Difference between revisions of "2022 USAJMO Problems"
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[[2022 USAJMO Problems/Problem 1|Solution]] | [[2022 USAJMO Problems/Problem 1|Solution]] | ||
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===Problem 2=== | ===Problem 2=== | ||
Let <math>a</math> and <math>b</math> be positive integers. The cells of an <math>(a + b + 1)\times (a + b + 1)</math> grid are colored amber and bronze such that there are at least <math>a^2+ab-b</math> amber cells and at least <math>b^2+ab-a</math> bronze cells. Prove that it is possible to choose <math>a</math> amber cells and <math>b</math> bronze cells such that no two of the <math>a+b</math> chosen cells lie in the same row or column. | Let <math>a</math> and <math>b</math> be positive integers. The cells of an <math>(a + b + 1)\times (a + b + 1)</math> grid are colored amber and bronze such that there are at least <math>a^2+ab-b</math> amber cells and at least <math>b^2+ab-a</math> bronze cells. Prove that it is possible to choose <math>a</math> amber cells and <math>b</math> bronze cells such that no two of the <math>a+b</math> chosen cells lie in the same row or column. | ||
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[[2022 USAJMO Problems/Problem 2|Solution]] | [[2022 USAJMO Problems/Problem 2|Solution]] | ||
===Problem 3=== | ===Problem 3=== | ||
− | Let <math>b\geq2</math> and <math>w\geq2</math> be fixed integers, and <math>n=b+w</math>. Given are <math>2b</math> identical black rods and <math>2w</math> identical white rods, each of side length 1. | + | Let <math>b\geq2</math> and <math>w\geq2</math> be fixed integers, and <math>n=b+w</math>. Given are <math>2b</math> identical black rods and <math>2w</math> identical white rods, each of side length <math>1</math>. |
We assemble a regular <math>2n</math>-gon using these rods so that parallel sides are the same color. Then, a convex <math>2b</math>-gon <math>B</math> is formed by translating the black rods, and a convex <math>2w</math>-gon <math>W</math> is formed by translating the white rods. An example of one way of doing the assembly when <math>b=3</math> and <math>w=2</math> is shown below, as well as the resulting polygons <math>B</math> and <math>W</math>. | We assemble a regular <math>2n</math>-gon using these rods so that parallel sides are the same color. Then, a convex <math>2b</math>-gon <math>B</math> is formed by translating the black rods, and a convex <math>2w</math>-gon <math>W</math> is formed by translating the white rods. An example of one way of doing the assembly when <math>b=3</math> and <math>w=2</math> is shown below, as well as the resulting polygons <math>B</math> and <math>W</math>. | ||
− | + | <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 <math>B</math> and <math>W</math> depends only on the numbers <math>b</math> and <math>w</math>, and not on how the <math>2n</math>-gon was assembled. | Prove that the difference of the areas of <math>B</math> and <math>W</math> depends only on the numbers <math>b</math> and <math>w</math>, and not on how the <math>2n</math>-gon was assembled. | ||
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[[2022 USAJMO Problems/Problem 6|Solution]] | [[2022 USAJMO Problems/Problem 6|Solution]] | ||
− | + | ==See Also== | |
− | + | {{USAJMO box|year=2022|before=[[2021 USAJMO Problems]]|after=[[2023 USAJMO Problems]]}} | |
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{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 18:22, 6 October 2023
Contents
Day 1
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
For which positive integers does there exist an infinite arithmetic sequence of integers and an infinite geometric sequence of integers satisfying the following properties?
is divisible by for all integers ;
is not divisible by .
Problem 2
Let and be positive integers. The cells of an grid are colored amber and bronze such that there are at least amber cells and at least bronze cells. Prove that it is possible to choose amber cells and bronze cells such that no two of the chosen cells lie in the same row or column.
Problem 3
Let and be fixed integers, and . Given are identical black rods and identical white rods, each of side length .
We assemble a regular -gon using these rods so that parallel sides are the same color. Then, a convex -gon is formed by translating the black rods, and a convex -gon is formed by translating the white rods. An example of one way of doing the assembly when and is shown below, as well as the resulting polygons and .
Prove that the difference of the areas of and depends only on the numbers and , and not on how the -gon was assembled.
Day 2
Problem 4
Let be a rhombus, and let and be points such that lies inside the rhombus, lies outside the rhombus, and . Prove that there exist points and on lines and such that is also a rhombus.
Problem 5
Find all pairs of primes for which and are both perfect squares.
Problem 6
Let be complex numbers, and define
for all nonnegative integers .
Suppose that for all . Prove that Solution
See Also
2022 USAJMO (Problems • Resources) | ||
Preceded by 2021 USAJMO Problems |
Followed by 2023 USAJMO Problems | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.