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.

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.

Number Theory and problem solving

I'll try to take you through the steps of solving a basic number theory problem:

Find the smallest positive integer whose cube ends in 888.


How do we begin? And keep in mind that this is a non-calculator question, although you should be expected to be able to pump out some basic arithmetic without one.

Perhaps we can start by looking at an easier problem: whats the smallest integer whose cube ends in 8? Hopefully it didn't take you more than a few seconds to see that it's 2. Next you should ask yourself, does this mean that any integer whose cube ends in 8 has to end in 2? No other single digit cubed ends in 8 (you can check). Also, the unit digit in a multiplication will only be affected by the unit digit of the factors. So this should mean that it has to end in 2.

Now lets work our way up - whats the smallest integer whose cube ends in 88? At first this seems a lot harder, but we can use what we already figured out to make it easier. Realize that the units and tens digit will only be affected by the units and tens digit of the products. So technically, we should be able to actually do out the multiplication of cubing "x2", where x is our unknown digit. You'll have to keep track of the tens digits while adding by considering them mod 10 only. I'll try to draw it out as best as I can here:

x 2
* x 2
------------
2x 4
2x
------------
4x 4
x 2
-------------
8x 8
4x
-------------
12x 8

Now our problem is just, what is the smallest digit x such that 12x is congruent to 8 mod 10. Which is just 4. So the smallest integer cubed who ends in 88 is 42.

Now lets try applying this method to 888. We can't assume that the tens digit is 4, but we can assume that the ones digit is still 2. I won't draw out everything here, but letting the number be x y 2 yields:

(12x + 6y^2) (12 y) (8)

This time, when we find a number for y, we can't just take it mod 10- we'll have to consider what is carried over and add it to the left.

Let's deal with y first. If y is 4 or 9, then 12y is congruent to 8 mod 10. Lets try 4 first:

(12x + 6*16) (48) (8)
(12x + 6*16 + 4) (8) (8)
(12x + 100) (8) (8)

So the smallest x such that 12x + 100 is 8 mod 10 is 4. That yields 442 as a possible number. However, trying out 9 might yield a smaller hundreds digit, so lets try it.

(12x + 6*81) (108) (8)
(12x + 486 +10)(8) (8)

(12x + 496) (8) (8)

So the smallest x such that 496 is 8 mod 10 is 1, yielding 192. Certainly much smaller than 442. Can we say for sure that these are the smallest possible numbers? We showed that any number whose cube ends in 888 must end in these numbers. We also explored the different cases, and picked the smallest from each, so there are no other possibilities for a smaller number. Our final answer is thus 192.

Keep in mind that this isn't the only way to do this. We could also have analyzed the expansion of (100a + 10b + 2)^3. Personally, I preferred the previous method because I'm not exceptional at expanding anything past binomials, but it would be good practice. In fact, I'll try it later tonight.

If anybody happens to be reading this, heres a problem for you - Find the smallest positive integer whose cube ends in 69. And do it WITHOUT brute forcing! (Hint, expand it, and take it mod 100).

By the way, it really isn't worth "memorizing" this trick. The most important thing is being able to solve it on your own, without having seen a very similar problem. That doesn't mean you shouldn't try to relate to elementary problems or other ones that you HAVE memorized, though!

What I'm trying to say is, that if you had never seen a problem like this, and after looking at it for a minute, you thought "eh, this is too hard, whatever." then you're NOT TRYING HARD ENOUGH. Here's a prime example: If you've ever looked at the USAMTS problems, they're hard. Maybe not quite USAMO level, or if you don't look at each one for more than an hour, depending on its difficulty, you probably won't solve it. I have a friend, who is very smart and good at math, and I asked him to try USAMTS this year. His response was:

"Nah, I looked at the problems last night and I couldn't get any, so it's probably too hard for me."

I'm going to try to convince him again, but a key part of problem solving is to not back down from a problem. You have to take the time to explore, play around with, and change the problem until you have a good idea of what is going on, before you can actually solve it. Just because you don't see the solution right away doesn't mean it's not there.

The Extreme Principle and Contradiction

The extreme principle is a simple yet very effective way of analyzing many problems. It basically says that, if possible, focus on the "largest" and "smallest" elements of the problem. However, sometimes it is not obvious where this simple insight can be applied, and it requires a lot of creativity to see it.

