complex bash lol

Trivial by vectors.

Is this right? Assume $b \ge w$ and induct on $w.$ We prove with induction on $w$ for all $w\ge 1,$ treating a $2$-gon as having zero area. Base case trivial. For the inductive step, it suffices to prove that by turning a white side and its partner black, the difference between the polygons increases by a fixed amount. This amount is the area gained by the $2b$-gon plus the area lost by the $2w$-gon. Call this side $s$ and its partner $s'.$ The $2w$-gon is composed of $w$ pairs of sides that are parallel and opposite each other, one of which are the translated copies of $s$ and $s'.$ By removing $s$ and $s'$ from the set of white sides, the $w-1$ sides on one side of $s$ and $s'$ get shifted by $1$ unit in the direction of $s$ towards the other $w-1$ sides to form the new resulting white $2w-2$-gon. So the area lost by the white polygon can be broken up into $w-1$ rhombuses of side length $1.$ The multiset of acute angles within these $w-1$ rhombuses coincide with the multiset of acute angles that the $w-1$ opposite pairs of white sides besides $s$ and $s'$ make with $s$ and $s'.$ Similarly, by adding $s$ and $s'$ to the set of black sides, $b$ consecutive sides of the $2b$-gon get shifted by $1$ unit in the direction of $s$ away from the other $w-1$ sides to form the new resulting black $2b+2$-gon. So the area gained can be broken up into $b$ rhombuses of side length $1.$ The multiset of acute angles within these $w-1$ rhombuses coincide with the multiset of acute angles that the $b$ pairs of opposite black sides make with $s$ and $s'.$ The union of these two multisets coincide with the multiset of acute angles that all $n-1$ pairs of opposite sides in the $2n$-gon besides $s$ and $s'$ make with $s$ and $s'.$ This is a constant only dependent on $n$ by symmetry, so it's easy to see the result by rhombus area formula. $\blacksquare$

Almost the same as #6

I feel like this one will have like 4 or 5 valid solutions. A very creative problem for sure

Imagine getting one point off for not drawing a diagram tho ... Anyways, it suffices to prove that switching the colors of two adjacent sides (and their opposite sides accordingly) does not change the difference. To do this, cut off the rest of the sides from the $2n$-gon to get a hexagon, and do a similar thing for $B$ and $W$ to get parallelograms (both the "before" and "after" versions with respect to switching the colors of the adjacent sides). The cut off part does not change, so we need only consider the differences of the resulting parellolgrams. To do this, we rearrange the equation to make it so that we are proving two sums of areas are equal. We now put the parallelograms together to form two (concave) hexagons, and we have to prove the areas of the hexagons are equal. To do this, we transform both of the hexagons into parallelograms by using shears on each of the parallologram parts (that were put together to form the hexagons). The resulting parallelograms are congruent, and this is seen by tracking vectors through this entire process.

This is also doable with law of sines, the sine area formula, and a bit of perseverance through the (straightforward) bash

There's also a nice solution with cross products!

Solved with p_square Change white and black to red and blue. Call a polygon nice if it has opposite sides parallel and of equal lengths (basically it can be tiled with parallelograms) and say its side vectors are $v_1, v_2, \cdots, v_n$ followed by $-v_1, -v_2, \cdots, -v_n$. Lemma Given a nice polygon with side vectors $v_1, v_2, \cdots, v_n$, then its area is $$\sum_{1 \le i<j \le n} v_i \times v_j$$ Proof: Draw a diagonal, find area of half of the polygon to be the above by induction and then double it. $\square$ Let $R, B$ be the sets of red and blue sides. Relabel such that $v_1, v_2$ are adjacent and are of different colors, now consider what happens when we swap $v_1, v_2$. The difference of change in area is $$(v_2 - v_1) \times \left (\sum_{i \in R, i \neq 1} v_i \right) - (v_1 - v_2) \times \left(\sum_{j \in B, j \neq 2} v_j \right)$$$$= (v_1 - v_2) \times \left(\sum_{i > 2}^n v_i \right) = 0$$where the last equality is true since the two vectors have the same direction as the polygon is regular, because if the vertices are $A_1, A_2, \cdots, A_{2n}$ if $v_i = \overline{A_iA_{i+1}}$, then the first vector is along the angle bisector of $\angle A_1A_2A_3$ and the second is along $A_3A_{n+1}$, which are parallel. Since we can reach any coloring of $b$ black and $w$ white using the above swaps, the difference in area only depends on $b,w$, as desired. $\blacksquare$

Flip the horizontal black edge. Then $[B]$ decreases by the sum of the blue lengths while $-[W]$ decreases by the sum of the red lengths, regardless of $b$ and $w$. Now, if we start with all edges colored black and flip the colors of some fixed number of opposite edges to white, we arrive at the same value of $[B]-[W]$ regardless of our edge choice.

