Math stuff and other things...but mostly math.: KRITTER

Honestly, Mario Strikers Charged will be the death of me. The cause of death will be a mix of a blood clot in my brain from screaming too loudly and loss of blood from cutting my leg after putting it through the TV, all happening the next time Kritter misses the sudden death breaking goal because they lobbed it, and he tripped backwards, letting it roll in.

Here's a problem from my new book (solution by myself, not that somebody else out there hasn't already done it:

There are 2000 points on a circle, and each point is given a number that is equal to the average of the numbers of its two nearest neighbors. Show that all the numbers must be equal.

2000 is obviously just a number chosen because this problem showed up in the year 2000. They think it's so fucking clever to do that. (here's an idea, if you're getting ready for a math competition like AMC and AIME, practice with problems using the number of that year. I'm sure they've thought of that when writing them, but if they put one in, you might be familiar enough with the properties of that number to get an advantage).

Suppose that not all of the numbers are the same. Call them $a_1, a_2, \cdots a_{2000}$ . WLOG, assume that since they are not all the same, $a_1$ is the smallest number. Then, according to the problem:

$a_1 = \frac{a_2 + a_{2000}}{2}$

$2 \cdot a_1 = a_2 + a_{2000}$

However, since $a_1$ is the smallest, it is assumed that:

$a_1 < a_2$ and $a_1 < a_{2000}$

Adding these yields:

$2 \cdot a_1 < a_2 + a_{2000}$

Which directly contradicts our previous equality. The equality was given by our problem, but the inequality was given by our assumption that there is a smallest number, which means it must be incorrect. Since if there is not a smallest number, then all of the numbers are equal, then it follows that all of the numbers are equal. Quite elegantly done, and we're done, which was what we wanted, just as planned. I think if I ever become a famous mathematician, I'm going to introduce Just as Planned (JAP) as an acceptable way to end a proof.

So right now it's fucking 4:30 in the morning. My sleep schedule is really messed up. It's not even that I just keep sleeping in and going to bed late (although it's usually that), but it keeps switching around. Anyway, here's the problem I'm working on now (don't have a solution yet):

On a large, flat field, n people (n > 1) are positioned so that for each person, the distances to all the other people are different. Each person holds a water pistol and at a given signal fires and hits the person who is closest. When n is odd, show that there is at least one person left dry. (Canada 1987)

Woah there, I sourced a problem! I'm going to fight temptations and solve this one on my own. It would be the first time I technically solved an Olympiad problem on my own. I feel like I almost have it, too. I'll keep working on it...when I found the pencil I just lost. Shit stained balls.

ALSO, USAMTS Round 1 Problems are coming out soon! Of course I can't discuss them here, hehe. But I'll be sure to give them some sort of mention on here afterwards.