Contradiction is another powerful tool. When we wish to prove something, we instead assume the opposite, and show that this "contrapositive statement" implies something which we show to be impossible. A simple example of this would be, prove there is no largest integer. Suppose such an integer exists. Call it n. Obviously, we can simply add 1 to that integer to obtain an even larger integer, contradicting that we had the largest one.

Let B and W be a finite set of black and white points, respectively, in the plane, with the property that every line segment that joins two points of the same color also passes through a point of the opposite color. Prove that both sets must lie on a single line segment.

Assume they are not all on the same line segment. Therefore, it must be possible to form at least one or more triangles by drawing all possible line segments. Consider the smallest area possible when connecting three of these points, regardless of what set they are from. Clearly, at least two of the points must be the same color, as we have three points and only two colors. Therefore, there must be another point of the opposite color in between them. However, by using this point and two of the others, we can construct a SMALLER triangle, contradicting that there exists a smallest triangle. This means either one of two things: there are no triangles, or triangles which keep getting infinitely small. Since there are a finite number of points, it is obviously the first one, and if it is impossible to draw a triangle from them, they must all be on the same line. Just as planned.


Consider finitely many points in the plane such that, if we choose any three points A,B,C among them, the area of triangle ABC is always less than 1. Show that all of these points lie within the interior or on the boundary of a triangle with area less than 4. (Korea 1995)

Consider the largest of these triangles, and call it ABC. It is given that [ABC] < style="font-weight: bold;">medial triangle of ABC, which means that A, B, C are the midpoints of that triangle. It is given from basic geometry that [LMN] = 4[ABC] < style="font-style: italic;">P is outside of the triangle. Then we can connect it to two of the vertexes of ABC to create a triangle of area greater than [ABC]. However, this contradicts that [ABC] is the largest triangle. Therefore, all of the points lie inside LMN, and [LMN] < 4. Just as planned.

Greatest Integer Function

Contest writers LOVE the greatest integer function (or GIF for short, although I've never seen it written like that, but I'll be calling it that from now on. Also, the problem will denote whether I'm refering to the floor or the ceiling version). Why do they love it so much? It's certainly not as curvy as its related fractional part function, if you know what I mean. But mostly because it's an odd function, and if you aren't familiar with it / have the creativity to know how to mess around with it to get an idea on how it is used in the problem, then you won't have much luck with that problem.

How many of the first 1000 positive integers can be expressed in the form:



where x is a non-negative real number?


Hm, where to begin? First off, you should note a few things about gif. For when the expression inside of the function is an integer, then it's equivalent to there being no floor function at all. Also, if it is in the form where n is an integer, then it is equivalent to . I'm not sure how to formally prove it, but its basically because the "floor" of the number cuts off the fractional part, so if you think of the number geometrically, you can cut off an integral length from the number, floor it, then reattach it. By no means a rigorous proof. =)

Another interesting thing about the floor function is the way it suddenly jumps from one integer to the next. In other words, if we call it f(x), and n is an integer, then



Anyway, back to the problem, we have multiplication inside of the function. All that means is that specific function advances quicker than the others, if you know what I mean (you can relate this to your study of the trig functions, such as the graph of sin(2x)). However, it turns out to be an important part of the problem, since we are adding several of these together.

For the sake of the problem, define a new function f(x) := that whole expression the problem gives (I'm not retyping it, latex needs to invent an easier way to do the floor function). f(0) = 0, and f(1) = 20 . We need to figure out what values it can take on inbetween x=0 and x=1. Once we do that, we can generalize up to f(50) = 1000.

To determine when f(x) jumps values, you have to consider when each individual part jumps values. Since I know you wont, I'll do it for you (I'm not doing latex for the fractions, it's too tedious):



1/2, 1


1/4 , 1/2 , 3/4 , 1


1/6, 1/3, 1/2, 2/3, 5/6, 1


1/8, 1/4, 3/8, 1/2, 5/8, 3/4, 7/8, 1

Now we can analyze how f(x) behaves. Consider f(1/8). Clearly, the last part of the function jumps up to 1, but the rest dont, so f(1/8) = 1. But what about 1/4? The floor of 8x jumps up, but so does the floor of 4x!, meaning we get 0 + 1 + 0 + 2 = 3. Likewise, when these "jumps" are shared by multiple parts, the function skips integers. For example, when a value is shared by n of them, then from the previous attainable value, it goes up by n, skipping n-1 of the integers in between