This wasn't a "traditional" geometry problem in that there's no way to draw a "large in-scale diagram" because it's so general. Do you think that the 1 point for large diagram rule still applies on this problem?

I think it does ... but if you didn't submit a solution, they won't give you -1 points, right?

This isn't a geo problem

I felt this was easier than U1/J2.

can someone make asymptote(PLEASETELLMEYOUAREDRAWINGTHIS)

Let $A(B)$ denote the area of $B$ and $A(W)$ denote the area of $W$. Let the vertices of $B$ be labeled $B_1,B_2,\cdots,B_{2b}$: label the vertices of $A$ analogously. We show the result by induction on $b$, starting with $b=1$ for which the problem is trivial. Take a particular vector $\vec v$ such that it and $-\vec v$ are moved from $W$ to $B$. For vectors $\vec v$ and $\vec w$, let $P(\vec v, \vec w)$ denote the area of the parallelogram with sides $\vec v,\vec w,-\vec v,-\vec w$ in that order. WLOG $\vec v$ is added between $B_{2b}$ and $B_1$ and $-\vec v$ is added between $B_b$ and $B_{b+1}$. Thus $A(B)$ increases by $P(\vec v, \vec{B_1B_b})$. WLOG suppose that $\vec v=\vec{W_{2w}W_1}$ and $-\vec v = \vec{W_wW_{w+1}}$. Then $A(W)$ decreases by $P(\vec v,\vec{W_1W_w})$. Remark then that we want to show $P(\vec v,\vec{B_1B_b}) + P(\vec v,\vec{W_1W_w})$ is fixed. We do this by putting the two parallelograms together: create the shape $LMNOPQ$ so that $\vec{LM}=\vec{NQ}=\vec{OP} = \vec v$, $\vec{MN} = \vec{LQ} = \vec{W_1W_w}$, and $\vec{NO} = \vec{QP} = \vec{B_1B_b}$. Remark that $\triangle MNO\cong \triangle LQP$, so it is enough to show $LMOP$ is fixed. But this is clear, so we are done.

