来个大家都爱的爆算题
做一个矩形，满足其内部能放下5个如图相切的圆，（相连处全都相切），题目出自Ander大佬 Anapple 发表于 2025-2-3 17:17有根式解哦
圆贼多，费神：
本帖最后由 qjchen 于 2025-2-3 09:40 编辑
挺有趣的题目，用python编了好一段代码才求出来，大概有14位的精度
随手改编下两道题，把此题右边的线改为45度线，直角梯形：），
结果又如何
这些圆圆的相切球，想起了当年做Joselin版主的许多切圆的题目，集合如下，快乐的年轻时光 ：）
101-200
[公告]典型问题及其解法大集中！ - 几何算法 - AutoCAD论坛 - 明经CAD社区 - 第2页 - Powered by Discuz! (mjtd.com)
1-100
[公告]典型问题及其解法大集中！ - 几何算法 - AutoCAD论坛 - 明经CAD社区 - Powered by Discuz! (mjtd.com)
有根式解哦 a=100;
b=86.850301679939548081238453054628...;
R1=31.090998694367849448184538993009...。
本帖最后由 yimin0519 于 2025-2-9 20:42 编辑
qjchen 发表于 2025-2-2 23:20
挺有趣的题目，用python编了好一段代码才求出来，大概有14位的精度
矩形的情况只有一种（所谓不同唯镜像矣），如为凸四边形则可能有多种情况。
高一定、“45°”直角梯形一角放置一“恒星”圆(其它四圆拱卫)，“唯一”性情况共有四种(且诸圆互不相等)，见下图：
有时间看能否“爆算”一下，看能付诸尺规作图的又有凡几。
本帖最后由 qjchen 于 2025-2-11 07:40 编辑
代码修改一下，倒是不难算，迟些有空整理好了再把代码放上来。
顺便开个与此题相类似的题目
几何作图题111
左上角以X轴镜像之后的求解代码如下（不少是AI辅助写的），此时，以基准圆半径为1进行计算，没啥技术含量，也没啥通用性，姑且放到这里
import numpy as np
from scipy.optimize import fsolve
import math
def line_circle_intersection(p1, p2, center, radius):
x1, y1 = p1
x2, y2 = p2
h, k = center
# 如果直线是垂直于 x 轴的直线（x = 常数）
if x1 == x2:
x_line = x1
discriminant = radius**2 - (x_line - h)**2
if discriminant < 0:
return []# 无交点
y1_intersect = k + np.sqrt(discriminant)
y2_intersect = k - np.sqrt(discriminant)
return [(x_line, y1_intersect), (x_line, y2_intersect)]
A = y2 - y1
B = x1 - x2
C = x2 * y1 - x1 * y2
distance = abs(A * h + B * k + C) / np.sqrt(A**2 + B**2)
if distance > radius:
return []
a = A**2 + B**2
b = 2 * (A * C + A * B * k - B**2 * h)
c = C**2 + 2 * B * C * k - B**2 * (radius**2 - h**2 - k**2)
discriminant = b**2 - 4 * a * c
if discriminant < 0:
return []
# 计算 x 坐标
x_intersect1 = (-b + np.sqrt(discriminant)) / (2 * a)
x_intersect2 = (-b - np.sqrt(discriminant)) / (2 * a)
# 计算 y 坐标
y_intersect1 = (-A * x_intersect1 - C) / B if B != 0 else (-A * x_intersect1 - C) / 1e-10
y_intersect2 = (-A * x_intersect2 - C) / B if B != 0 else (-A * x_intersect2 - C) / 1e-10
# 返回交点
return [(x_intersect1, y_intersect1), (x_intersect2, y_intersect2)]
# 定义方程组
def equations(vars, circle1, circle2, line):
x, y, r = vars
x1, y1, r1 = circle1# 示例圆1的参数
x2, y2, r2 = circle2# 示例圆2的参数
x3, y3, x4, y4 = line# 示例直线的参数
# 圆1的约束
eq1 = (x - x1)**2 + (y - y1)**2 - (r + r1)**2
# 圆2的约束
eq2 = (x - x2)**2 + (y - y2)**2 - (r + r2)**2
# 直线的约束
A = y4 - y3
B = x3 - x4
C = x4 * y3 - x3 * y4
eq3 = (A * x + B * y + C)**2 - r**2 * (A**2 + B**2)
return
# 二分法寻找 assumeX 使得 y5 = r5
def find_assumeX():
low, high = 0.4, 0.5
tolerance = 1e-16
while high - low > tolerance:
assumeX = (low + high) / 2
p1 = (assumeX, 0)
p2 = (assumeX, 3)
circle1center = (1, 1)
circle1radius = 1 + assumeX
intersections = line_circle_intersection(p1, p2, circle1center, circle1radius)
if intersections:
max_y_intersection = max(intersections, key=lambda point: point)
x_value, y_value = max_y_intersection
else:
x_value, y_value = 0, 0
# 初始猜测值
initial_guess =
circle1 =
circle2 =
line =
# 求解方程组
solution = fsolve(equations, initial_guess, args=(circle1, circle2, line))
x3, y3, r3 = solution
initial_guess =
circle11 =
circle13 =
# 求解方程组
solution = fsolve(equations, initial_guess, args=(circle11, circle13, line))
x4, y4, r4 = solution
initial_guess =
circle11 =
circle14 =
line1 =
# 求解方程组
solution = fsolve(equations, initial_guess, args=(circle11, circle14, line1))
x5, y5, r5 = solution
print(f"当前猜测值: {assumeX}, 解得: x = {x5}, y = {y5}, r = {r5}")
if abs(y5 - r5) < tolerance:
return assumeX
elif y5 > r5:
low = assumeX
else:
high = assumeX
return (low + high) / 2
# 找到合适的 assumeX
optimal_assumeX = find_assumeX()
print(f"找到的 assumeX: {optimal_assumeX}")
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