This is a problem from the extreme USAMO contest, which is 6 problems, 2 days, here's another one if you want another one:
Let ABC be a triangle with angle A=90degrees. Points D and E lie on AC and AB, respectively, such that angle ABD=angle DBC and angle ACE=angle ECB. Segments BD and CE meet I. Determine whether or not it is possible for segments AB, AC, BI, ID, CI, IE to all have interger side lengths.
Here's a random problem I found:
An integer is assigned to each vertex of a regular polygon so that the sum of the five integers is 2011. A turn of a solitaire game consists of subtracting an integer m from each of the integers at two neighbouring vertices and adding 2m to the opposite vertex, which is not adjacent to either of the first two vertices.( The amount m and the vertices chosen can vary from turn to turn.) The game is won at a certain vertex if, after some number of turns, that vertex has the number 2011 and the other four vertices have the number 0. Prove that for any choice of the initial integer, there is exactly one vertex which the game can be won.
(Note: it's not like I know how to do it, it's just for fun, because this would challenge most mathematicians.)
I guess you're right, I must've forgot the statistics, another thing that interests me is paradoxes for example:" No one goes to that restaurant; it's too crowded", probably hear people say this before, but they never notice they're actually contradicting themselves.
Unfortunately, I have to wait another 362 days to do AMC 8 again, but by then, I'll do much better than this time, probably know all of trigonometry by then, and I find the birthday problem quite interesting, it takes only 23 people to be 50% sure that at least 2 people share the same birthday,and when there are 50 people, the probability meets 97% which is incredible, you can even do a survey on just 50 people and 97 out of 100 times, at least 2 people will same the same birthday, this is the type of math problems that interest me the most of all, another will be if a 4% is the probability that a fighter will not get shot down, then what is the probability that after 50 missions, you won't get shot down? The probability is a whooping 97%, almost to 100%, you can get this using Euler's Number.
My favourite thing to do is math contests, because most of the time, whenever the contest results are handed out, I always get the highest score, except the Fibonacci contest, because then I get more bragging rights to my friends, I'm gonna fail AMC for sure, 25 questions, only 40 minutes, time is too tight, and the problems are extra hard, only 96 seconds a problem.
OH-NO, sorry! I got the whole problem wrong! It's suppose be 2010, not 2012 and 2010^2 has 81 factors, I can't believe that just a difference of 2 can make such a huge difference, now I hope that this problem is much more clear! I actually though that 5 was a factor of 2012!