umm overexplained and nonrigorous solution below... Change white and black to red and blue, rename $b$ and $w$ as $r$ and $b.$ We claim if we swap the colors of two adjacent sides, along with their opposite counterparts, the difference of areas will remain the same. Notice this is sufficient to solve the problem as we can achieve all arrangments of the rods by swapping adjacent sides. If we let $x$ and $y$ be the origial red and blue regions, and $x'$ and $y'$ the regions after the swap, it suffices to prove $x+y'=y+x'.$ Claim: We can connect some of the vertices of $x$ to form a parallelogram $X$ with at least one unit side length. If we do the same for $x',$ we can find $X$ and $X'$ such that $x-X=x'-X'.$ Proof. Notice when we swap two consecutive sides the $2n$-gon, the original and transformed shapes have $2r-2$ red sides in the same positions, with $r-1$ on each half of the $2n$-gon. Then, when we seperate the red sides into their own shape $x,$ section off the consecutive sides that were fixed, so we have two congruent sections (call these shapes mundane) and one quadrilateral that has opposite sides of length one. Notice the quadrilateral is a parallelogram because the mundane sections are congruent. Let $X$ be this parallelogram. Similarly, we split $x',$ and notice that the mundane sections from $x$ and $x'$ are congruent. Hence, we know $x-X=x'-X'.$ $\blacksquare$ Consider the altitude $a_X$ of the parallelogram $X$ that is perpendiular to a side with length one. Similarly define $a_{X'},a_Y,$ and $a_{Y'}.$ Notice $a_X\parallel a_{Y'}$ because the side of the parallelogram $a_X$ is perpendicular to recolors into the blue side that $a_{Y'}$ is perpendicular to. Also draw one of the lines perpendicular to the side that recolors from red to blue, and passes through one vertex of the side; call this $\ell.$ Finally, draw altitudes from the vertices of one of the mundane regions to $a_X,$ and draw altitudes from the vertices of the fixed region (of the $2n$-gon) to $\ell.$ We show a diagram of the $2n$-gon, $x,$ and $y'$ below, with all the extra lines drawn; the fixed and mundane regions are shaded. [asy][asy] import geometry; size(10cm); pair x1,x2,x3,x4,x5,x6,x7,x8,x9,x10; x1=dir(0); x2=dir(36); x3=dir(72); x4=dir(108); x5=dir(144); x6=dir(180); x7=dir(216); x8=dir(252); x9=dir(288); x10=dir(324); fill(x1--x2--x3--x4--cycle,lightgrey); draw(x10--x1--x2--x3,red); draw(x3--x4--x5,blue); draw(x5--x6--x7--x8,red); draw(x8--x9--x10,blue); int k=-3; pair y1,y2,y3,y4,y5,y6; y1=x8-k; y2=x7-k; y3=x6-k; y4=x5-k; y5=x2-x3+x5-k; y6=x1-x3+x5-k; fill(y4--y5--y6--cycle,lightgrey); draw(y1--y2--y3--y4--y5--y6--cycle,red); draw(y4--y6); draw(y1--y3); draw(foot(y3,y1,y6)--y3,purple); draw(y6--foot(y3,y1,y6),red+dotted); draw(foot(y5,foot(y3,y1,y6),y3)--y5); draw(foot(y5,y4,y3)--y5,dotted); draw(foot(x2,x3,x8)--x2,dotted); int l=-4; pair z1,z2,z3,z4; z1=x8-l; z2=x9-l; z3=x1-x10+x9-l; z4=x5-x6+x8-l; draw(z1--z2--z3--z4--cycle,blue); draw(foot(z4,z2,z3)--z4,orange); draw(x5--foot(x4,x1,x5)); draw(foot(x4,x1,x5)--foot(x3,x1,x5),orange); draw(foot(x3,x1,x5)--x1,purple); draw(foot(x2,x1,x5)--x2); draw(foot(x3,x1,x5)--x3); draw(foot(x4,x1,x5)--x4); [/asy][/asy] Claim: The sum of $a_X$ and $a_{Y'}$ is equal to the section of $\ell$ that is bounded by perpendiculars from the fixed section (grey section) of our $2n$-gon. That is, the purple and orange segments are equal to eachother in the diagram. Consider a section of $a_X$ bounded by two adjacent altitudes from mundane vertices to $a_X.$ Notice we can form a right triangle that has side lengths of the section of the altitude and a mundane side by sliding the section across the two parallel altitudes. Take the corresponding side of the $2n$-gon to the mundane side; then, draw the altitude from one of the vertices of that side to an adjacent altitude to $\ell$ to form a right triangle. Because all their sides a parallel and they have a hypotenus of one, the triangles are congruent. Hence, the part of $\ell$ bounded by the altutides from the side of the $2n$-gon corresponding to our mundane side is equal to the section of $a_X$ bounded by the altutides from the mundane side. We do this for all the mundane sides, including the mundane sides of $y',$ and find $a_X+a_{Y'}$ is equal to the section of $\ell$ bounded by the altitudes from the fixed sides of the $2n$-gon. $\blacksquare$ We do the same for $y$ and $x',$ and find $a_Y+a_{X'}$ is also equal to a segment with the same length as $\ell$ bounded by perpendicular lines from the fixed sides of the $2n$-gon. Since the sum of the altitudes of $x,y'$ are equal to the sum of the altitudes of $x'y,$ we know $$x-x'=X-X'=Y-Y'=y-y'.$$$\square$

What is the motivation to try swapping the colors of opposite edges?

samrocksnature wrote: What is the motivation to try swapping the colors of opposite edges? You're making small polygons out of the same oriented edges of the 2n-gon: the best thing to do is try to find some properties that different polygons share, since proving the condition directly by computing the area doesn't seem obvious. The two closest polygons you can find are ones that differ by a single pair of edges: that is, the black polygon upon adding in or removing a single pair of opposite edges. In other words, swapping colors of opposite edges.

Actually we can find that the area different of $B$ and $W$ is $\frac{b-w}{n} \cdot S$, where $S$ is the area of the original $2n$-gon. But I use Stokes' theorem to calculate it (but the calculation is not hard). And I really wonder if there exists an elementry method to find the area of $B$ and $W$ directly... My Answer.

pog wrote: can someone make asymptote(PLEASETELLMEYOUAREDRAWINGTHIS) I edited the first post with the asymptote that was used to draw the figure on the exam.

