I need to find the distance from the barycenter of an equilateral triangle to the edge in a given angle.

Here's a little sketch:

A little sketch

Given the outer radius of the triangle, the angle and the rotation (assuming the rotation in the picture would be $0$), I need to find the distance from the point on the edge (marked as red in the sketch) to the center.

Any help is appreciated!

  • $\begingroup$ Where do you count the angle from? $\endgroup$
    – user
    May 3, 2018 at 19:48
  • $\begingroup$ Counter-clockwise from the right, so 30° would be orthogonal to the right side of the triangle. $\endgroup$
    – DJ Coco
    May 3, 2018 at 19:51
  • 1
    $\begingroup$ It would be much simpler to count it either from direction to a (certain) vertex or from the direction to the midpoint of a (certain) edge. $\endgroup$
    – user
    May 3, 2018 at 19:53
  • $\begingroup$ I wouldn't know, really. I don't really have any idea on how to go about this. $\endgroup$
    – DJ Coco
    May 3, 2018 at 20:43
  • $\begingroup$ So you start the angle parallel to the bottom? $\endgroup$
    – ericw31415
    May 3, 2018 at 20:49

1 Answer 1


This expression works for your particular choice of the angle: $$d=\frac{R}{2\cos(\arccos(\sin(3\alpha))/3)},$$ where $R$ is the radius of circumscribed circle (I assume this was meant by "outer radius of the triangle").

  • $\begingroup$ Thank you so much! Just what I was looking for. $\endgroup$
    – DJ Coco
    May 3, 2018 at 20:56
  • $\begingroup$ @DJCoco You're welcome. $\endgroup$
    – user
    May 3, 2018 at 20:59
  • $\begingroup$ How would you calculate the distance when the triangle is isosceles instead of equilateral? $\endgroup$ Nov 1, 2019 at 6:52
  • $\begingroup$ @1FpGLLjZSZMx6k What kind of "central" point do you choose for isosceles triangle? $\endgroup$
    – user
    Nov 1, 2019 at 15:22
  • $\begingroup$ Let's say the center of gravity $\endgroup$ Nov 2, 2019 at 16:34

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