# AIME Math Contest Preparation

The American Invitational Mathematics Exam (AIME) is a math contest in the United States. This page outlines the contest details and topics covered, providing relevant wikis and quizzes for training prep and practice.

#### Contents

## Contest Information

The American Invitational Mathematics Exam (AIME) is a 15 question, 3 hour exam, administered twice each year. Students who do well on the AMC 10 or AMC 12 exams are invited to participate in the AIME, and a high score on AIME can lead to qualification for the United States of America Mathematics Olympiad (USAMO).

Each answer to the AIME is an integer number from 0 to 999. No calculators are allowed.

## Basic Topics

These are the basic topics covered by the AIME, which usually appear in the first 5 problems. This is a great place to start learning if you’re new to the AIME!

$\large \textbf{Algebra}$

$\large \textbf{Geometry}$

ChapterWikiQuizQuadrilaterals Volume Circle Properties Area of Triangles Trigonometric Functions

$\large \textbf{Combinatorics (Counting and Probability)}$

$\large \textbf{Number Theory}$

ChapterWikiQuizNumber Bases Introduction to Recursion GCD/LCM Prime Factorization and Divisors

## Intermediate Topics

These are the intermediate topics covered by the AIME, which usually appear in the middle 5 problems. This is a great place to hone your skills if you are already comfortable with solving the first few problems on the AIME.

$\large \textbf{Algebra}$

$\large \textbf{Geometry}$

$\large \textbf{Combinatorics (Counting and Probability)}$

$\large \textbf{Number Theory}$

## Advanced Topics

These are the advanced topics covered by the AIME, which usually appear in the last 5 problems. This is a great place to learn to solve the hardest problems on the AIME if you're shooting for a perfect score!

$\large \textbf{Algebra}$

$\large \textbf{Geometry}$

$\large \textbf{Combinatorics (Counting and Probability)}$

$\large \textbf{Number Theory}$

## A Difficult Sample Problem Showing General Tips and Tricks

There are many tricks to the AIME that don't show up in a "standard school curriculum." To illustrate, here is an example problem from 2007:

2007 AIME 1, Problem #14Let $m$ be the largest real solution to the equation

$\frac{3}{x-3} + \frac{5}{x-5} + \frac{17}{x-17} + \frac{19}{x-19} = x^2 - 11x - 4.$

There are positive integers $a,$ $b,$ and $c$ such that $m = a + \sqrt{b+\sqrt{c}} .$ Find $a+b+c .$

Note that this is in the "hard" set, yet this problem is entirely solvable with regular algebra and even follows the standard logic wanting to isolate $x$ by writing the problem as

$\text{(some factored polynomial)} = 0 .$

However, the specific tricks are unusual; finding them requires practice. This one makes clever use of **rearrangement**, **substitution**, and **symmetry**.

We incidentally can find right away that $x = 0$ is a solution:

$-1 + -1 + -1 + -1 = -4,$

so if we find all other solutions are negative, $0$ is our answer. $\big($However, we are essentially guaranteed in a competition context given the setup of $m = a + \sqrt{b+\sqrt{c}}$ that there will be a positive root.$\big)$

Constants tend to be one of the easiest terms to get rid of; either they can be transformed with substitution (as you'll see later) or they can be "shuffled into" some other part of the algebra. While there's not much possible with the $-4$ on the right-hand side of the equal sign, is there some potential on the left? Let's see:

$\begin{aligned} \frac{3}{x-3} + \frac{5}{x-5} + \frac{17}{x-17} + \frac{19}{x-19} &= x^2 - 11x - 4 \\\\ \frac{3}{x-3} + \frac{5}{x-5} + \frac{17}{x-17} + \frac{19}{x-19} + 4 &= x^2 - 11x. \end{aligned}$

There are four terms exactly to go with the 4, so what could happen if did **rearrangement** and split the 4 up to give 1 to each of the terms? We have

$\begin{aligned} \frac{3}{x-3} + 1 + \frac{5}{x-5} + 1 + \frac{17}{x-17} + 1 + \frac{19}{x-19} + 1 &= x^2 - 11x \\\\ \frac{3+x-3}{x-3} + \frac{5+x-5}{x-5} + \frac{17+x-17}{x-17} + \frac{19+x-19}{x-19} &= x^2 - 11x \\\\ \frac{x}{x-3} + \frac{x}{x-5} + \frac{x}{x-17} + \frac{x}{x-19} &= x^2 - 11x. \end{aligned}$

This is a fairly common competition trick: writing $\text{(some fraction)} + 1 ,$ $\text{(some fraction)} -1 ,$ or $1 - \text{(some fraction)}$ can produce useful results.