Assuming that each side the regular $2n$-polygon $P$ has a length of $1$. Let $H$ be heigth of $P$. Denote pairwise distinct unit vectors, which are parallel to sides of $P$, $u_1,u_2,\hdots,u_b,v_1,v_2,\hdots,v_w$ where $u_i$s belongs to the polygon $B$ and $v_j$s belongs to $W$. Let $h_{B,i}$ be the height of the polygon $B$ when consider $u_i$ as a base. Define $h_{W,j}$ analogously. Claim 1: $\text{area}(B) = \frac 12\sum\limits_{i=1}^{b} h_{B,i}$ and $\text{area}(W) = \frac 12\sum\limits_{i=1}^{w} h_{W,i}$. Proof. Dissect the polygon $B$ by segments joining its center to each of its vertices. The area of triangles are $\frac{1}{4}h_{B,i}$ where each $i$ appears twice, yielding the desired results. Define $h(x,y)$ a heigth of a unit rhombus of which sides are parallel to vectors $x,y$. Claim 2: For $x\in\{u_1,u_2,\hdots,u_b,v_1,v_2,\hdots,v_w\}$, $\sum\limits_{i=1}^{b} h(x,u_i)+\sum\limits_{j=1}^{w} h(x,v_j)=H$. Proof. Consider a side $x,x'$ of $P$. Choose consecutive $n$ sides which are between $x,x'$. Then translate all of those by $x$, and then connect those $n$ new points. We can now see $n-1$ rhombuses (and a degenerate one) of which heights $h(x,u_1),\hdots,h(x,u_b),h(x,v_1),\hdots,h(x,v_w)$ are stacking in the overall heigth $H$. We are done. By using the same idea, we have $\sum\limits_{i=1}^{b} h(u_k,u_i)=h_{B,k}$ and $\sum\limits_{j=1}^{w} h(v_k,v_j)=h_{W,k}$. Now consider $$\sum\limits_{i=1}^{b}\sum\limits_{j=1}^{w} h(u_i,v_j) = \sum\limits_{i=1}^{b}\left(H-\sum\limits_{j=1}^{b} h(u_i,u_j)\right)=bH-\sum\limits_{i=1}^{b} h_{B,i}=bH-2\cdot\text{area}(B)$$Similarly, we have $bH-2\cdot\text{area}(B)=wH-2\cdot\text{area}(W)$. Thus, difference between area of $B$ and $W$ is $\frac{b-w}{2}H$ which is independent to the way of coloring as desired.

Let $\mathcal{P}(t)$ be the regular $2n$-gon formed if the white rods have length $1-t$ and the black rods have length $t$ and their order is fixed. Also let the area of $\mathcal{P}(t)$ be $A(t)$. So this can be seen as a continuous transformation from $W$ (which is $\mathcal{P}(0)$) to $B$ (which is $\mathcal{P}(1)$) Claim: $A(t)$ is a quadratic expression for $t\in [0,1]$. If you take any triangle that has as a side among the sides of the polygon (the base $b$ ). Then it is easy to see that the length of $b$ is linear wrt $t$ (it is either $t$ or $1-t$) and so is the height from the third point of the triangle to $b$.The latter is the sum of some normal components to $\vec{b}$ of some other sides of the polygons and the relative angles are constant. So the area of the triangle is quadratic and triangulizing the polygon into such triangles you get that the area of $\mathcal{P}(t)$ is $A(t)=at^2+bt+c$ for some $a,b,c$ The difference of the areas of $B$ ,$W$ is $A(1)-A(0)=a+b$ which is precisely $\frac{dA}{dt}(\frac{1}{2})$.Of course $\mathcal{P}(1/2)$ is the regular $2n$ -gon whose sides are $1/2$ and small perturbations around $t=1/2$ dilate the black sides and contract the white ones. Now we are going to consider two versions of the polygon $\mathcal{P}_1(t),\mathcal{P}_2(t)$ where the difference between them is that two consecutive sides that have different colours (and their corresponding opposite sides) have been toggled (from black $\to$ white and vice versa). We are going to show that $A_1'(\frac{1}{2})=A_2'(\frac{1}{2})$ thereby showing that the order of the black, white sides is irrelevant as we can reach by performing such toggles a fixed state.So let the consecutive sides be $a_0a_1$, $a_1a_2$ and their opposite sides be $b_0b_1$ , $b_1b_2$. Draw the rectangle $a_0a_2b_0b_2$ (figure) and let it's sides be $a_0a_2=x(t)$ and $a_2b_0=y(t)$ : all elements of the polygon are pairwise congruent in the two versions (in particular the lengths $x(t),y(t)$ are common) except the parallelogram and in that the two versions differ only in the angle $\widehat{a_0a_2b_0}=\theta_1(t)$ in the first version and $\theta_2(t)$ in the second version.Note that $\theta_1(\frac{1}{2})=\theta_2(\frac{1}{2})=\pi/2$. It suffices to show that the derivatives of the areas of the parallelograms at $1/2$ are equal but these are $\frac{d}{dt}(xy\sin(\theta_1))_{t=1/2}=x'(1/2)y(1/2)+x(1/2)y'(1/2)+x(1/2)y(1/2)\theta_1'(1/2)\cos(\theta_1(1/2))\\ =x'(1/2)y(1/2)+x(1/2)y'(1/2)=\frac{d}{dt}(xy\sin(\theta_2))_{t=1/2}$ done. [asy][asy]size(5cm); 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); draw(dir(36)--dir(144)); draw(dir(144)--dir(216)); draw(dir(216)--dir(324)); draw(dir(324)--dir(36)); [/asy][/asy]

