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Revision as of 18:24, 6 October 2023

Day 1

Problem 1

Let $a$ and $b$ be positive integers. The cells of an $(a+b+1)\times (a+b+1)$ grid are colored amber and bronze such that there are at least $a^2+ab-b$ amber cells and at least $b^2+ab-a$ bronze cells. Prove that it is possible to choose $a$ amber cells and $b$ bronze cells such that no two of the $a+b$ chosen cells lie in the same row or column.

Solution

Problem 2

Let $b\geq2$ and $w\geq2$ be fixed integers, and $n=b+w$. Given are $2b$ identical black rods and $2w$ identical white rods, each of side length 1.

We assemble a regular $2n$-gon using these rods so that parallel sides are the same color. Then, a convex $2b$-gon $B$ is formed by translating the black rods, and a convex $2w$-gon $W$ is formed by translating the white rods. An example of one way of doing the assembly when $b=3$ and $w=2$ is shown below, as well as the resulting polygons $B$ and $W$.

[asy] size(10cm); real w = 2*Sin(18); real h = 0.10 * w; real d = 0.33 * h; picture wht; picture blk; draw(wht, (0,0)--(w,0)--(w+d,h)--(-d,h)--cycle); fill(blk, (0,0)--(w,0)--(w+d,h)--(-d,h)--cycle, black); // draw(unitcircle, blue+dotted); // Original polygon add(shift(dir(108))*blk); add(shift(dir(72))*rotate(324)*blk); add(shift(dir(36))*rotate(288)*wht); add(shift(dir(0))*rotate(252)*blk); add(shift(dir(324))*rotate(216)*wht); add(shift(dir(288))*rotate(180)*blk); add(shift(dir(252))*rotate(144)*blk); add(shift(dir(216))*rotate(108)*wht); add(shift(dir(180))*rotate(72)*blk); add(shift(dir(144))*rotate(36)*wht); // White shifted real Wk = 1.2; pair W1 = (1.8,0.1); pair W2 = W1 + w*dir(36); pair W3 = W2 + w*dir(108); pair W4 = W3 + w*dir(216); path Wgon = W1--W2--W3--W4--cycle; draw(Wgon); pair WO = (W1+W3)/2; transform Wt = shift(WO)*scale(Wk)*shift(-WO); draw(Wt * Wgon); label("$W$", WO); /* draw(W1--Wt*W1); draw(W2--Wt*W2); draw(W3--Wt*W3); draw(W4--Wt*W4); */ // Black shifted real Bk = 1.10; pair B1 = (1.5,-0.1); pair B2 = B1 + w*dir(0); pair B3 = B2 + w*dir(324); pair B4 = B3 + w*dir(252); pair B5 = B4 + w*dir(180); pair B6 = B5 + w*dir(144); path Bgon = B1--B2--B3--B4--B5--B6--cycle; pair BO = (B1+B4)/2; transform Bt = shift(BO)*scale(Bk)*shift(-BO); fill(Bt * Bgon, black); fill(Bgon, white); label("$B$", BO); [/asy]

Prove that the difference of the areas of $B$ and $W$ depends only on the numbers $b$ and $w$, and not on how the $2n$-gon was assembled.

Solution

Problem 3

Let $\mathbb{R}_{>0}$ be the set of all positive real numbers. Find all functions $f:\mathbb{R}_{>0} \to \mathbb{R}_{>0}$ such that for all $x,y\in \mathbb{R}_{>0}$ we have \[f(x) = f(f(f(x)) + y) + f(xf(y)) f(x+y).\]

Solution

Day 2

Problem 4

Find all pairs of primes $(p, q)$ for which $p-q$ and $pq-q$ are both perfect squares.

Solution

Problem 5

A function $f: \mathbb{R}\to \mathbb{R}$ is essentially increasing if $f(s)\leq f(t)$ holds whenever $s\leq t$ are real numbers such that $f(s)\neq 0$ and $f(t)\neq 0$.

Find the smallest integer $k$ such that for any 2022 real numbers $x_1,x_2,\ldots , x_{2022},$ there exist $k$ essentially increasing functions $f_1,\ldots, f_k$ such that\[f_1(n) + f_2(n) + \cdots + f_k(n) = x_n\qquad \text{for every } n= 1,2,\ldots 2022.\]

Solution

Problem 6

There are 2022 users on a social network called Mathbook, and some of them are Mathbook-friends. (On Mathbook, friendship is always mutual and permanent.)

Starting now, Mathbook will only allow a new friendship to be formed between two users if they have at least two friends in common. What is the minimum number of friendships that must already exist so that every user could eventually become friends with every other user?

Solution

See Also

2022 USAMO (Problems • Resources)
Preceded by
2021 USAMO Problems 		Followed by
2023 USAMO Problems
1 • 2 • 3 • 4 • 5
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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