } { 4}$
• $D + E = \pi$
• $2A + B + D = \pi$
• $2C + E + F = 4 - \pi$

As such, we do not have a unique system. We have to put in additional work, which in this case is splitting $E=f-g$ .