Similiar to the some of the solutions above, ( literally the same thing as post #2 I believe) We can show sides of regular $n-gon$ as vectors in roots of unity. Let $z=e^{\frac{2\pi i}{n}}$. Then black sides are vectors in form $\{ \pm z^{x_1}, \pm z^{x_2},..., \pm z^{x_b} \}$, similiarly, white sides are $\{ \pm z^{y_1},..., \pm z^{y_w} \}$. Negative/positive implies parallel and they all have length $1 $by definition To find the area of $B$, we can decompose the polygon as bunch of triangles such that with choosing any vertex and drawing diagonals from that vertex. It is sufficient to sum up these triangles.Naming the sides of $B$ clockwise as vectors $v_1,v_2,...,v_{2b}$, we can show these triangles' sides as $v_j,\sum_{i=1}^{j-1} v_i,\sum_{i=1}^j v_i$. By complex sholace formula, area of triangle with sides $p,q,p+q$ is $$ \frac{i}{4} \begin{vmatrix} 1&1&1\\ p&q&p+q\\ \overline{p}&\overline{q}&\overline{p+q} \\ \end{vmatrix}= \frac{i}{4}\begin{vmatrix} 0&0&1\\ p&q&p+q\\ \overline{p}&\overline{q}&\overline{p+q} \end{vmatrix}= \frac{i}{4} ( p\overline{q}- \overline{p}q)$$Hence, area of any triangle with side $z^{x_{j+1}}$ is $$ \frac{i}{4} \left( \left( \sum_{i=1}^{j} z^{x_i} \right) \overline{z^{x_{j+1}}}- \left( \sum_{i=1}^{j} \overline{ z^{x_i}} \right) z^{x_{j+1}} \right) $$Since $ 1 \le j \le b-1$ ( $\{b,2b \}$ is repeat o and $|z^i|=1$ the area of $B$ is ( from now on, we ignore i/4 and we will look at the expression only), \begin{align*}& \sum_{j=1}^{b-1} \left(\left( \sum_{i=1}^{j} z^{x_i} \right) \overline{z^{x_{j+1}}}- \left( \sum_{i=1}^{j} \overline{ z^{x_i}} \right) z^{x_{j+1}} \right) =\\ & \ \sum_{j=1}^{b-1} \left( \sum_{i=1}^j \left( \frac{z^{x_i}}{z^{x_{j+1}}}- \frac{z^{x_{j+1}}}{z^{x_i}} \right) \right)= \sum_{1 \le j<k \le b} \left( \frac{z^{x_j}}{z^{x_{k}}}- \frac{z^{x_{k}}}{z^{x_j}} \right) \end{align*}Similiarly, derive area of $A$. It is enough to show that after changing some colors of $2n$-gon, the difference stays same. Change color of any such sides. Then by rotation, WLOG let them represent $z^{x_1},z^{y_1}$. Let the changed versions of polygons be $B',W'$. We have: \begin{align*}&B-W=z^{x_1} \left( \sum_{i=2}^b \frac{1}{z^{x_i} }\right) - \frac{1}{z^{x_1}} \left(\sum _{i=2}^{b} z^{x_i}\right)-z^{y_1}\left( \sum_{i=2}^w \frac{1}{z^{x_i} }\right) + \\ &\frac{1}{z^{y_1}} \left(\sum _{i=2}^{w} z^{x_i}\right)+\sum_{2 \le j<k \le b} \left( \frac{z^{x_j}}{z^{x_{k}}}- \frac{z^{x_{k}}}{z^{x_j}} \right)-\sum_{2 \le j<k \le w} \left( \frac{z^{x_j}}{z^{x_{k}}}- \frac{z^{x_{k}}}{z^{x_j}} \right) \end{align*}and \begin{align*} &B'-W'=z^{y_1} \left( \sum_{i=2}^b \frac{1}{z^{x_i} }\right) - \frac{1}{z^{y_1}} \left(\sum _{i=2}^{b} z^{x_i}\right)-z^{x_1}\left( \sum_{i=2}^w \frac{1}{z^{x_i} }\right) + \\ &\frac{1}{z^{x_1}} \left(\sum _{i=2}^{w} z^{x_i}\right)+\sum_{2 \le j<k \le b} \left( \frac{z^{x_j}}{z^{x_{k}}}- \frac{z^{x_{k}}}{z^{x_j}} \right)-\sum_{2 \le j<k \le w} \left( \frac{z^{x_j}}{z^{x_{k}}}- \frac{z^{x_{k}}}{z^{x_j}} \right) \end{align*}The last thing we have to show is the difference of equations is $0$. Which equals To $$z^{x_1} \left( \sum_{i=2}^n \frac{1}{z^{x_i}}\right)- \frac{1}{z^{x_1}} \left( \sum_{i=2}^n z^{x_i}\right) - z^{y_1} \left( \sum_{i=2}^n \frac{1}{z^{x_i}}\right)+ \frac{1}{z^{y_1}} \left( \sum_{i=2}^n z^{x_i}\right)=0$$ By geometric sum formula $\sum_{i=2}^n z^{x_i} = \frac{z(z^n-1)}{z-1}- z^{x_1}, \sum_{i=2}^n \frac{1}{z^{x_i}} = \frac{ ( \frac{1}{z^n}-1)}{1-z} - \frac{1}{z^{x_1}}$. Using these, our equation becomes \begin{align*}& \frac{ ( \frac{1}{z^n}-1)}{1-z} ( z^{x_1} - z^{y_1})- \frac{z(z^n-1)}{z-1}( \frac{1}{z^{y_1}}- \frac{1}{z^{x_1}})=0 \Rightarrow \\ & \frac{1}{z^n}-1 - z(z^n-1)( \frac{1}{z^{y_1}z^{x_1}})=0 \end{align*}By Euler formula we have $z^n=1$, Hence our last equation is right, therefore they have the same difference, hence we have proven our claim.

