American Invitational Mathematics Examination

The competition consists of 15 questions of increasing difficulty, where each answer is an integer between 0 and 999 inclusive. Thus the competition effectively removes the element of chance afforded by a multiple-choice test while preserving the ease of automated grading; answers are entered onto an OMR sheet, similar to the way grid-in math questions are answered on the SAT. Leading zeros must be gridded in; for example, answers of 7 and 43 must be written and gridded in as 007 and 043, respectively.

Concepts typically covered in the competition include topics in elementary algebra, geometry, trigonometry, as well as number theory, probability, and combinatorics. Many of these concepts are not directly covered in typical high school mathematics courses; thus, participants often turn to supplementary resources to prepare for the competition.

One point is earned for each correct answer, and no points are deducted for incorrect answers. No partial credit is given. Thus AIME scores are integers from 0 to 15 inclusive.

Some historical results[3] are:

A student's score on the AIME is used in combination with their score on the AMC to determine eligibility for the USAMO or USAJMO. A student's score on an AMC exam is added to 10 times their score on the AIME to form a USAMO or USAJMO index.

Since 2017, the USAMO and USAJMO qualification cutoff has been split between the AMC A and B, as well as the AIME I and II.[4] Hence, there will be a total of 8 published USAMO and USAJMO qualification cutoffs per year, and a student can have up to 2 USAMO/USAJMO indices (via participating in both AMC contests). The student only needs to reach one qualification cutoff to take the USAMO or USAJMO.

During the 1990s it was not uncommon for fewer than 2,000 students to qualify for the AIME, although 1994 was a notable exception where 99 students achieved perfect scores on the AHSME and the list of high scorers, which usually was distributed in small pamphlets, had to be distributed several months late in thick newspaper bundles.

The AIME began in 1983. It was given once per year on a Tuesday or Thursday in late March or early April. Beginning in 2000, the AIME is given twice per year, the second date being an "alternate" test given to accommodate those students who are unable to sit for the first test because of spring break, illness, or any other reason. However, under no circumstances may a student officially participate both competitions. The alternate competition, commonly called the "AIME2" or "AIME-II," is usually given exactly two weeks after the first test, on a Tuesday in early April. However, like the AMC, the AIME recently has been given on a Tuesday in early March, and on the Wednesday 8 days later, e.g. March 13 and 20, 2019. In 2020, the rapid spread of the COVID-19 pandemic led to the cancellation of the AIME II for that year. Instead, qualifying students were able to take the American Online Invitational Mathematics Examination, which contained the problems that were originally going to be on the AIME II. 2021's AIME I and II were also moved online., 2022's AIME I and II were administered both online and in-person, and starting from 2023, all AIME contests must be administered in-person. [5]

• Given that

${\displaystyle {\frac {((3!)!)!}{3!}}=k\cdot n!,}$

where ${\displaystyle k}$ and ${\displaystyle n}$ are positive integers and ${\displaystyle n}$ is as large as possible, find ${\displaystyle k+n.}$ (2003 AIME I #1)

Answer: 839

• Find the number of ordered pairs of integers ${\displaystyle (a,b)}$ such that the sequence

${\displaystyle 3,4,5,a,b,30,40,50}$

is strictly increasing and no set of four (not necessarily consecutive) terms forms an arithmetic progression. (2022 AIME I #6)

Answer: 228

• If the integer ${\displaystyle k}$ is added to each of the numbers ${\displaystyle 36}$, ${\displaystyle 300}$, and ${\displaystyle 596}$, one obtains the squares of three consecutive terms of an arithmetic series. Find ${\displaystyle k}$. (1989 AIME #7)

Answer: 925

• Complex numbers ${\displaystyle a}$, ${\displaystyle b}$ and ${\displaystyle c}$ are the zeros of a polynomial ${\displaystyle P(z)=z^{3}+qz+r}$, and ${\displaystyle |a|^{2}+|b|^{2}+|c|^{2}=250}$. The points corresponding to ${\displaystyle a}$, ${\displaystyle b}$, and ${\displaystyle c}$ in the complex plane are the vertices of a right triangle with hypotenuse ${\displaystyle h}$. Find ${\displaystyle h^{2}}$. (2012 AIME I #14)

Answer: 375

[6]

• American Mathematics Competitions
• List of mathematics competitions
• Mandelbrot Competition

• Official website
• Complete Archive of AIME Problems and Solutions