olymmath

USAMO 2022

United States of America Mathematical Olympiad 2022.

Day 1

Problem 1

  • Day 1

Let aa and bb be positive integers. The cells of an (a+b+1)×(a+b+1)(a+b+1)\times (a+b+1) grid are colored amber and bronze such that there are at least a2+ab−ba^2+ab-b amber cells and at least b2+ab−ab^2+ab-a bronze cells. Prove that it is possible to choose aa amber cells and bb bronze cells such that no two of the a+ba+b chosen cells lie in the same row or column.

Answer, solutions and source

Problem 2

  • Day 1

Let b≥2b\geq2 and w≥2w\geq2 be fixed integers, and n=b+wn=b+w . Given are 2b2b identical black rods and 2w2w identical white rods, each of side length 1.

We assemble a regular 2n−2n- gon using these rods so that parallel sides are the same color. Then, a convex 2b2b -gon BB is formed by translating the black rods, and a convex 2w2w -gon WW is formed by translating the white rods. An example of one way of doing the assembly when b=3b=3 and w=2w=2 is shown below, as well as the resulting polygons BB and WW .

[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 + wdir(36); pair W3 = W2 + wdir(108); pair W4 = W3 + wdir(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(" WW ", WO); / draw(W1--WtW1); draw(W2--WtW2); draw(W3--WtW3); draw(W4--WtW4); */

// Black shifted real Bk = 1.10; pair B1 = (1.5,-0.1); pair B2 = B1 + wdir(0); pair B3 = B2 + wdir(324); pair B4 = B3 + wdir(252); pair B5 = B4 + wdir(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(" BB ", BO);[/asy]

Prove that the difference of the areas of BB and WW depends only on the numbers bb and ww , and not on how the 2n2n -gon was assembled.

Proposed by Ankan Bhattacharya

Answer, solutions and source

Day 2

Problem 5

  • Day 2

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

Find the smallest integer kk such that for any 2022 real numbers x1,x2,…,x2022,x_1,x_2,\ldots , x_{2022}, there exist kk essentially increasing functions f1,…,fkf_1,\ldots, f_k such that f1(n)+f2(n)+⋯+fk(n)=xnfor every n=1,2,…2022.f_1(n) + f_2(n) + \cdots + f_k(n) = x_n \text{for every } n= 1,2,\ldots 2022.

Answer, solutions and source

Problem 6

  • Day 2

There are 20222022 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?

Answer, solutions and source