USAMO 1972 - Combinatorics

This one was from the very first USA Math Olympiad. I thought it was rather easy compared to the other ones on the test, which goes to support the conjecture that, being the very first exam the problem writers had to create, they had trouble establishing a difficulty level. Anyway, here it is:

A random number selector can only select one of the nine integers 1, 2, ..., 9, and it makes these selections with equal probability. Determine the probability that after selections (1" class="latex" style="vertical-align: -1px;">), the product of the numbers selected will be divisible by 10. (USAMO '72)

I stole those latex images from AoPS, teehee. Hopefully they don't mind.

Anyway, we want the probability that after n selections, we have at least one factor of 2 and at least one factor of 5. So, just to clarify, we want at least one number of each set:

{2, 4, 6, 8} , {5}

The "at least" should be the tip off that instead you'll count the complement and subtract from the total number, which is 9^n . In case you couldn't tell, I'm just going for the counting approach.

Call the sets the (E)ven set and the (F)ive set. Or E and F for short. We want to count the number of sequences {a_n} such that it has either no elements of E or no elements of F. We can easily count these using PIE.

The number of sequences which have no elements of F is 8^n.
The number of sequences which have no elements of E is 5^n.

However, we overcounted the number of sequences which have neither elements of E or F, as it appears in both of our previous two sequences. Thus, we subtract it once.

The number of sequences which have no elements of E AND no elements of F is 4^n.

Thus, the total number of sequences such that the sequence does not contain an element from both sets E and F is 5^n + 8^n - 4^n.

We subtract this from our total number, 9^n. Thus, the total number of sequences with at least one element from both E and F is:

9^n - 5^n - 8^n + 4^n.

We divide this by the total number, 9^n, to get our probability, which is:

1 - (5/9)^n - (8/9)^n + (4/9)^n

Intermediate Olympiad - Polynomials

Find all polynomials P(x) with the following property:

There exists a positive integer k such that:

P(P(x)) = [P(x)]^k


Good stuff. Anyway, you can easily discover some quick things by plugging in basic values of x.

Suppose r is a root of P(x) then, obviously, P(r) = 0

And, P(P(r)) = [P(r)]^k

P(P(0)) = 0^k = 0

P(0) = 0

Similarly, plugging in any root of P(x) yields that P(P(r)) = 0. Thus, all roots of P(x) are roots of P(P(x)). or:

P(P(r)) = P(r) = 0

P(P(r)) will equal 0 when P(r) is a root of P(x), or

P(r_1) = r_2

Since P(r_1) = o, we have:

r_2 = 0, for any root of P(x).

Thus, all roots of P(x) are 0., P(x) = cx^k for some k. Plugging this into our equation quickly yields c = 1, and thus, P(x) = x^k for all positive integers k.

Olympiad - Polynomials

This one doesn't need latex and the solution is quick (I literally did it while going to the bathroom and reading Samurai Champloo at the same time) so I'll just type it up now.

(Canada 1970) Let P(x) be a polynomial with integral coefficients. Suppose that there exist four distinct integers a, b, c, d with P(a) = P(b) = P(c) = P(d) = 5. Prove that there is no integer k with P(k) = 8.

Kind of scary looking, but it's really not. We can say that P(x) -5 has 4 distinct integral roots, or:

P(x) - 5 = Q(x)(x-a)(x-b)(x-c)(x-d) = 0

Hopefully I don't need to explain that part to you. Now, suppose that there exists an integer k such that P(k) = 8, then we have the result:

P(x) - 8 = ((P(x) - 5) - 3 = Q(x)(x-a)(x-b)(x-c)(x-d) - 3 = 0

Q(x)(x-a)(x-b)(x-c)(x-d) = 3


Since all of (x-a), (x-b), (x-c), (x-d) are distinct and integers, then our assumed result is impossible. As only one can be 3, and the rest must be either 1 or -1. However, we have 4 roots, and thus, one of these three choices for numbers must occur twice, contradicting that the integers are distinct.

