This is a classical Sangaku problem, also known as old Japanese geometry problems, that I found out just recently. The figure shows a semicircle with a smaller circle and an equilateral triangle inscribed inside it. Note that the semicircle can be *any* general semicircle, but in this case it has a radius of $1$ unit. Also note that one of the vertices of the equilateral triangle lies on the center of the semicircle. I have solved this problem and I'll post my approach as an answer, in order to not clutter up this space. Please share your own approaches as well!

[![enter image description here](https://i.stack.imgur.com/48a0g.png)](https://i.stack.imgur.com/48a0g.png)
['https://math.stackexchange.com/questions/4583801', 'https://math.stackexchange.com', 'https://math.stackexchange.com/users/1092912/']
A slightly quicker step (2) skips the quadratic formula: Observe that the hypotenuse shared by your two 30-60-90 triangles has length $\frac 2{\sqrt 3} r$ (since the side lengths are proportional to $1:\sqrt 3:2$), hence
$$r + \frac 2{\sqrt 3}r = 1$$
which yields $r=2\sqrt 3-3$.
Here's my approach for the problem:

Please note that my answer uses a lemma that can be proven easily, that is, [the radii of two externally or internally tangent circles are collinear](https://youtu.be/0whTkdGBNn0).

[![enter image description here](https://i.stack.imgur.com/R518H.png)](https://i.stack.imgur.com/R518H.png)

1.) First, we join the center of the semicircle with that of the circle, and draw a perpendicular from the center of the circle to the tangent points (semicircle and the vertex of the equilateral triangle). Notice that the two triangles that are formed are congruent via the SSS property, therefore we can conclude that the line connecting two centers is an angle sector. This proves that the two triangles are in fact "$30-60-90$" triangle.

2.) Now, using the fact proven above, we can say that the base of those triangles (i.e. the side opposite to the $30^\circ$ angle) is $\frac{r}{\sqrt3}$. Finally, we can finish off the problem:

$$(1-r)^2=r^2+(\frac{r}{\sqrt3})^2$$

$$1-2r+r^2=r^2+\frac{r^2}{3}$$

$$r^2+6r-3=0$$

Since this isn't factorable, we can use the quadratic formula:

$$r=\frac{-6±\sqrt{48}}{2}$$

Since $\frac{-6-\sqrt{48}}{2}$ would be negative, we must reject that. Therefore:

$$r=-3+2\sqrt{3}$$