Most days when I go to work, I ride a matatu (this is a Kenyan word. I am not in Kenya but the concept is the same). When I get off the matatu, I need to walk around the circle to get to the street that my office is on. And on occasion while I am making this trip, I think about whether it is shorter to get off before the light and walk around the smaller circle, or to get off after the light; in reality, it is probably not a perfect circle, but for my purposes I'm simplifying. In the image below, I looking to find the breakeven angle in terms of traveling to the blue point where the distance around the inner circle plus the distance between the circles (crossing the street) is equal to the distance walking around the outer circle. (The green point is my office)
Today I actually took to calculating the answer to this.
The way I am setting up the equation:
r1 * x * PI + (r2 - r1) = r2 * x * PI
The inner radius times x radians times PI plus the difference between the radius (to cross the street) equals the outer radius times x radians time PI.
In which case the answer is 1/PI radians, right?
I just wanted to confirm that my math is still working and that this is the right answer
I still need to make an assumption: You may get off at either of the red circles and need to get to the blue one.
I will use $\theta$ for the angle between the two radii as that is a very common choice for angles. I will assume that it is in radians. So, travelling around a circle of radius $r$ for $\theta$ radians is a distance of $r \theta$.
The inner circle has radius $r_1$ and the outer $r_2$.
If you get off at the inner blue circle then you will walk $r_1 \theta + r_2 - r_1$. If you get off at the outer blue circle then you will walk $r_2 \theta$. The break even will be when $$r_1 \theta + r_2 - r_1 = r_2 \theta$$
So, indeed $\theta = 1$. Surprisingly neat and simple. A nice problem.