Every Friday from 3-4 the Grand Valley Math Department holds a "problem solving seminar." Really its just a small group of math students that gets together to work on tough problems from old copies of national and local math competitions. Sometime we go the whole hour without solving a thing. Most of the time we still end up talking about interesting math or at least trying several different intriguing yet ultimately unsuccessful attempts. Today i got one right which turned out to be from the Putnum which is usually one of the hardest test. I'll let you try it.
Proove (give a convincing argument) that given a sphere and five points on the surface of the sphere, four of those points must lie on the same closed hemisphere. That is, we can always find a hemisphere of the sphere that contains at least 4 of the 5 points.Posted by Matthew at February 16, 2007 10:55 PM
Hi, may you give me the answer of this question. I found this very interesting, but I am not able to find the answer.
Thank you
Fred
Here is the solution we came up with:
Let's say you are placing five points on a sphere and are trying to make sure that no hemisphere contains 4 points. First you put two (distinct) points down anywhere on the surface. These two points define a great circle on that sphere -- that is, they determine exactly one circle that has its center at the center of the sphere. We will consider this circle to be like the equator on the earth. Now, we can place one point in the northern hemisphere. There are now 3 points in the northern hemisphere because the closed hemisphere contains the equator and its two points. Since we can't place any more points on the northern hemisphere without making 4 points there, we place the next point in the southern hemisphere. Now the southern hemisphere has three points as well. Now no matter where we place the fifth point (northern hemisphere, southern hemisphere, or equator), there will be at least 4 points in one of the hemispheres.
Posted by: Matthew at April 13, 2007 07:17 PM