There is a semi-circular drive on the west side of the campus, and I usually bike or walk up it on my way in to school every day. Most pedestrians walk on the sidewalk on the outer edge. Let r denote the radius to the sidewalk on the inner edge of the drive, and r’ the width of the road. Then a pedestrian on the outer edge walks a distance of (π/2) r + r’ to reach the east side of the top of the drive, whereas a pedestrian on the inner edge walks a distance of (π/2) (r + r’). Clearly the inner path is shorter, yet fewer people take it.
As an extra credit problem, why does it make sense for me to take the outer path anyway?
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WARNING: Male-Answer-Syndrome (and boredom) rearing it’s ugly head
I think something is screwy with your spatial geometry because 1 < (π/2) => (π/2) r + r’ < (π/2) (r + r’). Based on what you wrote, this means that the outer path is actually shorter.
(A gripe: I just spent way too much time making a little ASCII diagram of how I am envisioning the situation, only to discover that your comments section doesn’t support the <tt> tag.)
Anyway, if I’m visualizing it right, I calculate the distances as follows:
Outer loop: π (r + r’) — half the circumference of a circle, radius (r + r’).
Inner loop: π r + 2r’ — half the circumstance of a circle, radius r, plus two crossings of the road.
And so the inner loop really is shorter, just like you said.
The answer to the extra credit problem seems to be that you don’t remember geometry. (And you don’t like crossing the road.)
Because you’re heading east anyway.
I think you take the outer loop because there’s more hot chicks to ogle on that side. But maybe that’s just me. 🙂
Ari is the closest. Because I’m heading northeast anyway, and coming from the southwest, we have to measure the paths starting from the same point. I gave the inner loop path an unfair advantage of starting r’ farther north. If we take that into account, the path difference is (2 – π/2) r’ in favor of the outer loop.
Okay, first of all, not defining the starting point and (still) not defining which direction is East is unfair.
Second, is this only a quarter-circle? Because how else are you getting (π/2) factors? Isn’t the circumference of a circle 2πR so the distance around a semi-circle is πR?
I should probably spend less time thinking about this.
Yes, indeed it is a quarter loop. Think of it like a hunt puzzle — you have to think to go look at the Berkeley map or something…
You called it a SEMI-circular drive! That means half a circle!
Your puzzle is broken.
Although it’s interesting to note that being a quarter-circle vs being a semi-circle makes the difference as to which path is longer.