After much moping about the house, I'm finally back in my math groove! Except....now my LaTeX image generator isn't working anymore!!! T_T. Luckily this one doesn't require much latex, so I'll just type it out:
A function f is defined by f(z) = (4+i)z^2 + az + y, where a and y are complex numbers. Given that f(1) and f(i) are both real, what is the minimum value of |a| + |y|?
They also mentioned something about how i^2 = -1. If you didn't know that then I would recommend you close out the browser and go open a middle school math textbook.
So how should we start? I'll tell you something I've noticed. The AMC problems are meant to look intimidating. They really aren't that bad. So we are given that certain values have no imaginary component. This is a clue to split a and y into real and imaginary components:
a := a + bi
y := y + zi
I used := because I mean that we're completely redefining them now.
So lets look at f(1) and f(i):
f(1) = 4 + i + a + bi + y + zi
From this we can conclude that b + z = -1, as we need all the i terms to go away
f(i) = -4 - i + ai - b + yi - z
Again, we conclude that a + y = 1
This is where most people would get stuck. How can we possible find the minimum value of |a| + |y| given this information? Here, I turned to a geometric interpretation using vectors. In AP Physics we've been doing vector component stuff. If you interpret a and y as the x components and b and y and z as the y components, then we get a picture like this:
I thought that would be too big but now it looks too small.
Well anyway, the point is that if we look at the complex numbers as vectors, their x and y components both add up to 1. Well, technically one of them adds up to -1, but it doesn't matter. So the absolute value of each will be the length of each vector added together. However, since we know how their components add up, we know that if we add the vectors together using that silly head to tail method, they go to the point (-1,1). What's the shortest path from 1 point to another? A straight line. The distance from the origin to (-1,1) is the square root of 2. Problem solved!
I'm back!
College stress
So apparently all the schools I'm looking at require an SAT II in science.
Which means I'm screwed. Royally.
I'm pretty damn good at math, and really just terrible at all sciences. My friend told me that the Physics one is pretty hard, but that's the only one I would have a shot at. Honestly, It was a miracle that I didn't fail biology and chemistry. I somehow pulled off high C's when I clearly deserved much less.
Scholarships, financial aid, deadlines, my retarded french guidance counselor. I'm just feeling overwhelmed. I was expecting my parents to help me out with most of this and apparently they were expecting me to stick up for myself and do most of the work.
I am now thoroughly depressed and stressed out. Don't expect any math problems in the near future. T_T
Applying to MIT + Number theory problem
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)
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.