USAMO 2022 Problem 2

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 + 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(" $W$ ", 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(" $B$ ", 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.

Proposed by Ankan Bhattacharya