1/4 - shared by 2, 1 integer is skipped
1/2 - shared by 4, 3 integers are skipped
3/4 - shared by 2, 1 integer is skipped
1 - shared by 4, 3 integers are skipped

From this, we can conclude that for positive numbers not greater than 1, f(x) takes on 12 distinct values, seeing as there are 20 possible values, and 8 of them are skipped. We can generalize this for all values of x, using the property previously mentioned, to determine what values are skipped. In other words, for every 20 integers, only 12 of them are representable in the desired form. It's not coincidence that 1000 is a multiple of 20. Our final answer is thus:



Oomph. So how are you doing? Yesterday I stayed up till 6, and still couldn't sleep, so became reduced to watching reality tv shows of retired musicians on MTV. It was a good time. Tonight I'll try to go to bed earlier though.

Random quote and an easy problem

"Guys have the romantic idea that you don’t poop. I doubt that shattering that dream would ruin an otherwise great relationship too terribly."

Hehe.

Anyway, I keep getting stumped by problems out of my books so I needed a breather. Here's an easier one that I quickly made up.

Prove that infinitely many of the Fibonacci numbers are multiples of 3

Really all you have to do is look at them mod 3:

1, 1, 2, 3, 5, 8, ....

In mod 3, goes:

1, 1, 2, 0, 2, 2, 1, 0, 1, 1, ....

And as you can see, this sequence repeats, making infinitely many numbers congruent to 0 mod 3.

It is now 5:51 in the AM. I really need to even out my schedule. And still no luck with that other problem. I haven't given up though.

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 . WLOG, assume that since they are not all the same, is the smallest number. Then, according to the problem:





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

and

Adding these yields:



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.

A fresh look at an old problem

A while ago I did the following problem:
Consider all 5 digit numbers (from 10,000 to 99,999). Call a 5 digit number weakly increasing if the first two digits (ten-thousands and thousands) are strictly increasing from left to right, and the last 3 digits (hundreds, tens. and ones) are strictly increasing from left to right. How many weakly increasing 5 digit numbers are there

If you read my solution, you'll notice that it was tedious and had a LOT of casework. However, I'm a bit more experienced now and have a much simpler solution.

Notice that for any arrangement of digits, there is exactly one way to arrange them in increasing order. So for the first two digits, (only 9 digits because we can't use 0, it will always be the first digit and that makes it a 4 digit number), and for the last three digits, . So our final answer is

The lesson to be learned from this is, even when you've solved a problem, try to solve it in a different way. If you have time, it's the best method to check your work, and it can give you new insights too.

Summer

Haven't posted in this for a while...(not that anybody reads it). But anyway, of course it's summer now. So far mine has consisted of the following:

Math studies - The AIME class is going well, but the problems are getting a lot harder. Most of them I'll have to save and look at later in the year when I'm more capable of handling them. I'm considering asking my parents to sign me up for an olympiad training class in the fall, since I didn't ask for anything for my birthday. I also bought some new books, but amazon is taking a while to ship them, so I won't have all of them for probably 2 or so more weeks.

Summer homework - I've barely started this. I've glanced over the calculus and physics homework, and it didn't seem too bad, but I should probably dedicate at least a day to each chapter. The law will probably be the thing that kills me. We have to read a ton of stuff. I'll admit, it's not that much, but I hate reading. I'm just reading a bit at a time, but I'm trying to get it out of the way before August.

Video games - Ahhhh. I haven't played good video games since I was obsessed with Diablo II during the winter. I'm glad to say I've gotten back into them:
Fire Emblem: Radiant Dawn - I need to finish my file, I got to the very end and then forgot to finish =P
House of the Dead 2 - I have an AWESOME boss time for the magician. Seriously, it's amazing. 48 seconds.
Final Fantasy VII - I consider myself to be a final fantasy fan, and yet I've never gone all the way through FF7. I think I tried to once, but my discs were messed up. I guess it's time to try again.
I might have to reinstall Counter Strike too...

Trolling - There's not much to say about this. We go on anime forums (Ironically, I'm an anime fan), and try to piss off people who only want to role play. It's a pretty good time. Last night we took over a thread for 59 pages. I think night-time trolling is easier because that's when the troll-able material comes out.

Not much else is going on. I just remembered I need to study vocab for SAT in the fall since I didn't do that well on that section when I last took it. No problem because I'm very tired and should probably take a nap before class tonight.