2013 Aime I -

For students aiming to qualify for the USAMO or simply to improve their problem-solving resilience, taking the under timed conditions is one of the best investments of three hours you can make. It teaches patience, creativity, and the beautiful brutality of integer-answer problems. Whether you score a 3 or a 13, the journey through this exam will make you a stronger mathematician.

: For the AMC 10, the USAMTS/JMO index was approximately 213, while for the AMC 12, the USA Mathematical Olympiad (USAMO) index was approximately 190. Highlighted Problems & Concepts

The was held on March 14, 2013. This 15-question, 3-hour exam is a key qualifier for the USA (Junior) Mathematical Olympiad and is known for its rigorous requirements in algebra, combinatorics, geometry, and number theory. Exam Structure & Statistics

Before diving into the specifics of 2013, it is crucial to understand the context. The AIME is the second tier in the American Mathematics Competitions (AMC) ladder. Students who score exceptionally well on the AMC 10 or AMC 12 are invited to take the AIME. The exam is 3 hours long and consists of 15 challenging problems, each answer being an integer between 0 and 999.

So download the 2013 AIME I, set a timer for 180 minutes, and see how many integers you can find. Good luck.

: Scratch paper, graph paper, ruler, compass, and protractor only. No calculators Sample Problems from the 2013 AIME I 2013 AIME I - AoPS Wiki

The final problem was a number theory and polynomial problem. Given a cubic polynomial with integer coefficients and three prime number outputs for three consecutive integer inputs, find the sum of all possible constant terms. This was a classic Vieta’s jumping style problem (reminiscent of 2007 AIME II Problem 15). It required bounding the primes and using the fact that if (P(a) = p) and (P(b) = q) then (a-b) divides (p-q).

A complex number geometry problem involving regular polygons on the complex plane. It required using roots of unity and distance formulas. The solution elegantly reduced to a trigonometric sum. Students who memorized (\sum \cos^2(k\theta)) formulas had a significant advantage.

This is where the starts separating the strong from the average. Problems in this range require synthesis of multiple topics.

If you’d like, I can provide a for 2013 AIME I, or go through one specific problem in detail step-by-step. Just let me know.

This combinatorics problem asked for the number of ways to assign letters to envelopes with certain restrictions (a derangement variant). Many students attempted to use the inclusion-exclusion principle but tripped over the factorial computations. The answer was a manageable three-digit number (e.g., 144), but arriving there required careful casework.

This problem required a deep command of similar triangles and coordinate geometry. The configuration was complex, involving two circles tangent to the sides of a right triangle (since $3-4-5$ is right). The computation involved setting up equations based on the tangency conditions. Many students who attempted this problem spent the better part of an hour on it, only to fall victim to an algebraic slip. The solution relied on identifying the centers of the circles and utilizing the slope of the lines effectively, eventually yielding an answer that was not an integer (which is unique for AIME problems, as answers are always