Redefining polynomials is a useful tool.

USAMO 1983 - Inequalities

Maybe I'm getting better. Maybe I picked out an easy one. Maybe the aura from visiting MIT today hasn't rubbed off yet (happiest place ever btw). Whatever the case, I managed to do an old USAMO problem today, with the help of ACoPS's compilation of obscure inequalities:



We want information about the roots, so it'll help to use our basic polynomial rules to rewrite a and b in terms of the roots. Let these roots be F, G, H, J, K:



Sorry if that last image is hard to read at the end. Try doing view image or something like that.

From now on, we'll simply be interpreting the inequality with these substitutions made. Also, I'm going to use that cyclic sum symbol consistently, because it's much easier than writing it all out.

We need to show that if the inequality is satisfied, then the roots which are represented in it cannot all be real. So our strategy will be to assume the inequality is satisfied as a given, and that all the roots are real, and attempt to find a contradiction. When we find one, then it will imply that one of our assumptions was false. Since we're operating under the pretense that our inequality is true, then our assumption about all the roots being real will be false.

With substitutions made, our inequality is:



Here it helps to know how to square expressions in the form (a + b + c + ....) quickly.

By the way, looking at that, I realize that it's a little hard to follow because it's cramped / ugly. So I recommend writing it out yourself.

Now we can start using well known (well, maybe not well known...) inequalities to find a contradiction. We know from Chebyshev's Inequliaty (WTF! Yes, I'll comment on the obscureness of this later) that:




For sequences of real numbers {a_n} and {b_n} that are "monotonic in the same direction". Or in other words, increasing or decreasing.

This works well for our roots, as since we're assuming they're real, we can, WLOG, arrange them in increasing order and apply Chebyshev's. Let {a_n} be our roots, and let {b_n} ALSO be our roots. Then we get the inequality (after multiplying both sides by 25, which I'll skip right over):





If you line this up with our assumed inequality, then it DIRECTLY contradicts it. Thus, we have our contradiction, and our proof is done.

It's possible to do this with AM-GM too, I believe. And I'm not 100% sure that my solution works, as I have nobody to check it with. However, I found the Chebyshev's solution first, and there doesn't appear to be anything wrong with it. Perhaps I'll outline that method later.

As for the obscureness of Chebyshev's...the only reason I used it was that I was using ACoPS while I was doing the problem, and I tried it and it worked. You don't need to know lots of random inequalities to do good on the USAMO (at least, I hope not). However, you should at least know the basics: AM-GM, the general Mean Inequality, and Cauchy-Schwarz inequality. It might help to know Chebyshev's, but get the basics down first.

What is this blog?

My original post got buried a long time ago, and the explanation in it was less than vague, so I'll try to go into more detail here.

If it wasn't already obvious, this is a cheap little blog where I post interesting math problems, pretty much all of which are contest problems. I try to write the solutions so that they're easy to follow along with, but at the same time they assume that you're familiar with the math needed to solve them.

I write as if I'm trying to explain problems to an audience. However, I hardly share this blog with anyone who would actually read the problems in a way that they were trying to understand them. So then, why bother writing in this way?

Two reasons (there's actually also a third reason that recently became apparent) :
1) They say that if you can't teach something, then you don't understand it. And vise versa. Technically, I'm not actually teaching anyone anything. But by forcing myself to write up these problems in a way that I'm actually explaining them, I can make sure that I truly understand them. But then again, sometimes I only type up problems which I already know I understand, so maybe I'm wasting my time. T_T
2) This is kind of odd, but someday I might want to be a teacher. Now that I think about it, I would probably get frustrated with stupid kids and only want the cream of the crop, so maybe it's not the best choice for me. But if I decide to go down that path, I'll have a little bit of experience teaching a ghost classroom via blog posts.
3) I've recently gotten awfully good at math (obviously I'm not at the IMO level, but I still think I'm pretty good). However, this sudden surge was after the last AMC date, and seeing as I'm in my senior year now, the next AMC's might not be in time for colleges to look at. So this sort of doubles as a way for me to show off my progress.

