USAMO

Conducting Body: Mathematical Association of America (MAA)
Who It Is For: Top-scoring high school students from the AIME
Subjects: Algebra, Combinatorics, Geometry, Number Theory
Question Type: Proof-based problems
Duration: 9 hours (over two days)
Format: 6 problems total (3 per day)
Frequency: Annually
Qualifies For: Mathematical Olympiad Program (MOP)

The United States of America Mathematical Olympiad (USAMO) is the final stage of the American Mathematics Competitions series for high school students. It is an invitation-only exam for the top performers from the American Invitational Mathematics Examination (AIME), selected based on a combined index of their AMC 12 and AIME scores. The USAMO serves as the primary selection test for the team that represents the United States at the International Mathematical Olympiad (IMO).

The exam is a two-day, nine-hour contest. On each day, students are given three challenging problems and four and a half hours to solve them. Unlike the multiple-choice or single-answer formats of earlier rounds, the USAMO requires students to write full, rigorous proofs for their solutions. The questions are drawn from four main areas: algebra, combinatorics, geometry, and number theory.

The practical difficulty of the USAMO comes from the depth and creativity required to solve the problems. The questions are designed to be unfamiliar and demand significant insight beyond standard textbook methods. Success depends on a student's ability to develop original arguments and present them in a clear, logical, and complete proof under intense time pressure.

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From a mathematics paper rather than USAMO, worked the way every question on this site is. USAMO questions are being added.

JEE Advanced 2016

Consider a pyramid OPQRS located in the first octant ( $x \geq 0$ , $y \geq 0$ , $z \geq 0$ ) with O as origin, and OP and OR along the x–axis and the y–axis, respectively. The base OPQR of the pyramid is a square with $OP = 3$ . The point S is directly above the mid-point, T of diagonal OQ such that $TS = 3$ . Then

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$\text{Given: pyramid OPQRS in first octant, } O = \text{origin}$

OP ∥ x-axis, OR ∥ y-axis, OPQR square

$OP = 3;\quad S \text{ above } T \text{ (midpoint of } OQ),\quad TS = 3$

$\text{To check: statements (1), (2), (3), (4)}$

(angle OQ-OS, plane OQS, dist(P, OQS), dist(O, RS)) $O(0,0,0),\quad P(3,0,0),\quad R(0,3,0)$ $Q = P + R = (3,3,0),\quad T = \text{mid}(OQ) = \left(\tfrac32,\tfrac32,0\right)$ $S = \left(\tfrac32,\tfrac32,3\right)$ $(1)\ \cos\theta = \dfrac{\vec{OQ}\cdot\vec{OS}}{|\vec{OQ}|\,|\vec{OS}|}$ $(2)\ \text{normal of plane } OQS = \vec{OQ}\times\vec{OS}$ $(3)\ \text{dist}(P,\text{plane }OQS) = \dfrac{|\vec{OP}\cdot\vec{n}|}{|\vec{n}|},\quad (4)\ \text{dist}(O,RS) = \dfrac{|\vec{RO}\times\vec{RS}|}{|\vec{RS}|}$ $\vec{OQ} = (3,3,0),\qquad \vec{OS} = \left(\tfrac32,\tfrac32,3\right)$ $\vec{OQ}\cdot\vec{OS} = \tfrac92+\tfrac92+0 = 9$ $|\vec{OQ}| = 3\sqrt2,\qquad |\vec{OS}| = \sqrt{\tfrac{27}{2}} = \dfrac{3\sqrt3}{\sqrt2}$ $\cos\theta = \dfrac{9}{3\sqrt2\cdot\frac{3\sqrt3}{\sqrt2}} = \dfrac{1}{\sqrt3}$ $\theta \approx 54.7^\circ \neq \dfrac{\pi}{3} \Rightarrow (1)\ \text{false}$ $\vec{n} = \vec{OQ}\times\vec{OS} = (9,\ -9,\ 0) \parallel (1,\ -1,\ 0)$ $\vec{n}\cdot(x,y,z) = 0 \Rightarrow x - y = 0 \Rightarrow (2)\ \text{true}$ $\text{dist}(P,\, x-y=0) = \dfrac{|3-0-0|}{\sqrt{1^2+(-1)^2}} = \dfrac{3}{\sqrt2} \Rightarrow (3)\ \text{true}$ $\vec{RS} = \left(\tfrac32,\ -\tfrac32,\ 3\right),\qquad \vec{RO} = O - R = (0,\ -3,\ 0)$ $\text{dist}(O, RS) = \dfrac{|\vec{RO}\times\vec{RS}|}{|\vec{RS}|}$ $\vec{RO}\times\vec{RS} = (-9,\ 0,\ \tfrac92),\qquad |\vec{RO}\times\vec{RS}| = \sqrt{81+\tfrac{81}{4}} = \dfrac{9\sqrt5}{2}$ $|\vec{RS}| = \sqrt{\tfrac94+\tfrac94+9} = \sqrt{\tfrac{27}{2}} = \dfrac{3\sqrt6}{2}$ $\text{dist}(O, RS) = \dfrac{9\sqrt5/2}{3\sqrt6/2} = \dfrac{3\sqrt5}{\sqrt6} = \dfrac{\sqrt{30}}{2} = \sqrt{\dfrac{15}{2}} \Rightarrow (4)\ \text{true}$ $\boxed{\text{Answer: (2), (3), (4)}}$

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