Art of Problem Solving
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2022 USAMO Problems/Problem 2: Difference between revisions

Newer edit →
Wzs26843545602 (talk | contribs)
editor
28 edits
Created page with "==Problem== 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> ident..."
 
Pog (talk | contribs)
editor
197 edits
Line 1: Line 1:
==Problem==
==Problem==
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.
==Solution==
==Solution==
[WIP]
[WIP]

Revision as of 18:40, 26 May 2022

Problem

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

[WIP]

See also

2022 USAMO (Problems • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2022_USAMO_Problems/Problem_2&oldid=174042"