Let b≥2 and w≥2 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