Robert Eisele
Engineer, Systems Architect and DBA

Misc: 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?


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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