Let the length of each rod be 1. Consider starting with all black rods, and flipping pairs of opposite rods until we get the desired configuration. For convention, define the area of an empty set of rods as 0, and the area of a pair of opposite rods and no other rods of that color is also 0. Align the polygon so that the edges we are flipping from black to white are the horizontal ones. After this flip, the polygon $W$ gains an interior region that is a parallelogram whose base is 1 and height is the sum of the vertical components of each white rod on one of the two sides. However, the polygon $B$ loses an interior parallelogram whose base is 1 and height is the sum of the vertical components of each black rod on one of the sides. Therefore, the increase in $W-B$ is equal to the sum of the vertical components of all rods on one side, which depends only on $n$. Therefore, as long as we do the same number of flips, the end result of $W-B$ is the same regardless of which edges we actually flipped, so we are done.

yofro wrote: This is also doable with law of sines, the sine area formula, and a bit of perseverance through the (straightforward) bash *cough cough* totally not me

A year late but whatever Our first goal is to find a nice way to compute the area of both the $2b$-gon and $2w$-gon. Label the sticks $1,2,\ldots, 2n$ in clockwise order. Claim 1: If the $2b$-gon is made up of the sticks $s_1 < s_2 < \ldots < s_b < s_1+n < s_2+n < \ldots< s_b+n$, then the area of the entire $2b$-gon is: \[\sum_{1\leq i<j\leq b} \sin\left(\frac{360}{2n} (s_j-s_i)\right)\]where $s_{b+j} = n + s_j$. Proof: Note that since the $2b$-gon is convex, we know that the sticks must be placed together in increasing order. Then, we may triangulate, and let $B$ be equal to double the area of the first $b$ triangles by symmetry: \begin{align*} B&= 2\cdot \sum_{i=1}^{b-1} \text{area}[0, v_1+\cdots v_i, v_1 + \cdots v_{i+1}] = 2\sum_{i=1}^{b-1} \text{area}[0, v_1+\cdots v_i, v_{i+1} ]\\ &= 2\cdot \frac12 \sum_{i=1}^{b-1} (v_1+\cdots v_i)\times v_{i+1} = \sum_{i=1}^{b-1} v_1\times v_{i+1} +\cdots v_i\times v_{i+1} \\ &= \sum_{1\leq j<i\leq b} v_j\times v_i = \sum_{1\leq j<i\leq b} \sin(\frac{360}{2n} (s_j-s_i)) \end{align*}where the $v_i$ represent the vector of the stick $s_i$. Note that the second equality from above comes from the fact that the height and base are the same. Then, we convert to cross products, and derive the claim. $\square$ If we consider the set of sticks $1,2,\ldots, n$, note that they have to be divided between the white and black polygon. Then, we claim that the overall difference is Claim 2: We may exactly compute the difference between the polygons as \[B-W = (b-w)\cdot \sum_{k< \frac{n}{2}} \sin\left(\frac{360}{2n} k\right) +\begin{cases} \frac{b-w}{2}& \text{n is even} \\ 0 & \text{n is odd}\end{cases} \] Proof: Note that for all $k<n$, the value $(b-w) \cdot \sin\left(\frac{360}{2n} k\right)$ captures the number of terms from Claim 1 with either $s_j-s_i = k$ or $s_j-s_i = (n-k)$ since \[\sin\left(\frac{360}{2n} (n-k)\right) = \sin\left(180 - \frac{360}{2n} k\right)= \sin\left(\frac{360}{2n} k\right).\] But, note that we can split up $1,2,\ldots, n$ into cycles each with step size $k$ (note that if $\gcd(n,k)=1$, we just have one big cycle). Then, within this cycle, the contribution to $B-W$ is the number of (black, black) pairs minus (white, white) pairs along the path, which for cycles of length $\geq 3$ is just equal to the difference between the number of blacks and whites. For 2-cycles (the $i=\frac{n}{2}$ case), it's half that since a $B\leftrightarrow B$ only contributes one. Thus, the claim is clearly true, and very clearly only depends on $(b-w)$ and we're done. $\blacksquare$.

