# Calculating the break-even point between traveling around the inner / outer circle

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


I just wanted to confirm that my math is still working and that this is the right answer

• Could you please specify the start and end? Before we find the shortest path from A to B we should know what A and B are! – Unit Jun 17 '19 at 13:55
• @Unit there are actually 2 point A's: the intersection between line (street) going straight up on the inner circle and the outer circle. Point B is at the intersection between the outer circle and the line that's offset. I am interested in finding the angle at which the distance from both point A's to point B is the same... The break even angle – ControlAltDel Jun 17 '19 at 14:28
• Where is your office? On one of those radii, I guess but on the inner circle, the outer circle, or in between? You have a choice of two places to get out, I guess that they are on the other radius. Are the choices, the two intersections with the circles? If you could add labels to the diagram, it would help a lot. – badjohn Jun 17 '19 at 14:44
• Also, if $x$ is in radians then you probably don't want to multiply it by $\pi$. $x r$ is already the distance for $x$ radians around a circle of radius $r$. – badjohn Jun 17 '19 at 14:47
• @badjohn so that would make the answer 1 radian, or? – ControlAltDel Jun 17 '19 at 15:02

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.

• Something else I think is neat about this: since 1 radian is irrational, in real life one of the two options must always be better (never mind that we don't walking in infinitely thin lines, nor that in real life this is an oval not a circle) – ControlAltDel Jun 17 '19 at 17:00
• $1$ radian is irrational in terms of degrees but rational in radians (it is $1$). Whether it is irrational or not depends on your ruler. Think of ordinary straight lines. You make a ruler and I make one. Mine happens to match the diagonal of a $1 \times 1$ square drawn with yours. Lines that I think have a rational length will seem irrational to you. – badjohn Jun 17 '19 at 18:11
• Except that you can define 1 inch or 1 centimeter as X number of wavelengths of a certain light coming from X atom. Whereas root 2 (or 1 radian) is irrational (cannot be defined with a certain number of wavelengths of anything) when you've already defined 1 inch/centimeter using wavelengths – ControlAltDel Jun 17 '19 at 19:43
• What if we use different isotopes or transitions for our length standard? Would they necessarily have a rational relationship? We could never know since we could never perfectly compare them. – badjohn Jun 17 '19 at 20:24