## Wednesday, October 21, 2009

### What do I think about while running? Trapezoids!

Though I'm not particularly fast, I would still call myself a runner. Many non-runners wonder how runners can possibly entertain themselves over the course of a 5-mile run. To be honest, I don't normally remember what goes through my head, but today it was trapezoids.

Specifically, given trapezoid ABCD (where AB || CD), how can you prove that CA + AB + BD is greater than CD?

It seems obvious that the sum of the other three sides should be longer (even if CA and BD are really short), but that doesn't mean it doesn't merit a proof! Drawing a single diagonal lends itself to a simple proof using the triangle inequality:

This yields the following two inequalities:

(1) CA + AB > BC
(2) BC + BD > CD

If we add BD to both sides of (1) we have:

(3) CA + AB + BD > BC + BD

Combining with (3) we have:

(4) CA + AB + BD > BC + BD > CD

So this proves that CA + AB + BD > CD!

I was thinking about this while running around the Charles River this morning, wondering whether it would ever be shorter to run past a long bridge (CD) to cross the river at a shorter bridge (AB) and then come back, but now we've proved that is never the case!

I only used a trapezoid because that more closely matched the geometry of what I was running, but I think it's trivial to extend the proof to any quadrilateral (though ones that are not convex may not cooperate).

1. This is one of the many reasons I'm happy to have you as my friend. Thanks for the smile.

2. Wow. Just. Wow.

3. the proof works for any polynomial

4. Here's another proof.

1. On a Euclidean surface, the shortest distance between two points is a straight line.
2. Therefore, CD is the shortest distance between C and D.
3. CA + AB + BD is not a straight line between C and D.
4. Therefore, CA to AB to BD is not the shortest distance between C and D.
5. Therefore, CD is shorter than AB + BC + CD.

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