Find the two points for an equilateral triangle inscribed inside a circle

Question

I made up the following problem and I'd appreciate some hints for how to approach it.

I have a circle of known radius $ 10 $, with the origin at $(0,0)$ and I want to determine three points that would determine the vertices of an equilateral triangle inside the circle. I arbitrarily decide that the first vertex, $ P_1 $, is $ (0, 10) $.

I have created the following system of equations that determines the constraints for each point based on the following two premises:

• Each vertex is at an equal distance, $ D $, from the other vertices
• Each vertex is at the same distance, $ R $, from the origin. This distance is the radius of the circle.

The system is as follows:

\begin{cases} (p_{2x})^2 + (p_{2y})^2 = 10^2 \\ (p_{3x})^2 + (p_{3y})^2 = 10^2 \\ (p_{2x})^2 + (10-p_{2y})^2 = D \\ (p_{3x})^2 + (10-p_{3y})^2 = D \\ (p_{2x}-p_{3x})^2 + (p_{2y}-p_{3y})^2 = D \\ \end{cases}

The first two equations determine the distance of the remaining vertices, $ P_2, P_3 $ to the centre. The remaining three are the distances between the vertices.

I guess my first question is, can I solve this equation system to get the coordinates of each point? There are 4 variables and 5 equations so it should be possible.

If so, I have the feeling that knowledge of matrices would help me to solve this? The usual method by elimination/substitution seems a little bit painful, at first sight, for this type of system.

Thanks.

Answer 1

You are making this much too difficult. For the unit circle centered at the origin on a Cartesian coordinate plane, the set of points $$(x_k, y_k) = \left(\cos \frac{2\pi k}{n}, \sin \frac{2\pi k}{n}\right)$$ for $k = 0, 1, 2, \ldots, n-1$, describes the vertices of an inscribed regular $n$-gon. Since in your case $n = 3$, and you want a radius of $10$, and you want one vertex at $(0,10)$, all that needs to be done is multiply these coordinates by $10$ and switch the $x$- and $y$-axes values, which gives us $$(x_k, y_k) = \left(10 \sin \frac{2\pi k}{3}, 10 \cos \frac{2\pi k}{3}\right), \quad k = \{0, 1, 2\}.$$ Evaluating for each such $k$ yields $$\begin{array}{c|c c} k & x_k & y_k \\ \hline 0 & 0 & 10 \\ 1 & 5 \sqrt{3} & -5 \\ 2 & -5 \sqrt{3} & -5 \\ \end{array}$$