0
$\begingroup$

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.

$\endgroup$
2
  • 1
    $\begingroup$ 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. $\endgroup$
    – KReiser
    Aug 10 '20 at 23:07
  • $\begingroup$ Thanks for letting me know, sorry about that $\endgroup$
    – Jon
    Aug 10 '20 at 23:11
2
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ Woah, definitely a better approach. Thank you very much for the insight! $\endgroup$
    – Jon
    Aug 10 '20 at 23:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.