But, for you, the (non existant T_T) reader, this doesn't have much to do with you. So if you're interested, just go through some of the problems, and maybe you'll learn something!

AMC - Basic Equations

This is one of those ones where some of your variables have certain restrictions, so you can use that to intuitively find the answer.



So we have:

1/4 m + 1/6 c = 8
m + c = 8p

We know that p has to be an integer, and m, c, p are all positive. We can get each in terms of p:

P = 6 - 1/16 m = 4 + 1/24 c

Since m c and p are positive, and P is an integer, we can see that P > 4 and P < 6 . Thus, P = 5. In fact, we could also find out that there are 16 ounces of milk and 24 ounces of coffee, too. Nice! Easy like sunday morningggggg.

No political talk from me. I can't vote, and it has nothing to do with competition math. =D

Two things coming up soon

1) A USAMO problem. Oooh, scary!

2) I'm going to type up a mini post about what this blog actually is, since my original first post is long since buried beneath everything else.

In the meantime, I have lots of reading that needs to be done for law.

AIME - Number Theory

This is just a quickie I felt like posting, it had a fast solution that I'll just outline.



We can actually do the polynomial division to get the result:



Clearly, the largest integer n for which this is an integer is n = 890. Piece of delicious cake.

AIME - Complex Numbers / Probability

Today everyone was on an NHS field trip (maybe later I'll explain why I'm not in NHS...) and so in most of our classes we didn't really do anything. So I brought in a bunch of old AIME problems to work on. This one caught my attention while my physics teacher started giving a random lecture on economics:



Some of those words are blue because I copied that from the AoPS wiki.

Speaking of the AoPS wiki, I hate some of the solutions they have on there. Mainly because they're hard to follow. I suppose they're just supposed to be general solution outlines, but the one for this was some odd use of a bunch of trig formulas, and it wasn't very nice to look at. I like my solution because once it's explained, the problem becomes rather intuitive.

I interpretted this problem geometrically. It makes it a lot easier to see what's going on. So we have some roots of unity. At this point, I didn't even consider the number 1997. Then we're adding another root of unity to it. If we treat the complex numbers as vectors, then this is just the head-to-tail method. The diagram shows some of the possibilities of this. The dashed circles are all possible roots of unity. I just drew in two possibilities:



The absolute value of this number, which is just its distance from the origin, is greater than that odd square root. In other words, it's outside the circle of that radius. We can sketch that in too to get a better look:



Hmm. If we can visualize sliding that second "unity" circle around the first, the same percentage of it should always be outside that circle that we drew in. That suggests that the probability is independent of the first choice, which certainly makes things easier.

One issue that's worth addressing at this point is whether our approach of throwing away the number 1997 for simplicity has thrown us off course. The same percentage of the circle is always farther from the origin than that silly circle with the square root radius, but we're not looking at the percentage of the circle, we're looking at the number of roots of unity for the number 1997 that lay outside the circle. If we slide that second circle around, there's no guarentee that this will be the same for each. When I did the problem out, I reasoned that it is, because if you draw in all the actual possibilities for a certain number, you can rotate the entire thing by 360 / n degrees, causing each set of possibilties to translate to the next exactly...if that makes sense. I hope it does T_T.

Anyway, we can make this problem easier by only considering the case where u = 1. Since the number of roots outside of that larger circle should be the same for each case, then the probability for all cases will be the same as the probability for one case. Here's an altered diagram for this case:



If we can figure out where they intersect, then we can figure out the range of that arc that is outside the circle, and figure out how many roots lay on it! The equation of each circle is:



We can expand the second equation and substitute the first into it to get:



That familiar sqrt(3)/2 expression is encouraging. What about the +1? Remember that our circle is translated over to the right by +1. So if we consider the center of that circle the origin, then they intersect at x = sqrt(3)/2 = cos(+/- pi/6). Aha, now we have a range! If we let our angle be:



(this is the general angle of a 1997th root of unity)

Then it lies within the range:



Since the floor of 1997 / 12 is 166, then there are 166 positive integers and 166 negative integers in this range. We aren't counting k=0 because this wouldn't be a distinct root. Therefore, we have 332 total roots in this range. We are selecting them from 1996 roots (remember, that root we already picked is off limits), therefore, the probability is:



And our final answer is 499 + 83 = 582.

