en>Mecanismo: uniform structure for the intro to the eliptic and parabolic equations articles

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{{unreferenced|date=December 2012}}
The '''American Invitational Mathematics Examination (AIME)''' is a 15-question 3-hour test given since 1983 to those who rank in the top 5% (or score at least 100) on the [[American Mathematics Contest|AMC 12]] high school mathematics contest (formerly known as the AHSME), and starting in 2010, those who rank in the top 2.5% (or score at least 120) on the [[American Mathematics Contest|AMC 10]].

The AIME is the second of two tests used to determine [[United States of America Mathematical Olympiad#Qualification|qualification]] for the [[United States of America Mathematical Olympiad]] (USAMO), the first being the [[American Mathematics Competitions|AMC]].

The use of calculators is not allowed on the test.

==Format and scoring==

The exam consists of 15 questions, where each answer is an integer between 0 and 999 inclusive. Thus the test effectively removes the element of chance afforded by a multiple-choice test while preserving the ease of automated grading; answers are entered onto an [[Optical mark recognition|OMR]] sheet, similar to the way grid-in math questions are answered on the [[SAT]].

Concepts typically covered on the exam 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 exam.

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 recent results are:

<center>

{| class=wikitable
|----- align="center"
| Year || Students<br>sitting || Mean<br>score
| Median<br>score || Students with<br>perfect scores
|----- align="center"
| 2006 || 22764 || 2.741 || - || 4
|----- align="center"
| 2005 || 12476 || 2.717 || 2 || 1
|----- align="center"
| 2003 || 13444 || 3.059 || 3 || 3
|----- align="center"
| 1999 || 11945 || 2.195 || 2 || 4
|}
</center>

A student's score on the AIME is used in combination with their score on the [[American Mathematics Competitions|AMC]] to determine eligibility for the [[USAMO]]. A student's score on the AMC is added to 10 times his/her score on the AIME. In 2006, the cutoff for eligibility in the USAMO was 217 combined points.

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.

==History==

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 take both exams. The alternate test, 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 15 days later, e.g. March 7 and 22, 2006.

==Sample problems==

*Given that

<center><math> \frac{((3!)!)!}{3!} = k \cdot n!, </math></center>

where <math>k</math> and <math>n</math> are positive integers and <math>n</math> is as large as possible, find <math>k + n.</math> (''2003 AIME I #1'')
:''Solution: 839''

*If the integer <math>k</math> is added to each of the numbers <math>36</math>, <math>300</math>, and <math>596</math>, one obtains the squares of three consecutive terms of an arithmetic series. Find <math>k</math>. (''1989 AIME #7'')
:''Solution: 925''

*Complex numbers <math>a</math>, <math>b</math> and <math>c</math> are the zeros of a polynomial <math>P(z) = z^3+qz+r</math>, and <math>|a|^2+|b|^2+|c|^2=250</math>. The points corresponding to <math>a</math>, <math>b</math>, and <math>c</math> in the complex plane are the vertices of a right triangle with hypotenuse <math>h</math>. Find <math>h^2</math>. (''2012 AIME I #14'')
:''Solution: 375''

==See also==
* [[American Mathematics Competitions]]
* [[List of mathematics competitions]]
* [[Mandelbrot Competition]]

== External links ==
* [http://www.unl.edu/amc/ The Official AMC Home Page]
* [http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45 Complete Archive of AIME Problems and Solutions]
[[Category:Mathematics competitions]]
[[Category:Recurring events established in 1983]]

Elliptic partial differential equation - Revision history

en>Mecanismo: uniform structure for the intro to the eliptic and parabolic equations articles