olymmath

USAJMO 2022

United States of America Junior Mathematical Olympiad 2022.

Day 1

Problem 1

  • Day 1

For which positive integers mm does there exist an infinite arithmetic sequence of integers a1,a2,...a_1, a_2, . . . and an infinite geometric sequence of integers g1,g2,...g_1, g_2, . . . satisfying the following properties?

  • an−gna_n - g_n is divisible by mm for all integers n≥1n \ge 1 ;
  • a2−a1a_2 - a_1 is not divisible by mm .

Holden Mui

Answer, solutions and source

Problem 2 · Also USAMO 2022 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 3 · Also USAMO 2022 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 4

  • Day 2

Let ABCDABCD be a rhombus, and let KK and LL be points such that KK lies inside the rhombus, LL lies outside the rhombus, and KA=KB=LC=LDKA = KB = LC = LD . Prove that there exist points XX and YY on lines ACAC and BDBD such that KXLYKXLY is also a rhombus.

Proposed by Ankan Bhattacharya

Answer, solutions and source

Problem 5 · Also USAMO 2022 Problem 4

  • Day 2

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

Answer, solutions and source

Problem 6

  • Day 2

Let a0,b0,c0a_0, b_0, c_0 be complex numbers, and define \begin{align*}a_{n+1} &= a_n^2 + 2b_nc_n b_{n+1} &= b_n^2 + 2c_na_n c_{n+1} &= c_n^2 + 2a_nb_n\end{align*}for all nonnegative integers n.n.

Suppose that max⁡{∣an∣,∣bn∣,∣cn∣}≤2022\max{\{|a_n|, |b_n|, |c_n|\}} \leq 2022 for all n.n. Prove that ∣a0∣2+∣b0∣2+∣c0∣2≤1.|a_0|^2 + |b_0|^2 + |c_0|^2 \leq 1.

Answer, solutions and source