python-3.x - 为平面中给定的两个点创建等边三角形 - Python

Question

我在笛卡尔平面上有两个点 X = (x1,y1) 和 Y=(x2,y2)。我需要找到第三个点 Z = (x,y) 使得这三个点构成一个等边三角形。

我正在使用以下代码示例计算两点之间的欧几里得距离:

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)

理论上，我需要将 XZ 和 YZ 的距离等同于 XY。这给了我们两个可能的答案，我需要它们两者也是。但是我很难在我的代码中启动 Z 点。有人可以帮我吗？以下是我尝试过的示例。

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)))

但这会返回以下错误:

 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

Answer 1

为了获得第三个顶点，您可以将点 (x2, y2) 围绕点 (x1, y1)< 旋转 60 度。另一个可接受的解决方案是通过 -60 度的旋转获得，即在相反的方向上。

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)