Great problem. (accidentally used $n$ instead of $2n$ oops) I claim that if we swap two adjacent sides of different colors (and the two sides on the opposite side of the polygon), then $B-W$ will remain the same. This implies the statement since we can then obviously make any polygon we want. Pick two adjacent sides of different colors, $A$, $B$. Say $A$ is ahead counterclockwise speaking. As of now, let's leave them uncolored, and their diametrically opposite sides uncolored too. Now consider the polygon formed by the rest of the black sides. It is formed of two congruent halves. If we choose $A$ to color black, then we want to insert this new side into the black polygon. To do this, we got to translate one half of the polygon such that the translation vector is the side we want to insert. The new area is now composed of a bunch of parallelograms: $x$ to be specific, where $x$ is the number of sides in one half of our polygon. The total, new area is \[ \sum_{}^{} \sin(\theta)/2 \]where $\theta$ is the angle our new side makes with each already chosen black side. The formula trivially follows from $ab\sin(C)/2$. Thus, we can work with this new formula. The value of $\theta$ is wholly determined by the number of sides between two given sides. Thus, instead of looking at $\theta$, let's look at the multiset of the number of sides from $A$ to all the existing black sides (0-indexed). Let this be $x$. Similarly let $y$ be the multiset of the number of sides from $A$ to the white sides, let $u$ be from $B$ to black sides, and $v$ be from $B$ to white sides. If we color in $A$ black and $B$ white, the added area corresponds to $x-v$, and if we do it the other way, the added area corresponds to $u-y$. We will show that these two values are equivalent. First, note that $a,n/2+a$ are always in the same set, and are only in one of $x,y$ and one of $u,v$ since they are opposite sides. Now note that $a$ and $n-a$ always form the same angle wrt a specific side, so they can be treated as equivalent. Also, note that $u_i=x_i+1$ and $v_i=y_i+1$ because $B$ is one side behind. In addition, note that $0,n-1,n/2-1,n/2$ aren't in $x,y$ corresponding to the four uncolored sides, and similarly $0,1,n/2,n/2+1$ aren't in $u,v$, Now we can finally look at $x-v,u-y$. Consider an element $t$. Assume $t$ is not $1$ or $n/2+1$, we will take care of those at the end. If $t$ is not in $x$ but in $v$, then $t-1$ and $t$ must be in $y$, but then $t-1$ isn't in $x$, so $t$ is also in $y$ but not $u$. If $t$ is in $x$ but not $v$, then $t-1$ and $t$ are in $x$, so $u$ as $t$ but $y$ hasn't. Finally, if $t$ is in both $x,v$, then since a number can appear at most in $2$ out of the $4$ sets, it is cancelled in $x-v$ and doesn't appear in $u-y$. Now we lastly look at $1,n/2+1$. The trick is to remember that $a \iff n-a$ so we can replace $1,n/2+1 \iff n-1,n/2-1$. Now they are in the same set because their difference is $n/2$. But from here, it's trivial: if they are both in $x$, they are either in $v$ which cancels, or they are in $u$ which causes both of them to appear in $x-v,u-y$, and similarly if both of them are in $y$. Thus, we are done.

