Creating an equilateral triangle for given two points in the plane - Python

Question

I have two points X = (x1,y1) and Y=(x2,y2) in the Cartesian plane. I need to find the third point Z = (x,y) such that these three points make an equilateral triangle.

I'm calculating the Euclidean distance between two points using the following code sample:

def distance(points, i, j):
dx = points[i][0] - points[j][0]
dy = points[i][1] - points[j][1]
return math.sqrt(dx*dx + dy*dy)

In theory, I need to equate the distances of XZ and YZ to XY.This gives us two possible answers and I need them both too. But I'm having a difficulty in initiating the point Z in my code. Can someone please help me with this? The following is a sample of what I tried.

L = [0, 6]  #known two points
d= distance(points, L[0], L[1])
x = Symbol('x')
y = Symbol('y')
newpoint = x,y   #coordintes of the third point of the triangle
f1 = distance(points, L[0], newpoint)
f2 = distance(points, L[1], newpoint)
print(nsolve((f1, f2), (x, y), (d,d)))

But this returns the following error:

 File "/Users/*.py", line 99, in <module>
    f1 = distance(points, L[0], newpoint)

  File "/Users/*.py", line 36, in distance
    dx = points[i][0] - points[j][0]

TypeError: list indices must be integers or slices, not tuple

Equilateral triangles have 60°angle. Use trigonometry. You have distance and angle, you can get position. — politinsa, May 26 '18 at 21:28
Note: `math.hypot(dx, dy)` is the equivalent to the `math.sqrt(...)` call. @politinsa you don't need trig for this problem because you know that the points are the perpendiculars from the middle of the XY vector at a distance of `math.sqrt(distance()**2 - (distance()/2)**2) == math.sqrt(3)*distance()/2` [Pythagoras theorem] — AChampion, May 26 '18 at 21:56
Why are you passing a `point` as the 3rd argument to `distance()`, it is meant to be an index into the list of points, assume you need to add `newpoint` to points and then pass the appropriate index, or change your `distance()` function. — AChampion, May 26 '18 at 21:58
@AChampion Thanks for your comment! I'm calculating the Euclidean distance of two sides and equating that to the known distance (d). ` newpoint ` is the coordinates of the third vertex of the triangle. — ccc, May 26 '18 at 21:59

ewcz · Answer 1 · 2018-05-26T22:18:36.857

In order to get the third vertex, you could just rotate the point (x2, y2) by 60 degrees around point (x1, y1). The other admissible solution would be obtained with a rotation by -60 degrees, i.e., in the opposite direction.

import math


def get_point(x1, y1, x2, y2):
    #express coordinates of the point (x2, y2) with respect to point (x1, y1)
    dx = x2 - x1
    dy = y2 - y1

    alpha = 60./180*math.pi
    #rotate the displacement vector and add the result back to the original point
    xp = x1 + math.cos( alpha)*dx + math.sin(alpha)*dy
    yp = y1 + math.sin(-alpha)*dx + math.cos(alpha)*dy

    return (xp, yp)


print(get_point(1, 1, 2, 1))
# (1.5, 0.1339745962155614)

AChampion · Accepted Answer · 2018-05-27T15:47:24.940

To find the third point of a equilateral triangle doesn't need anything really complicated, simply find the mid-point between X and Y, you know that this forms a right angle to point Z so just map to the origin, multiply by sqrt(3) (simplification of Pythagoras theory for equilateral triangles) and rotate 90 degrees in both directions (x,y => y,-x, x,y => -y,x), and map back, e.g.:

X, Y = (1,1), (2,1)
M = (X[0]+Y[0])/2, (X[1]+Y[1])/2            # Mid point
O = (X[0]-M[0])*3**0.5, (X[1]-M[1])*3**0.5  # Map to origin, multiply sqrt(3)

In []:
M[0]+O[1], M[1]-O[0]                        # Rotate 90 (y,-x), map back

Out[]:
(1.5, 1.8660254037844386)

In []:
M[0]-O[1], M[1]+O[0]                        # Rotate -90 (-y,x), map back

Out[]:
(1.5, 0.1339745962155614)

You can do this in numpy:

X, Y = np.array([1,1]), np.array([2,1])
M = (X + Y) / 2
O = (X - M) * 3**0.5
t = np.array([[0, -1], [1, 0]])   # 90 degree transformation matrix

In []:
M + O @ t

Out[]:
array([1.5      , 1.8660254])

In []:
M + O @ t.T                       # the transpose gives -90

Out[]:
array([1.5      , 0.1339746])

Creating an equilateral triangle for given two points in the plane - Python

2 Answers2