剛看了《最強大腦》中英對決,其中難度最大的項目需要選手先腦補泰森多邊形,再找出完全相同的兩個泰森多邊形。在驚呆且感嘆自身頭腦愚笨的同時,不免手癢想要藉助電腦弄個圖出來看看,閒來無事吹吹NB也是極好的。
今天先來畫畫外接圓和內切圓,留個大坑後面來填 :-]。
外接圓圓心:三角形垂直平分線的交點。
內切圓圓心:三角形角平分線的交點。
有了思路,就可以用萬能的python來計算了
import matplotlib.pyplot as plt
from scipy.linalg import solve
import numpy as np
from matplotlib.patches import Circle
'''
求三角形外接圓和內切圓
'''
# 畫個三角形
def plot_triangle(A, B, C):
x = [A[0], B[0], C[0], A[0]]
y = [A[1], B[1], C[1], A[1]]
ax = plt.gca()
ax.plot(x, y, linewidth=2)
# 畫個圓
def draw_circle(x, y, r):
ax = plt.gca()
cir = Circle(xy=(x, y), radius=r, alpha=0.5)
ax.add_patch(cir)
ax.plot()
# 外接圓
def get_outer_circle(A, B, C):
xa, ya = A[0], A[1]
xb, yb = B[0], B[1]
xc, yc = C[0], C[1]
# 兩條邊的中點
x1, y1 = (xa + xb) / 2.0, (ya + yb) / 2.0
x2, y2 = (xb + xc) / 2.0, (yb + yc) / 2.0
# 兩條線的斜率
ka = (yb - ya) / (xb - xa) if xb != xa else None
kb = (yc - yb) / (xc - xb) if xc != xb else None
alpha = np.arctan(ka) if ka != None else np.pi / 2
beta = np.arctan(kb) if kb != None else np.pi / 2
# 兩條垂直平分線的斜率
k1 = np.tan(alpha + np.pi / 2)
k2 = np.tan(beta + np.pi / 2)
# 圓心
y, x = solve([[1.0, -k1], [1.0, -k2]], [y1 - k1 * x1, y2 - k2 * x2])
# 半徑
r1 = np.sqrt((x - xa)**2 + (y - ya)**2)
return(x, y, r1)
# 內切圓
def get_inner_circle(A, B, C):
xa, ya = A[0], A[1]
xb, yb = B[0], B[1]
xc, yc = C[0], C[1]
ka = (yb - ya) / (xb - xa) if xb != xa else None
kb = (yc - yb) / (xc - xb) if xc != xb else None
alpha = np.arctan(ka) if ka != None else np.pi / 2
beta = np.arctan(kb) if kb != None else np.pi / 2
a = np.sqrt((xb - xc)**2 + (yb - yc)**2)
b = np.sqrt((xa - xc)**2 + (ya - yc)**2)
c = np.sqrt((xa - xb)**2 + (ya - yb)**2)
ang_a = np.arccos((b**2 + c**2 - a**2) / (2 * b * c))
ang_b = np.arccos((a**2 + c**2 - b**2) / (2 * a * c))
# 兩條角平分線的斜率
k1 = np.tan(alpha + ang_a / 2)
k2 = np.tan(beta + ang_b / 2)
kv = np.tan(alpha + np.pi / 2)
# 求圓心
y, x = solve([[1.0, -k1], [1.0, -k2]], [ya - k1 * xa, yb - k2 * xb])
ym, xm = solve([[1.0, -ka], [1.0, -kv]], [ya - ka * xa, y - kv * x])
r1 = np.sqrt((x - xm)**2 + (y - ym)**2)
return(x, y, r1)
if __name__ == '__main__':
A = (1., 1.)
B = (5., 2.)
C = (5., 5.)
plt.axis('equal')
plt.axis('off')
plot_triangle(A, B, C)
x, y, r1 = get_outer_circle(A, B, C)
plt.plot(x, y, 'ro')
draw_circle(x, y, r1)
x_inner, y_inner, r_inner = get_inner_circle(A, B, C)
plt.plot(x_inner, y_inner, 'ro')
draw_circle(x_inner, y_inner, r_inner)
plt.show()
下面看看兩個三角形的結果