Yes I'm Still Alive!

HEY!

Sorry I couldn't get back sooner... I tried to put up a post a couple times but it wouldn't save or publish the post. Anyhow, I'm back! I'll try to do a summary of last ones this time for yall.

Pigeonhole Principle-- the idea of putting pigeons in holes. If there are m pigeons in m holes, then––
a) either there is one pigeon per hole;
b) or there is an empty hole and a hole with more than one pigeon.

If there is m pigeons in n holes, and m>n, then there is at least one hole with more than one pigeon. If m<n, then there is at least one empty hole.

This is essentially the same thing as the socks from last week–– but the socks have magically transformed into birds (sheep or cotton plants would be much more fitting). You want at least two sheep in one of the holes-- so you make the number of sheep more than the number of holes.

Here's an interesting problem involving the pigeonhole principle that I found––


There are five points on the surface of a sphere. Prove that at least one hemisphere has four of the points in it.
In spherical geometry, any two points on the sphere determine a great circle that divides two hemispheres. According to the pigeonhole principle, at least two of the other points should be on one of the hemispheres. So, including the two on the great circle, there are four points on that hemisphere. Cool, right?


I also found a really cool logic/geometry problem in a book I read.

In a square swimming pool, there is a boy in the center. At one of the corners is a teacher that can run three times faster than the boy can swim. The boy can run faster than the teacher can. Logically, can the boy escape if the teacher must try to catch him? (Don't make assumptions!)
Because the boy is faster than the teacher on land, his goal is simply to reach the edge. If the boy tries to swim to the opposite corner, the teacher will arrive before he does. However, if he tries to swim to one of opposite edge midpoints, the teacher will arrive exactly as he does. If the teacher was a fraction slower, he would arrive before and escape. So, if he begins to travel towards one of the opposite midpoints, the teacher is forced to head there.
Consider the diagram.

If the boy swims n/3 units towards midpoint of BC, then the teacher is required to move to midpoint of AB because if they don't, the boy will escape that way. Then, the boy can swim n units towards  CD and escape. Easy as pi.

Well, I guess that was a good long one for everybody. Again, sorry I haven't posted.

STay coolio,
John