How to find coordinates of a 2d equilateral triangle in C?

Question

I have the coordinates (x,y) of 2 points. I want to build the third point so that these 3 points make an equilateral triangle.

How can I calculate the third point?

Thank you

Solution 1

After reading the posts (specially vkit's) I produced this simple piece of code which will do the trick for one direction (remember that there are two points). The modification for the other case shold be trivial.

#include<stdio.h>
#include<math.h>

typedef struct{
  double x;
  double y;
} Point;

Point vertex(Point p1, Point p2){
  double s60 = sin(60 * M_PI / 180.0);    
  double c60 = cos(60 * M_PI / 180.0);

  Point v = {
    c60 * (p1.x - p2.x) - s60 * (p1.y - p2.y) + p2.x,
    s60 * (p1.x - p2.x) + c60 * (p1.y - p2.y) + p2.y
  };

  return v;
}

Solution 2

You could rotate the second point 60° around first to find the location of the third point.

Something like this:

//find offset from point 1 to 2
dX = x2 - x1;
dY = y2 - y1;
//rotate and add to point 1 to find point 3
x3 = (cos(60°) * dX - sin(60°) * dY) + x1;
y3 = (sin(60°) * dX + cos(60°) * dY) + y1;

Solution 3

Let's call your two points A and B. Bisect AB, call this point C. Find the slope of AB (Y_A-Y_B / X_A-X_B), call it m. Find the perpendicular to that (-1/m) and call it m₂. Then compute a segment CD whose length is sin(60) * length(AB), at the slope m₂ (there will be two such points, one to each side of AB). ABD is then your equilateral triangle.

That, obviously, is a "constructive" method. You should also be able to do it by solving a set of linear equations. I haven't tried to figure out the right system of equations for this case, but this approach tends to be somewhat more stable numerically, and has fewer special cases (e.g., with the constructive version, a slope of 0 has to be dealt with specially).

Comments

[James McNellis]

FYI: There are two such points.

[MSN]

If the two points have the same x coordinate, there is no "upper point".

[BlueRaja - Danny Pflughoeft]

First one to do this without trig functions gets my upvote :)

[Escualo]

Isn't it [cos(60) sin(60); -sin(60) cos(60)]? I think your signs are wrong.

[vkit]

I think those signs are correct. See also: stackoverflow.com/questions/786472/rotate-a-point-by-an-angl‌​e

[vkit]

@arrieta Just tried this out, if you use the signs as you suggest it rotates the point the other way (clockwise/counter-clockwise) so depending on how your coordinate system is set up, one or the other might make more sense.

[Admin]

I said the same thing! And I didn't put any explicit code for a reason: graphics API can do rotations (and translations) much better(precision, trigonometric calculations etc etc) than we can ever hope to do ourselves. Since we do not know the API OP is using, all we can do is mention what to do. Oh well...