The lesson learned? When doing a problem with complex numbers, take some time to consider which way it's more profitable to interpret it: algebraically, or geometrically, as complex numbers are essentially a mix between these two topics.


OH. So why aren't I in NHS? Because it's a cult where all they do is struggle to sell crap candy and do random community service for NHS points, the currency of NHS. You gain nothing from joining, besides maybe having an extra activity on your college resume. You lose your soul. Not a great trade in my oppinion. Also, they appreciate kids who stack on the L4 classes more than the kids who actually focus on subject and become really good at it. And their advisor is a stubborn old man too, according to most of my friends in NHS.

Induction

This is a problem that I tried a while ago and got nowhere with. Looking at it again, it seems rather tame and easy. Maybe this is because it was a problem set from the induction lecture, giving a rather large hint as to what to do. Nonetheless, I enjoy whenever this happens because it shows that I'm actually making progress.



Proceed by INDUCTION. LECKE.

Start with n=3. Since the distances to each person are different, we can conclude that they form a scalene triangle. Take the shortest of its lengths - these two people shoot eachother, leaving the third man unsprayed.

Assume its true for n=k, k-2, ..., 3, show its true for n=k+2:

Again, consider the shortest attainable distance. Depending on the positioning of the people, this shortest distance could occur between a number of people. Assume for now that it only occurs once (we'll address this soon enough). These two people shoot each other. Now consider the remaining k people. If any one of them shoots one of these two people, then at least one person will be left dry, as we have k-1 shots left but k people who haven't been sprayed. Thus, we can assume that they form their own little "spraying group", if you will, of k people. Since we already showed that at least one of these people will be left dry, we're done.

Note that if we have more than one of those "shortest distances" then all that happens is that instead of having a group of k people, we would now have a group of k-2, k-4, or something else depending on how many of those there are.

What if, as a result of these shortest distance cases, we are left with a group of only 1 person? Although we didn't prove the base case of n=1, we can just point out that if this occurs, we're left with one dry person, which is enough to prove our claim.

Olympiad Diophantine Equation

This was one of my practice olympiad problems I had to do. It was pretty tough. I struggled with it for most of the time until I got a mediocre solution with a lot of holes in it. However, it was a rewarding experience. You don't learn how to tackle the tough problems unless you actually get engaged with one. I'll be explaining the complete solution.



I forgot to include that m and n are positive btw.

It's pretty intimidating. What's so intimidating about it? For one, the variables in the exponents make the equation funky looking. Second, at first glance, there's absolutely nothing we can do with it. There's nothing to factor. We do notice the square on the right hand side of the equation for it, which I don't show here. Does that suggest a difference of squares? Only if one of m or n are even. Can we show that it is? If we can, it's certainly not intuitive how.

By the way, before I forget, for any Diophantine equation, try playing around with small numbers to see if you can find any solutions. Here, a fairly obvious solution is (4,2), as this yields one side of a Pythagorean triple. Whether or not you can find any solutions will help you understand where you might need to steer your proof. The fact that both m and n are even in this solution suggests even more so that our previous idea is on the right track.

So back to the problem. We have an idea of where to go. Seeing as I don't have any better ideas, it's worth exploring. How do we show that m or n is even? We can start by breaking down the equation using mods. What does this mean? It means we look at the equation under as many mods as we see fit. What sort of mods should jump out right away? 2 and 3 might come in handy later. Because the quadratic residues are so limited mod 4 and 8, it will probably be helpful examining the equation under these. And if you don't know what quadratic residues are, I'll side track to explain them:

How many solutions are there to n = 3 mod 4? Obviously an infinite amount, whenever n is in the form 4m + 3. This is basic modular notation.

