AIME

I don't really feel like posting any problems or anything, I've just been kind of tired lately. However, I will say that I've signed up for an AIME prep course (I'm fairly confident that I can pass the AMC without taking a course). It's an online, weekly 90 minute class length course which discusses AIME problem solving, "aimed" (lolololol) towards specific test strategies, as well as going over techniques which will be helpful for approaching the different categories of problems. The schedule looks like this:

Week 1: Equations
Week 2: Complex Numbers and Polynomials
Week 3: Functions
Week 4: Inequalities, Optimization Problems, Sequences and Series
Week 5: Counting
Week 6: Probability
Week 7: Number Theory
Week 8: Number Theory (Algebraic Methods)
Week 9: Trigonometry and Analytic Geometry
Week 10: Euclidean Geometry
Week 11: Euclidean Geometry
Week 12: Extending Your Geometric Range

"Extending Your Geometric Range" looks kind of funny. Like they're trying to be sneaky and not give away even a scent of the "secrets of geometry" that they'll be sharing, so they just call it that to not give any hints. My first class is next week, so I'm looking forward to it. Looking at the schedule, the emphasis seems to be on counting, number theory, and geometry, starting at the second half of the class, while the first half is more general stuff. And I have a math test next week. I've been struck with a 98 streak, so hopefully I'll finally break free of it and get the triple digits.

Btw, I know that sounds absurd (not being happy with a 98), but a 100 is well within my reach, so it's more that I'm unhappy with myself not living up to what I'm capable of.

Inequalities and Optimization I

The following is a compilation of ideas and theorems on inequalities.

General Inequality:

The square of any real number will always be non-negative, or in other words,



for all real x.

Prove that for all real numbers x and y, that:



Typically when you find your solution, you'll want to work backwards, that is, start with what you want to prove, and turn it into something you know is true. Then, when you present your solution, take what you know to be true, and use it to show what you want to prove. I'll present the above with the second method.

From our general inequality:









Rearrane to make our factoring a bit more apparent:



This now factors into:



Tadaa.



Arithmetic-Geometric (AM-GM) Inequality:

For positive numbers , the arithmetic mean of them will be greater than or equal to their geometric mean, or in other words:




Find the maximum value of:



Call the maximum y. Rewriting this in terms of inequalities:



Invert both sides:



It's important to understand that when we invert both sides, the inequality sign is reversed. This is because the manual steps for inverting both sides ends up moving each term to the other side of the inequality.

Divide out the left side:



Rearrange the left side:



Applying AM-GM to our two x terms:



The two x terms under the radical cancel out and turns into 9, and the entire RHS simplifies to 3.





Looking back at:



The only way we can minimize the left side is if we minimize the x terms. Since we've just determined the minimal value of them is 6, we can now write:





And there's our minimum, 1/9 .

However, there's an inconsistency in this solution (at least one which we have not addressed). We used AM-GM to find the maximum, but AM-GM only works for positive numbers. This means that we assumed x and 9/x are positive values. The reason we can do this without any errors is because our maximum will occur for a positive value of x. This can be confirmed by use of a graphing calculator (which is a big no-no on math competitions lawl), or algebraically.

When x is positive, the entire fraction is positive. When x is 0, the entire fraction is 0. When x is negative, the numerator will be 0, but since the minimum value of the quadratic in the numerator is positive (check the vertex), then the entire fraction will be negative. Therefore, the maximum value only occurs when x is positive.

The next post will be a follow up problem which uses these methods for solving it.

Complex Numbers (AMC 12A - #23)

OMFG COMPLEX NUMBERS!!

The solutions of the equation are the vertices of a convex polygon in the complex plane. What is the area of the polygon?


Wow, at first look, that looks really complicated. It's not. Things you should know:

A good understand of polar form
How to find roots of complex numbers

Ok, one thing should pop out immediately, and thats the 1-4-6-4-1 form of the equation. I hate solutions to problems where they said "it should be apparent that...". Well if it was so fucking apparent, then I would've gotten it right. Unfortunately, that's what this comes down to. The reason 1-4-6-4-1 is important is because those are the general coefficients of the binomial expansion .