Since we're looking for non-zero roots, we can safely divide both sides of the equal sign by $x:$

$\frac{1}{x-3} + \frac{1}{x-5} + \frac{1}{x-17} + \frac{1}{x-19} = x - 11.$

Unfortunately, maneuvering the right side of the equal sign now isn't helpful, so let's look at the terms on the left side instead. There's a definite **symmetry** pattern going on: $-3$ and $-5$ are two apart, and $-17$ and $-19$ are two apart. The symmetry isn't quite written in a form useful to us, but we can use **substitution** to help bring it out.

Let $Q = x - 11 .$ Note that $-11$ is midway between $-3$ and $-19$ as well as $-5$ and $-17;$ we picked this number specifically to expose symmetry. It also (by "coincidence") happens to match the right side of the equal sign:

$\begin{aligned} x - 3 = x - 11 + 8 &= Q + 8 \\ x - 5 = x - 11 + 6 &= Q + 6 \\ x - 17 = x - 11 - 6 &= Q - 6 \\ x - 19 = x - 11 - 8 &= Q - 8. \end{aligned}$

Plugging these back into the original equation $($including the right side which has $x -11 ),$

$\frac{1}{Q + 8} + \frac{1}{Q + 6} + \frac{1}{Q - 6} + \frac{1}{Q - 8} = Q.$

It's tempting to note the matching of +8/-8 and +6/-6. Let's do some more rearrangement and combine the terms:

$\begin{aligned} \frac{1}{Q + 8} + \frac{1}{Q - 8} + \frac{1}{Q + 6} + \frac{1}{Q -6} &= Q \\ \frac{Q-8}{(Q + 8)(Q-8)} + \frac{Q+8}{(Q + 8)(Q-8)}+ \frac{Q-6}{(Q + 6)(Q-6)} + \frac{Q+6}{(Q + 6)(Q-6)} &= Q \\ \frac{2Q}{Q^2-64} + \frac{2Q}{Q^2-36} &= Q. \end{aligned}$

Not forgetting that $x = 11$ is a solution to the original problem, divide by $Q$ on both sides:

$\frac{2}{Q^2-64} + \frac{2}{Q^2-36} = 1.$

We can make the squared terms easier to deal with using another substitution; let's use $R=Q^2:$

$\frac{2}{R-64} + \frac{2}{R-36} = 1.$

Then get all terms on one side of the equal sign:

$\begin{aligned} 2(R-36) + 2(R-64) &= 1(R-64)(R-36) \\ 2R-72 + 2R-128 &= R^2 - 100R + 2304 \\ 0 &= R^2 - 104R + 2504. \end{aligned}$

Applying the quadratic formula here (and only worrying about the larger root) gets $R = 52 + \sqrt{200} .$ Remembering that $R = Q^2$ (and again only worrying about the larger root), $Q = \sqrt{ 52 + \sqrt{200} } .$ Finally, remembering that $Q = x -11 ,$

$x = 11 + \sqrt{ 52 + \sqrt{200} }.$

Almost done! The problem asks about $a + b + c$ in the root $m = a + \sqrt{b+\sqrt{c}} ,$ so our desired answer is $11 + 52 + 200 = 263 .$

Let's summarize what was used:

Rearrangement to move the $-4$ term, which caused each of the terms on the left hand of the equal sign to get an $x$ in the numerator.

Dividing the $x$ out.

Using a substitution $Q = x - 11$ to expose the symmetry that allowed the terms to be more easily combined.

Using another substitution $R = Q^2$ to avoid dealing with the squared term until necessary.

Doing standard algebra involving making a quadratic $ax^2 + bx + c = 0$ and then using the quadratic formula.

Back-substituting to solve for the largest real solution for $x .$

Note while the entire solution as a whole can be intimidating, each individual step is a piece of ordinary algebra. Once you start to spot these sorts of tricks more easily, you can solve even the hardest of AIME problems!

**Cite as:**AIME Math Contest Preparation.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/aime-math-contest-preparation/