圆半径关系推导：大圆内N个内切小圆的R与r关系求解

Alright, let's break down how to derive the relationship between the radius of the big circle ((R)) and the small circles ((r)) when (N) identical small circles are tangent to the big circle's circumference and tangent to their neighboring small circles (this tight packing is the only scenario that gives a unique, meaningful relationship).

Step 1: Map Key Distances #

First, since each small circle touches the big circle, the distance from the big circle's center to any small circle's center is exactly (R - r). Think about it: draw a line from the big center to the point where a small circle meets the big one—this line passes through the small circle's center, so it’s the big radius minus the small radius.

All small circle centers lie on a smaller concentric circle with radius (d = R - r).

Step 2: Form a Triangle with Adjacent Small Circles #

Take the big circle's center, plus the centers of two adjacent small circles. Connect these three points, and you’ll get an isosceles triangle with:

• Two equal sides: (d = R - r) (distance from big center to each small center)
• Base: (2r) (since the small circles are tangent, their centers are separated by (r + r))

Step 3: Apply Trigonometry #

The angle at the big circle's center in this triangle is (\frac{2\pi}{N}) radians (or (\frac{360^\circ}{N}))—since (N) small circles are evenly spaced around the big circle.

Use the law of cosines on this triangle:
[
(2r)^2 = (R - r)^2 + (R - r)^2 - 2(R - r)(R - r)\cos\left(\frac{2\pi}{N}\right)
]

Simplify the right-hand side:
[
4r^2 = 2(R - r)^2\left[1 - \cos\left(\frac{2\pi}{N}\right)\right]
]

Now use the trigonometric identity (1 - \cos\theta = 2\sin^2\left(\frac{\theta}{2}\right)). Here, (\theta = \frac{2\pi}{N}), so (\frac{\theta}{2} = \frac{\pi}{N}):
[
1 - \cos\left(\frac{2\pi}{N}\right) = 2\sin^2\left(\frac{\pi}{N}\right)
]

Substitute back into our equation:
[
4r^2 = 2(R - r)^2 \times 2\sin^2\left(\frac{\pi}{N}\right)
]
[
4r^2 = 4(R - r)^2\sin2\left(\frac{\pi}{N}\right)
]

Divide both sides by 4 (all values are positive, so we don’t need to worry about negative roots):
[
r^2 = (R - r)^2\sin2\left(\frac{\pi}{N}\right)
]
[
r = (R - r)\sin\left(\frac{\pi}{N}\right)
]

Step 4: Solve for (R) #

Rearrange the equation to isolate (R):
[
r = R\sin\left(\frac{\pi}{N}\right) - r\sin\left(\frac{\pi}{N}\right)
]
Bring the (r)-term to the left:
[
r + r\sin\left(\frac{\pi}{N}\right) = R\sin\left(\frac{\pi}{N}\right)
]
Factor out (r):
[
r\left[1 + \sin\left(\frac{\pi}{N}\right)\right] = R\sin\left(\frac{\pi}{N}\right)
]
Finally:
[
R = r \times \frac{1 + \sin\left(\frac{\pi}{N}\right)}{\sin\left(\frac{\pi}{N}\right)}
]

Or, for a more intuitive split:
[
R = r\left(1 + \frac{1}{\sin\left(\frac{\pi}{N}\right)}\right)
]

Quick Sanity Check #

Let’s test this with (N=6) (a common case of 6 circles fitting perfectly inside a big one):

• (\sin\left(\frac{\pi}{6}\right) = 0.5)
• (R = r\left(1 + \frac{1}{0.5}\right) = 3r)

This makes sense: each small circle’s center is (2r) away from the big center ((R - r = 3r - r = 2r)), and adjacent small centers are also (2r) apart—forming equilateral triangles, which fits perfectly with 6 evenly spaced circles.

内容的提问来源于stack exchange，提问作者Pulkit Gupta