Seeing the various negative signs and i coefficients, you should think of . Again, I hate saying that "you should think of", but once again, that's really what it comes down to. Actually expanding this is:





A little different from our actual equation, but we can easily alter to to fit it:







Now we can use De Moivre's Theorem to find the fourth roots of i + 1

First convert i + 1 to polar form:



Now find the four fourth roots:










So now what? We have a bunch of angles with which to attach a radius value to, AND once we convert all of them to rectangular, we have to subtract an i. Anybody can tell that the values are going to get very, very ugly. Luckily, we can take a shortcut. We only have to find the area of the polygon formed by the four solutions. That means:

The area will be unchanged if we translate it.
The area will be unchanged if we rotate it.

So how can we make our points easier to look at? Simple. First off, forget about subtracting i from each one. All that would do is move all of the points down one, which is translation, so it doesn't matter if we don't do it, we'll get the same area.

Second, we can subtract 11.25 from each of the angles. All this will do is rotate each of the points around the center of the polygon, keeping the area the same.

So now, the points of our polygon are:



Turn them into rectangular:



So now we have a square with its vertices on the axes. Using elementary properties of a square, we can easily solve for one of the sides of it, which is .

Squaring this, the area of our polygon is:

Word problem

Word problems can kill you if you aren't good at rewriting them in the form of an equation that can be solved. Here's an example:

Given three equally spaced candles, the middle candle is 3 times the height of the other two. From left to right, the candles burn down in 9, 6, and 18 hours respectively. After t hours, all of the tops of the candles are collinear. Determine the smallest possible value for t.


Is collinear spelled right? My spell checker says it is, but for some reason it looks odd to me. Oh well.

A couple of things you should acknowledge off the bat:

We are not given specific heights, just the relation between the heights of the candles. Just write the heights in terms of an unknown x, and when we solve for t, it will most likely divide out.

They are equally spaced. This helps, because we know all that's required for them to be collinear is the same difference in height (I'll explain this better soon. In fact right now).

Collinear means all of the points lay on the same line. Since a line is, well, a line, its slope is constant. That means, for the points to be collinear (for the sake of explanation, I'm going to start calling the points from left to right A, B, and C), the difference between the height of A and B should be the same as the difference between the height of B and C. We already know that their width is constant. A much more challenging problem would have something like slanted candles so the width AND height is changing. Why somebody would set up slanted candles, I don't know. It sounds like a fire hazard.

So now that we know what equality must exist for the points to be collinear, we have an objective: write this equality in terms of t, our variable for time. To make it even clearer, this equality is:

Difference between height of B and height of A = Difference between height of C and height of A.

Just to confirm: Does it matter if we write the difference as A - B = B - C, as opposed to B - A = C - B ? Not really. The equality holds up either way. If you aren't convinced, draw a line, and pick three points with the same distance horizontally between them, and see for yourself.

We can rewrite this again:

Height of B - Height of A = Height of C - Height of B

So we set out to find these heights in terms of t. If a candle is burning at a rate r, then the candle loses height rt after time t. OR, the height of the candle, in terms of its original height H, becomes:

H - rt

We have an original height for each, and a rate of burning, so we can write each of the heights of the candles in terms of t (and our unknown height x, but this will divide out, since both sides of the equation can factor out a X).

Height of A:
Original height: X
Rate: x / 9

If you're confused how we got the rate as x / 9, I'll explain. Using our formula
Distance = Rate * Time
And the fact that the entire candle of "distance" x (Think of it as how fast the top of the candle moves downwards), plus the fact the entire candle burns in 9 hours:

X = 9r
r = x / 9

Alright, there we go.

Plus these back in:



I was hoping that would look prettier.

The process for the others is exactly the same, except that the height for the middle one is 3x, because it's 3 times the height of the others.




Now substitute all of these back into our original equality:





Just like a promised, the x goes away. Divide each side by x:




Get rid of those parentheses and distribute negatives where needed:





Combine the constant terms:




Multiply by 18:




Like terms:




Now just solve for t:






And there's our answer.