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.

Combanatorics Identity

My blog coder is a piece of junk and won't do binomial notation for combinations (nevermind, I figured it out later), so I'll have to code that part on some other site.

Prove the following combinatorial identity algebraically:




Show that the RHS is equal to the LHS. Use our definition for a combination:





And we have to show that is the same as this. Expand both forms as above.



Factor out a n! :



Heres where things get tricky. Looking at our definition for
And

Meaning we can rewrite our original as

We can use this to alter each of the fractions to a common base.




And.




Substituting the last line of each into our fraction gives us a common base:




Add the two:



Multiply the n! back in:



Remembering our work before:


And:




So now we can rewrite it:



Which is the expression we were trying to equate it to, JUST AS PLANNED.

This identity can also be explained logically.

Consider some particular object of a group of n + 1 objects, and call it A. Then we can choose k objects which include A in only ways, because after including A, we must pick k-1 of the remaining n objects.

Also, consider picking k objects from the group which DON'T include A. This can be done in ways, because after removing A from our group, we have n objects to choose k amount from.

However, by counting all combinations which do include A and which don't include A, we have counted all possible combinations of picking k objects from a group of n+1 objects, leading to the conclusion that:



JUST. AS. PLANNED.

Logarithms

So here's some stuff about logarithms. I assume you know all of the basic properties of logs. If you don't, go look them up here:

http://www.artofproblemsolving.com/Wiki/index.php/Logarithms

Or anywhere else you see fit.

The following type of problem is incredibly common:

If and , express in terms of P and Q

In log problems where the bases are different, we usually have to use change of base formula. Remember, the base we are changing to doesn't HAVE to be the common base 10 (that's the way it is usually taught in school, so logs can be estimated on the calculator). It can be any base, as long as the bases are the same in the numerator and denominator of the changed expression.

Seeing that the expression for Q and the expression we are trying to simplify are both taking the log of 5, we should change to the base of Q, which is 3.



Notice that 10 = 5 * 2. We can rewrite the bottom logarithm to include another Q:




Noticing that the log contains a 3, changing to base 8 will allow us to include a P. Also, 2 is a power of 8, so we will be able to eliminate the log completely.




I know that last expression as the solution is a bit ugly, but I don't think it simplifies.

A lot of times, with those kind of log problems, you really do have to play around with it, or "guess and check" (I hate, hate, HATE guess and check). Obviously it's not all guess and check. Being able to recognize when certain properties can be used to simplify it is key. But usually you're not really heading for the answer right away, if you know what I mean. Rather, you're just sort of tinkering with it until it clicks, and then you start heading for it. I'm sure this method applies to other problems too, but I find it noticeably with these sort of "this is p, this is q, whats this in terms of p and q" problems.

You know what it is

So I figured I'd turn this into a sort of rant / journal / math blog. Unfortunately, I don't have a good rant for today. So here's the problem:

Oh, by the way, this problem is part of an entrance exam for a math camp which is STILL TAKING APPLICATIONS. So if any of you feel the desire to apply, heres the answer to #1.

Write 1,000,000 as a sum of a prime and a perfect square

Obviously this could be solved through guess and check(cough cough), but that's a lot of numbers to go through. Doing it algebraically is a lot faster:
Let p be the prime and n be the integer that is being squared.
Also realize that




We've seen this enough, it should be apparent. Factor the right side as difference of squares:



Now we can use our definition of a prime number to help us. The only factors of a prime number are 1 and itself. From our equation above, our two factors are (1,000 + n) and (1,000 - n). Which means one must be 1, and one must be a prime number. Let's say 1,000 + n is the prime, and 1,000 - n is 1. (Note: Obviously the only way 1,000 + n could be 1 is if n is negative, which makes 1,000 + (-n) become 1,000 - n, and makes 1,000 - (-n) become 1,000 + n. Since negative or positive n results in the same thing, we'll just have n be positive, it's easier to work with.)

So now we have:

which is easily solved as:


Now we have to check to see if 1000 + n is a prime.

Check this against a list of primes, and it fits. So,

Rewritten as a sum of a prime and a perfect square,


Not too difficult. Hopefully you're seeing a pattern. When you have two squares, you can usually factor them and then analyze the factors in some way.

I have another proof which I really want to type up, but I have to show some self discipline and go do homework, so I'll get to that some other time. Btw, it's also another answer to the entrance exam.