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)，使这三个点成为等边三角形。

我正在使用以下代码示例计算两点之间的欧几里德距离：

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，我也需要它们。但是我在代码中启动点Z时遇到了困难。有人可以帮我这个吗？以下是我试过的样本。

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)))

但是这会返回以下错误：

 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，您可以围绕点(x2, y2)将点60旋转(x1, y1)度。另一个可接受的解决方案是旋转-60度，即相反方向。

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)

Answer 2

要找到等边三角形的第三个点并不需要任何真正复杂的东西，只需找到X和Y之间的中点，你知道这形成了一个直角点Z所以只需映射到原点，乘以sqrt(3)（等边三角形的毕达哥拉斯理论的简化）并在两个方向上旋转90度（x,y => y,-x，x,y => -y,x），并映射回来，例如：

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)

您可以在numpy中执行此操作：

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])

为平面中的给定两点创建等边三角形 - Python

2 个答案: