Square inside a semicircle


You are given a semicircle with an inscribed square of side length \(s.\) Let \(a\) be the length of the diameter on either side of the square. Find the value of \(s / a.\)


Right away, you should notice that \(a\) and \(s\) are both dependant on the radius of the semicircle, let's call it \(r.\) Because we're looking for the value of \(s / a,\) we should start by looking for ways to express \(r\) in terms of \(s,\) \(a,\) or both.

We notice that there is a side of the square located on the diameter of the semicircle. Since the length of the diameters are equal on both sides of side s on the diameter, we can conclude that the midpoint on that particular side s is also the midpoint of the diameter of the circle. So the radius r of the semicircle must be \(r = (1/2)s + a.\)

We also know that a segment formed from the midpoint of the semicircle to the upper right vertex of the square is also the radius of the semicircle. Therefore, it is possible to form a right triangle, as seen here:

Now we can use the Pythagorean theorem to express \(r\) in another way:

$$\begin{align} r^2 &= s^2 + \left(\dfrac{1}{2}s\right)^2 \\[1em] r^2 &= s^2 + \dfrac{1}{4}s^2 \\[1em] r^2 &= \dfrac{5}{4}s^2 \\[1em] r &= \dfrac{\sqrt{5}}{2}s \end{align}$$

Now we have two ways of expressing \(r,\) so let's set them equal to one another:

$$r = \dfrac{\sqrt{5}}{2}s = \dfrac{1}{2}s + a$$

Now let's try solving for \(s / a:\)

$$\begin{align} a &= \dfrac{\sqrt{5}}{2}s - \dfrac{1}{2}s \\[1em] a &= \left(\dfrac{\sqrt{5 - 1}}{2}\right)s \\[1em] \dfrac{a}{s} &= \dfrac{\sqrt{5 - 1}}{2} \\[1em] \dfrac{s}{a} &= \dfrac{2}{\sqrt{5 - 1}} \end{align}$$

Source: Square inside a Semi-Circle by Hieu Huy Nguyen