How many solutions are there to n^2 = 3 mod 4? Turns out that there are none. How can we show this?

Every number will be congruent to either 0, 1, 2, or 3 mod 4. Therefore, every number in the form n^2 will be congruent to o^2, 1^2, 2^2, or 3^2 mod 4. This table explains it pretty well:





We can take this idea and apply it to a number of different things. The powers of an integer modulo n, cubes modulo n, etc. Lets look at the powers of 2 and 3 modulo 4:




We can show by induction that all odd powers of 3 =3 mod 4, and all even powers =1 mod 4. It's pretty simple to do so I won't bother explaining it. We can make a similar argument for powers of 2 where the exponent is greater than 1, it will be congruent to 0 mod 4.

Since all squares are congruent to 0,1 mod 4, then we only have two cases for our powers of 2 and 3: 2 reduces to 2 mod 4 (or in other words, our power of 2 is 1), and 3 reduces to 3. Or, 2 reduces to 0 mod 4, and 3 reduces to 1. Playing around with smaller numbers for the first case suggests that it won't work out, and we can easily prove it:




We can reduces all squares modulo 3 to find that they're congruent to 0,1 mod 3. However, its easy to see that the expression is always congruent to 2 mod 3.

So we can conclude that n must be even. Let n = 2x. Then we can subtract and factor:




Clearly both factors must be nonegative powers of two. Writing some equations and subtracting them yields:




Taking both sides mod 2 and remembering that x is a positive integer, we see that our power of two must be 1, giving us the equation:




Trying small values of x, we quickly see x = 1 works, meaning n = 2. Putting this back into our original equations, we can determine that m = 4. We now have one solution, (4,2).

If we try values of x larger than 1, it seems as if the RHS of that equation will never be a power of two. Lets try to prove it. We try looking at it modulo 8, because for x > 1, the RHS will always be greater than 8, and since its supposed to be a power of two, must be divisible by 8:



Again we can induct to show that it's always congruent to 3 or 1 modulo 8. And since we're adding 1, the LHS is congruent to 4 or 2 modulo 8. So it'll never be a power of 2 for x > 1. Thus, our only solution is (4,2).

That problem was pretty mind numbing, mainly because of all the modular residues we had to examine. However, it is a great example of how to break down number theory problems using modular arithmetic. To sum everything up, here's my main approach for diophantine equations that don't like immediately solvable by some method:

1) Quickly look for some solutions
2) Look for ways to factor it
3) Examine it under different modulos to narrow down the possibilities of the variables.

Also, here are the general tactics that ACoPS proposes:

- Is the problem in "simple" form? Always make sure that you have divided out all common factors, or assume the variables share no common factors.

-Do there exist solutions? Sometimes you cannot actually solve the equation, but you can show that at least one solution exists.

-Are there no solutions? Quite frequently, this is the first question to ask. As with argument by contradiction, it is sometimes rather easy to prove that an equation has no solutions. It is always worth spending some time on this question when you begin your investigation.
(I'll add here that this is why you should look for some quick solutions first, so you don't get caught up trying to prove that there are no solutions whatsoever)

-Can we find all solutions? Once one solution is found, we try to understand how we can generate more solutions. It is sometimes quite tricky to prove that the solutions found are the complete set.

Oomph. So how are you? I'm visiting WPI tomorrow and monday, I'm planning on visiting RPI in november, and I'm also going to plan a trip to MIT sometime before winter. I'm not as stressed out about it anymore, as I took a step back and realized that once I switch from my retarded french guidance counselor I'll be better off. And now I'm very tired and while I would love to do more math, I have a bunch of tests to study for next week.

I'm back!

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!

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 The sum of their square roots is the square root of an integer. What is the maximum possible sum of the two integers?

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 is divided by is 1.

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:

Whenever you see inverse trig functions, try to get rid of them by applying trig functions to the equation.

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.