Difference between revisions of "2022 USAMO Problems/Problem 2"
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==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 | + | 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 
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 .
![[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]](http://latex.artofproblemsolving.com/8/8/c/88c26c3ed57745f62327c44dc471fbe3c959a287.png)
Prove that the difference of the areas of and depends only on the numbers 
and 
, and not on how the -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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
