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Latest revision as of 12:47, 22 November 2023

Day 1

Note: For any geometry problem whose statement begins with an asterisk $(*)$, the first page of the solution must be a large, in-scale, clearly labeled diagram. Failure to meet this requirement will result in an automatic 1-point deduction.

Problem 1

Let $\mathbb{N}$ be the set of positive integers. A function $f:\mathbb{N}\to\mathbb{N}$ satisfies the equation \[\underbrace{f(f(\ldots f}_{f(n)\text{ times}}(n)\ldots))=\frac{n^2}{f(f(n))}\]for all positive integers $n$. Given this information, determine all possible values of $f(1000)$.

Solution

Problem 2

Let $ABCD$ be a cyclic quadrilateral satisfying $AD^2 + BC^2 = AB^2$. The diagonals of $ABCD$ intersect at $E$. Let $P$ be a point on side $\overline{AB}$ satisfying $\angle APD = \angle BPC$. Show that line $PE$ bisects $\overline{CD}$.

Solution

Problem 3

Let $K$ be the set of all positive integers that do not contain the digit $7$ in their base-$10$ representation. Find all polynomials $f$ with nonnegative integer coefficients such that $f(n)\in K$ whenever $n\in K$.

Solution


Day 2

Problem 4

Let $n$ be a nonnegative integer. Determine the number of ways that one can choose $(n+1)^2$ sets $S_{i,j}\subseteq\{1,2,\ldots,2n\}$, for integers $i,j$ with $0\leq i,j\leq n$, such that: for all $0\leq i,j\leq n$, the set $S_{i,j}$ has $i+j$ elements; and $S_{i,j}\subseteq S_{k,l}$ whenever $0\leq i\leq k\leq n$ and $0\leq j\leq l\leq n$.

Solution

Problem 5

Two rational numbers $\tfrac{m}{n}$ and $\tfrac{n}{m}$ are written on a blackboard, where $m$ and $n$ are relatively prime positive integers. At any point, Evan may pick two of the numbers $x$ and $y$ written on the board and write either their arithmetic mean $\tfrac{x+y}{2}$ or their harmonic mean $\tfrac{2xy}{x+y}$ on the board as well. Find all pairs $(m,n)$ such that Evan can write $1$ on the board in finitely many steps.

Solution

Problem 6

Find all polynomials $P$ with real coefficients such that \[\frac{P(x)}{yz}+\frac{P(y)}{zx}+\frac{P(z)}{xy}=P(x-y)+P(y-z)+P(z-x)\]holds for all nonzero real numbers $x,y,z$ satisfying $2xyz=x+y+z$.

Solution


See Also

2019 USAMO (Problems • Resources)
Preceded by
2018 USAMO Problems 		Followed by
2020 USAMO Problems
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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