Intersection of equilateral triangle and circle

Question

I am a math student and in my free time am tutoring highschool students in math. Today, while tutoring, I came across a problem which I can't seem to solve. In highschool, geometry has never been a strong suite of mine, which is still the case, so sometimes certain questions can give me a hard time, like this one.

The problem is stated as follows. Say we have a circle $c$ with centre $M$. Take two points $P$ and $C$ lying on the circle such that $\triangle MCP$ is equilateral with sides wich have length $10$. Now draw a perpendicular line from $C$ to $MP$ and let $A$ be the point that divides $MP$ in two. Let $B$ be a point outside of $c$ such that $\triangle ABC$ is also equilateral. Define point $S$ as the intersection of $c$ and $AB$. The question is: what is the length of $SB$? I have made a picture using GeoGebra to illustrate the problem, see below.

I have explained that the answer can be acqired by means of choosing $M$ to be the origin of $\mathbb{R}^2$, but this method is too tedious for the scope of the question, so I was wondering if anyone could point out a more elegant method (without the introduction of coordinates).

Answer 1

$PAS$ is a $30$ degree angle.

Lets drop the altitude from $AS$ to $MP$ and call the point of intersection $Q.$ Lets say the length of $AS = x\\ SQ = \frac 12 x\\ AQ = \frac {\sqrt 3}{2} x$

$(5+AQ)^2 + (SQ)^2 = 10^2\\ (5+\frac {\sqrt 3}2 x)^2 + (\frac 12 x)^2 = 10^2\\ x^2 + 5{\sqrt 3} x -75 = 0$

Quadratic formula

$x = \frac {-5 \sqrt 3 + \sqrt{375}}{2}\\ x = \frac {-5 \sqrt 3 + 5\sqrt{15}}{2}$

$AB = 5\sqrt 3\\ BS = 5\sqrt 3 - x = \frac {15\sqrt 3 - 5\sqrt {15}}{2}$

Answer 2

Here is a solution using complex numbers.

I will consider that $M$, $P$ and $C$ have affix $0$, $1$ and $e^{2i\pi/3}$ respectively.

Hence $S$ has affix :

$$s=\frac12+r\,e^{i\pi/6}$$

with $r>0$, and we also have $|s|=1$. This last equality is equivalent to :

$$\left(\frac{1+r\sqrt 3}{2}\right)^2+\left(\frac r2\right)^2=1$$

$$4r^2+2r\sqrt 3-3=0$$

The positive solution is :

$$r=\frac{\sqrt{15}-\sqrt3}{4}$$

We deduce :

$$SB=AB-AS=\frac{\sqrt3}2-r=\frac{3\sqrt3-\sqrt{15}}{4}$$

If we choose the unit length provided by the OP, we have to multiply this by $10$.