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Difference between revisions of "2022 USAJMO Problems"
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[[2022 USAJMO Problems/Problem 1|Solution]]
 
[[2022 USAJMO Problems/Problem 1|Solution]]
 +
 
===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>.
  
−
[image here]
+
<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.
Line 26: Line 86:
 
==Day 2==
 
==Day 2==
 
===Problem 4===
 
===Problem 4===
−
Let <math>ABCD</math> be a rhombus, and let <math>K</math> and <math>L</math> be points such that <math>K</math> lies inside the rhombus, <math>L</math> lies outside the rhombus, and <math>KA=KB=LC=LD</math>. Prove that there exist points <math>X</math> and <math>Y</math> on lines <math>AC</math> and <math>BD</math> such that <math>KXLY</math> is also a rhombus.
+
<math>(*)</math> Let <math>ABCD</math> be a rhombus, and let <math>K</math> and <math>L</math> be points such that <math>K</math> lies inside the rhombus, <math>L</math> lies outside the rhombus, and <math>KA=KB=LC=LD</math>. Prove that there exist points <math>X</math> and <math>Y</math> on lines <math>AC</math> and <math>BD</math> such that <math>KXLY</math> is also a rhombus.
  
 
[[2022 USAJMO Problems/Problem 4|Solution]]
 
[[2022 USAJMO Problems/Problem 4|Solution]]
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[[2022 USAJMO Problems/Problem 6|Solution]]
 
[[2022 USAJMO Problems/Problem 6|Solution]]
  
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{| class="wikitable" style="margin:0.5em auto; font-size:95%; border:1px solid black; width:40%;"
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==See Also==
−
| style="background:#ccf;text-align:center;" colspan="3" | '''[[2021 USAJMO]]''' ('''[[2021 USAJMO Problems|Problems]]''' • [http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=176&year={{{year}}} Resources])
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{{USAJMO box|year=2022|before=[[2021 USAJMO Problems]]|after=[[2023 USAJMO Problems]]}}
−
|-
 
−
| width="50%" align="center" rowspan="{{{rowsp|1}}}" | {{{beforetext|Preceded&nbsp;by<br/>}}}'''{{{before|[[2021 USAJMO]]}}}'''
 
−
| width="50%" align="center" rowspan="{{{rowsf|1}}}" | {{{aftertext|Followed&nbsp;by<br/>}}}'''{{{after|[[2023 USAJMO]]}}}'''
 
−
|-
 
−
| colspan="3" style="text-align:center;" | [[2022 USAJMO Problems/Problem 1|1]] '''•''' [[2022 USAJMO Problems/Problem 2|2]] '''•''' [[2022 USAJMO Problems/Problem 3|3]] '''•''' [[2022 USAJMO Problems/Problem 4|4]] '''•''' [[2022 USAJMO Problems/Problem 5|5]] '''•''' [[2022 USAJMO Problems/Problem 6|6]]
 
−
|-
 
−
| colspan="3" style="text-align:center;" | '''[[USAJMO Problems and Solutions | All USAJMO Problems and Solutions]]'''
 
−
|}<includeonly></includeonly><noinclude>
 
−
 
 
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 18:22, 6 October 2023

Day 1

$\textbf{Note:}$ 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

For which positive integers $m$ does there exist an infinite arithmetic sequence of integers $a_1,a_2,\cdots$ and an infinite geometric sequence of integers $g_1,g_2,\cdots$ satisfying the following properties?

$\bullet$ $a_n-g_n$ is divisible by $m$ for all integers $n>1$;

$\bullet$ $a_2-a_1$ is not divisible by $m$.

Solution

Problem 2

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 3

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

Day 2

Problem 4

$(*)$ Let $ABCD$ be a rhombus, and let $K$ and $L$ be points such that $K$ lies inside the rhombus, $L$ lies outside the rhombus, and $KA=KB=LC=LD$. Prove that there exist points $X$ and $Y$ on lines $AC$ and $BD$ such that $KXLY$ is also a rhombus.

Solution

Problem 5

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

Solution

Problem 6

Let $a_0,b_0,c_0$ be complex numbers, and define

\[a_{n+1}=a_n^2+2b_nc_n\] \[b_{n+1}=b_n^2+2c_na_n\] \[c_{n+1}=c_n^2+2a_nb_n\] for all nonnegative integers $n$.

Suppose that $\max{(|a_n|,|b_n|,|c_n|)}\leq2022$ for all $n$. Prove that \[|a_0|^2+|b_0|^2+|c_0|^2\leq 1.\] 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. AMC logo.png

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