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!
\[\large \textbf{Algebra} \]
\[\large \textbf{Geometry} \]
Chapter Wiki Quiz Quadrilaterals 
Volume Circle Properties Area of Triangles Trigonometric Functions
\[\large \textbf{Combinatorics (Counting and Probability)} \]
\[\large \textbf{Number Theory} \]
Chapter Wiki Quiz Number 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 #14
Let \(m\) 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:
\[\begin{align} \frac{3}{x-3} + \frac{5}{x-5} + \frac{17}{x-17} + \frac{19}{x-19} &= x^2 - 11x - 4 \\\\ \frac{3}{x-3} + \frac{5}{x-5} + \frac{17}{x-17} + \frac{19}{x-19} + 4 &= x^2 - 11x. \end{align}\]
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
\[\begin{align} \frac{3}{x-3} + 1 + \frac{5}{x-5} + 1 + \frac{17}{x-17} + 1 + \frac{19}{x-19} + 1 &= x^2 - 11x \\\\ \frac{3+x-3}{x-3} + \frac{5+x-5}{x-5} + \frac{17+x-17}{x-17} + \frac{19+x-19}{x-19} &= x^2 - 11x \\\\ \frac{x}{x-3} + \frac{x}{x-5} + \frac{x}{x-17} + \frac{x}{x-19} &= x^2 - 11x. \end{align}\]
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:
\[\begin{align} x - 3 = x - 11 + 8 &= Q + 8 \\ x - 5 = x - 11 + 6 &= Q + 6 \\ x - 17 = x - 11 - 6 &= Q - 6 \\ x - 19 = x - 11 - 8 &= Q - 8. \end{align}\]
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:
\[\begin{align}
\frac{1}{Q + 8} + \frac{1}{Q - 8} + \frac{1}{Q + 6} + \frac{1}{Q -6} &= Q \\
\frac{Q-8}{(Q + 8)(Q-8)} + \frac{Q+8}{(Q + 8)(Q-8)}+
\frac{Q-6}{(Q + 6)(Q-6)} + \frac{Q+6}{(Q + 6)(Q-6)} &= Q \\
\frac{2Q}{Q^2-64} + \frac{2Q}{Q^2-36} &= Q.
\end{align}\]
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:
\[\begin{align} 2(R-36) + 2(R-64) &= 1(R-64)(R-36) \\ 2R-72 + 2R-128 &= R^2 - 100R + 2304 \\ 0 &= R^2 - 104R + 2504. \end{align}\]
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!
