# Misc Solution: Inscribed Squares

A square is inscribed in a circle. A smaller square is drawn within the circle such that it shares a side with the inscribed square and its corners touch the circle. What is the ratio of the large square's area to the small square's area?

## Solution

Let's focus on the large square first. In order to get it's size we say the circle has radius $$r$$. The diameter and as such also the diameter of the square is thus $$2r$$. The resulting triangles inside the big square have its hypotenuse being the diameter and the catheti being the side length $$a$$ of the square. Using Pythagorean theorem we know $$(2r)^2 = 2a^2 \Leftrightarrow 4r^2=2a^2\Leftrightarrow a=r\sqrt{2}$$:

Now to the small square. Lets say the sides of the smaller square is $$b$$. We can now draw a line with length $$r$$ from the circle center to one of the corners of the small square sitting on the circle. We can then also draw a line from the middle of the circle to the middle of the small square, which is perpendicular to the side of the squares:

The length of the perpendicular line is $$\frac{a}{2}$$ plus $$b$$. We can use Pythagorean theorem again and state

$\begin{array}{rrl} & r^2 =& \left(\frac{b}{2}\right)^2 + \left(\frac{a}{2}+b\right)^2\\ & =& \left(\frac{b}{2}\right)^2 + \left(\frac{r\sqrt{2}}{2}+b\right)^2\\ & 0 =& \frac{1}{4}b^2+\frac{2}{4}r^2+rb\sqrt{2}+b^2-r^2\\ & = & \frac{5}{4}b^2 +rb\sqrt{2} -\frac{1}{2}r^2 \end{array}$

This can be used for the Quadratic equation:

$\begin{array}{rl} b &= \frac{-r\sqrt{2}- \sqrt{2r^2 + 4\frac{5}{8}r^2}}{2\frac{5}{4}}\\ \end{array}$

It follows that $$b=\frac{r\sqrt{2}}{5}$$ or $$b=-r\sqrt{2}$$. Since we are looking for the length of a side, the second solution isn't possible. The ratio of the areas is thus $$\frac{a^2}{b^2}=\frac{(r\sqrt{2})^2}{(\frac{r\sqrt{2}}{5})^2} \Rightarrow 25:1$$.

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