# Intersection of equilateral triangle and circle

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).

## 2 Answers

$PAS$ 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

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

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

• You have too many $B$'s here... Commented Apr 4, 2017 at 21:07
• @Aretino Thanks, lost track of my labels. Commented Apr 4, 2017 at 21:09
• ... and the last-but one formula should be $AB$, not $AS$. Commented Apr 4, 2017 at 21:11

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 $10$.

• Thank you for your answer! Unfortunately, I cannot accept this as an answer to my question, as you are introducing coordinates, albeit complex ones, which I explicitly want to aviod. Also, and this is something I haven't adressed in the question, my student is not familiar with complex numbers, so I cannot use this as an explanation in our sessions. Commented Apr 4, 2017 at 20:46
• Using the radius of the circle as $1, AB = \frac {\sqrt 3}{2}$ Commented Apr 4, 2017 at 21:08
• @DougM: Oops ! Of course. Thank you, editing ... Commented Apr 5, 2017 at 5:43