# Find this angle, in terms of variables

I have a geometry question.

Take a look at this figure:

It is 3 circles, symmetrically placed so that the arc length that is outside is equal for all sides. I tried to determine the angle that is tangent to two sides as they intersect (Theta in the drawing) in terms of $L$ (arc length) and $R$ (radius). Obviously, all $L_1=L_2=L_3$ and $R_1=R_2=R_3$ since the circles are the same.

I figured the angle to be $\dfrac{5\pi}{3} - \dfrac{L}{R}$.

Anyways, so my problem is: If I were to make a polygon, made of symmetrical circles where the vertices are intersection, e.q.

Is there a way to find the angle in term of the number of sides.

I am assuming it would look like N( ) - L/R, where N is the number of sides and ( ) is some angle expression.

I appreciate all helps!

• In diagram 1, things seem to be symmetric, and therefore theta = (π/2) + (π/6) = 2π/3
– Mick
Commented Jul 2, 2014 at 17:49
• @Mick I see where you are coming from. It could be 2pi/3 with the right combination of L and R, however, it will not always be this straight (still symmetric), it could very well have the circles a bit farther from each other, etc. Therefore the equation must include R and L in it. Commented Jul 2, 2014 at 18:16
• In that case, you are looking for 2 or even 3 different "theta"s.
– Mick
Commented Jul 2, 2014 at 18:29
• @Mick Not exactly, theta will be the same on all side due to symmetry. The value of theta will vary depending on the arc length (L) and radius (R) and the number of circles, but once you find one, the others should be equal. Commented Jul 2, 2014 at 19:24

Consider the regular $N$-gon formed by the centers of the circles: The outer angle of that polygon, the orange one, is

$$\alpha=2\pi-\frac{(N-2)\pi}{N}=\pi+\frac{2\pi}N$$

The green angle $\beta$ is related to the angle you're asking for. Due to a mistake in my formulation of this answer, I originally assumed $\beta=\frac\theta2$, but that is not the case even though it looks that way in the figure. Instead, the tangents are perpendicular to the radii, so we get

$$\beta=\frac\pi2-\frac\theta2=\frac{\pi-\theta}2$$

Therefore the purple angle is

$$\gamma=\frac\pi2-\beta=\frac{\theta}2$$

The angle corresponding to the outer arc, colored in cyan, is

$$\delta=\alpha-2\gamma=\pi+\frac{2\pi}N-\theta=\frac LR$$

So you can solve this for $\theta$ and obtain

$$\theta=\alpha-\frac LR=\pi+\frac{2\pi}N-\frac LR$$

This isn't exactly the shape you suggested, particularly since $N$ occurs in the denominator, not in the numerator. But for $N=3$ it now agrees with your result, thanks to the fix your comment triggered.

• @Koopa: I guess I found a mistake inmy formulation: I'm assuming that the green line are tangents to the circle, which they are not. So the $\beta=\theta/2$ part is incorrect. Will fix that when I find the time.
– MvG
Commented Jul 7, 2014 at 17:31
• @Koopa: Fixed my mistake, but for the $N=2$ case I still disagree. Take $L=2\pi$ and $R=1$, then the circles just touch, and you get $\theta=0$ from my updated formula, which looks correct now, but $2\pi$ from your formula, which doesn't make a lot of sense as an angle between two lines, does it?
– MvG
Commented Jul 7, 2014 at 18:05
• This makes more sense, thanks mate! Commented Jul 7, 2014 at 19:23
• hey, small question, what software did you use to draw your picture? Much appreciated. Commented Jul 7, 2014 at 21:10