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

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.

• Please read tags before applying them: the algebraic-geometry tag "should not be used for elementary problems which involve both algebra and geometry," per the tag description. Aug 10 '20 at 23:07
• Thanks for letting me know, sorry about that
– Jon
Aug 10 '20 at 23:11

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 $$\begin{array}{c|c c} k & x_k & y_k \\ \hline 0 & 0 & 10 \\ 1 & 5 \sqrt{3} & -5 \\ 2 & -5 \sqrt{3} & -5 \\ \end{array}$$

• Woah, definitely a better approach. Thank you very much for the insight!
– Jon
Aug 10 '20 at 23:47