Pi Day Square and Rectangle

Let the red area be $A_{\color{#D61F06}\text{red}}$ . Then the blue area $A_{\color{#3D99F6}\text{blue}} = 2\left(\pi - A_{\color{#D61F06}\text{red}}\right)$ and

$\begin{aligned} 2\left(\pi - A_{\color{#D61F06}\text{red}}\right) & = 6 A_{\color{#D61F06}\text{red}} \\ \pi - A_{\color{#D61F06}\text{red}} & = 3 A_{\color{#D61F06}\text{red}} \\ \implies A_{\color{#D61F06}\text{red}} & = \frac \pi 4 \end{aligned}$

Since the area of the square $A_\square = \pi$ , the side length of the square is $\sqrt \pi$ , which is also the long side length of the red rectangle. Therefore, the short side length of the red rectangle is $\dfrac {\frac \pi 4}{\sqrt \pi} = \dfrac {\sqrt \pi}4$ , which is also the short side length of the larger rectangle. And the long side length of the rectangle is $\dfrac \pi {\frac {\sqrt \pi}4} = 4 \sqrt \pi$ $\implies n = \boxed 4$ .

Let $B$ be the combined area of the blue sections and $R$ be the area of the red section. Since both rectangle and square have an area of $\pi$ , $B + 2R = 2\pi$ , and since the area of the blue sections are $6$ times the area of the red sections, $B = 6R$ . These two equations solve to $R = \frac{1}{4}\pi$ .

Since the square has an area of $\pi$ , its sides are $\sqrt{\pi}$ , so the red section has this side and a side equal to $\frac{R}{\sqrt{\pi}} = \frac{\frac{1}{4}\pi }{\sqrt{\pi}} = \frac{1}{4}\sqrt{\pi}$ , and the rectangle with an area of $\pi$ has this side and a side equal to $\frac{\pi}{\frac{1}{4}\sqrt{\pi}} = 4\sqrt{\pi}$ . Therefore, $n = \boxed{4}$ .

Let $a, b$ be the longer and shorter lengths of the rectangle, $c$ be the side length of the square, and $B$ and $R$ be the total area of the blue and red rectangles, respectively.
Let $n$ be such that $a = n\sqrt{\pi}$ . Then since $a*b = \pi$ , $b = \frac{\sqrt{\pi}}{n}$ . Since the square has area $\pi$ , $c = \sqrt{\pi}$ .

Notice that $R = b*c$ , i.e., $R = \frac{\sqrt{\pi}}{n} \times \sqrt{\pi} = \frac{\pi}{n}.$

We can use inclusion-exclusion to get the sum of the areas of the blue rectangles: it is equal to the original rectangle plus the area of the square (each of which is $\pi$ ) minus twice the area of the red rectangle, i.e.

$B = 2*\pi - 2*\frac{\pi}{n} = \frac{2(n-1)\pi}{n}$

Since $B = 6R$ , we have $\frac{2(n-1)\pi}{n} = \frac{6 \pi}{n}$ , which simplifies to $2*(n-1) = 6$ , i.e. $n=4$ .

Let the side of the square is c, and the rectangle are a, and b (the longer side). ab=c^2= π Then a(b – c) + c(c – a) = 6 ac. or ab + c^2 – 8 ac = 0, a=(1/4) √π. b = 4√π