So I've decided to include a link to this blog on my application to MIT..it's definitely a stretch, but my heart is in the right place. That means two things. 1) I'll be buffing this thing up with problems in the next few months. 2) No more rants riddled with swears. I'll also have to go clean up those old ones too, sorry folks!
What folks, nobody reads this....
And if you're from MIT, please let me into your school and shower me with financial aid. I truly deserve it. =D And I need a cat friendly dorm for my sweet kitty!!!
Ok, here's a problem:
Find all positive integers (x,y) such that
I've seen so many of these problems that I'm sick of them. If there's a Diophantine equation on the USAMO this year and I get it wrong, I'm going to smack myself upside the head.
So one factor is 71 and the other is 1 (71 is prime). It doesn't matter how we set them equal...adding them together still yields:
(remember, only positive integers)
Substituting this into the previous ones gives two answers for y, we're only interested in the positive one. It turns out to be 35. Our final solution is (6,35)
Applying to MIT + Number theory problem
AIME - Number Theory
font style="font-weight: bold;">Two positive integers differ by 
I got rid of one of the conditions because it just made more busy work, but it's really still the same problem, just with a different answer.
Call the integers x and x + 60. Then:
At this point, it was rather intuitive to me that $x(x+60)$ must be a perfect square for this to be possible, since I had already done a lot of work with adding radicals. usamts??? I've said too much ;) . But you can show this by squaring both sides of the equation.
Now let z = x+30:
I'm telling you, that difference of squares is key. One of the 2006 IMO problems was a tough looking number theory problem - until you see the difference of squares. Even Alex Zhai didn't see it immediately! And he's like...omnipotent, when it comes to math. It's like he's THERE when they're writing the problems. Just goes to show you that you can never overlook simple tools.
Since we will maximize x by maximizing z, we'll go through the cases, starting with where z is the largest:
Solving for z yields a number that is not an integer, so we throw it out.
Solving for z yields z = 226, so x = 196 .Thus, our desired sum is:
If you're curious, the condition that I excluded was that our root sum cannot be an integer, which this answer doesn't fit. I took it out because I just would've had to type up another case, and I'm sure you got the idea.
"Dying is the day worth living for"
So I finally started watching At World's End. I don't know why everybody said it was awful. I mean, granted, it was no Dark Knight (which was absolutely mind blowing), but it was worth a watch.
I also started watching a new anime called Monster. The premise is that this doctor saved the life of a small boy who grew up to be a serial killer. So the killer in a way idealizes the doctor, while the doctor is trying to track him down. It reminded me of Death Note a bit, only without the magic notebooks or flying shinigami.
I'm also officially enrolled in WOOT now. I'm wondering if it'll be enough to keep me competitive. After all, all of the MOSP participants automatically get into it. Then again, I think I'm progressing rather quickly. WOOT should help me out with that. We'll see.
Made up number theory problem
Prove that, for all positive integers x and r, the remainder when 
This is pretty easy if you interpret it as:
Which, according to the binomial theorem, is:
If we expand this out, then every term except for the first will be divisble by (x-1). The first term is just 1, so the remainder is 1. Nifty, right? I thought so.
WOOT
No, that is NOT the douchalicious phrase that people say online when something causes them to orgasm all over their keyboard. Sorry, I'm a bit angry right now. Anyway, it stands for Worldwide Olympiad Online Training. And I've managed to convince one of my parents (guess which one) to sign me up for it. I'm REALLY happy, even if you can't tell from my constant unintentionally unsatisfied mood. It's basically a USAMO prep class.
I've been wondering, would it be possible for me, somebody who had never heard of the AMC less than a year ago, to make it to the IMO? It's certainly a HUGE stretch. I was looking at the winning USAMO scores from this year, they were all in the 33 - 37 range (two people scored 37, the highest). That would mean that I would probably have to solve EVERY problem at least somewhat. Several of them would have to be perfect. I guess I just have to ask myself one thing: Do I think I have what it takes to practice, learn what needs to be learned, and, when the time comes, shit in the mouth of 6 math problems? Of course I can. I'm going for it.
Wish me luck.
AIME - Trigonometry
Find a positive integer n such that:
Keep in mind that, since this is a non calculator test, frantically entering this into your calculator wont help you, as its miles away in the Mojave Desert being cocooned by military spiders. So you best prepare to figure it out on your own.
First off, here's a tip:
With that said, lets take the tangent of both sides:
The LHS looks like it can be simplified using our tangent sum formula. If you don't know it (and you should), it's:
However, we don't have two terms, we have four. So the best thing to do is split the sum of the four into a sum of two. Just add some parenthesis in to see what I mean:
This can start to get a little messy, so lets split it into two variables:
Calculate each using the sum formula:
Now plug these back in:
(Here I made the substitution right away, rather than type out that whole ugly expression. Resubstitute in our original expressions for x and y if you want to see what I was avoiding.)
The rest is simply the algebra of solving for n:
Tadaa. If you're practicing outside of the AIME, you can plug that into your calculator and check your answer.
Now, what would've we done if we didn't get a positive integer for an answer? Well, it's the AIME, so if they say solve for n, you can be sure it's a positive integer. Not to mention they right out say it, too. You should go back and check your work for answers. Work your away up instead of starting at the top, checking for arithmetic errors first. If you still cant find it, then check your logic for errors too. If you STILL cant find it, well then, I'd say skip that one for a bit and attempt it again later.
Coming soon to a desolate math blog near you
Geometry problems! I've shied away from geometry problems here because if I don't provide a diagram, then it's just confusing and hard to follow. However, I recently downloaded a neat free geometry program, so perhaps I could get some going. And no, it's not geometer's sketchpad. Not only is that not free, but for some reason I've always found it to be a bitch to use. Perhaps I'll give it another go though (I own it).
And the sun just rose. I really need to get my sleep in order. I keep nagging my parents to buy some red bull so I can even it out, but they wont. I think it's time to ask Nick to procure some for me.