This isn't actually too bad if you just do the obvious. Let the side length of the rods be $1$. The idea is to look at what changes when you swap the color of one pair of parallel edges. Label the edges (as vectors) $v_1 = [1, 0]$, $v_2 = [x_2, y_2]$, and so on. WLOG we may assume by rotation that we are swapping the color of $v_1$ and $v_{n+1}$. Let $$M = \sum_{1 < i \leq n, \text{$v_i$ black}} |y_i| \text{ and }N = \sum_{1 < i \leq n, \text{$v_i$ white}} | y_i|.$$Assume that $v_1$ and $v_n$ are both colored black. Claim. Upon swapping the colors, $[B]$ decreases by $M$, and $[W]$ increases by $N$. Proof. We can partition the new polygon $W$ into two polygons that piece into the original $W$, and a parallelogram with opposite sides $v_1$ and $v_n$. The distance between these sides is equal to $M$ by definition of orientation. Similarly, by working in reverse, if we had swapped the colors again, $[B]$ would have increased by $N$, thus $[B]$ decreases by $N$ in the original swap. $\blacksquare$ Thus the value $[W] - [B]$ changes by $M+N$. On the other hand, $M+N$ is equal precisely to the distance between the opposite edges, which is constant. As a result, we can show any two configurations yield the same areas by 1) swapping edges until all edges are black, and 2) swapping edges from the black picture until we get the second configuration.

Very nice problem, hard to wrap your mind around it though. Check what happens when you swap the color of one pair of parallel edges. Take the edges as vectors v_1 = [1,0], v_2 = [x_2, y_2], etc. By rotation, only change the color of v_1 and v_{n+1}, where we start in all black and go to white, which we'll denote by v_i\equiv\{0,1\}. Let P = \sum_{1 < i \leq n,v_i\equiv0} |y_i| and Q = \sum_{1 < i \leq n,v_i\equiv1} | y_i|. In this way, each time we switch from 0 to 1 for the bottom and top vectors, B decreases by a parallelogram of base 1 and height P which is area P, while W increases by Q (this also works vice versa). Indeed, now [W] - [Black] changes by P+Q each time, which is equal to the distance between the opposite edges (constant). By suitably switching colors we can then get a configuration which is only constant on the distance from opposite edges (in fact, this only depends on 2n).

I should remark that this is extremely similar to 2006 IMO 6. I think this is a much cleaner problem due to the regular polygon and pairs of opposite sides are monochromatic conditions. A solution to the IMO problem that is very similar to the vectorial ones given above in the thread can be found here, written by neergard.

It is enough to show that swapping two adjacent sides of different colors preserves this difference, since we may look at only one half of the polygon and repeatedly swap pairs of adjacent edges consisting of a black edge clockwise of a white edge. This must terminate since each step, a white edge moves clockwise one step, so assigning weight $1$ to the edge furthest clockwise, weight $2$ to the second furthest clockwise and so on, the sum of weights must decrease yet remain an integer at all times. Thus, when it is terminated we can see that the configuration must consist of all white edges at the clockwise end and all black edges at the counterclockwise end of our half-polygon. We may reduce any configuration to this, and since all of these final configurations are rotations of each other, their difference of areas must all be equal, which would finish. To show this, we consider swapping some pair of white edges $w$ with an adjacent pair of black edges $b.$ Notice that by drawing the parallelogram given by the four vertices of $w,$ we divide $W$ into three regions: the parallelogram $w,$ and two other congruent possibly degenerate polygons which we will each call $P.$ Notice that all edges of $P$ are unaffected by swapping, since the two swapped white edges are in $w.$ Thus $P$ has the same area before and after the swap, and similarly for the black edges. Then, if we let $w_1,w_2$ be $w$ before and after swapping, respectively, and similarly define $b_1,b_2$ then we want to show that $[w_1]-[b_1]=[w_2]-[b_2],$ which is equivalent to $[w_1]+[b_2]=[w_2]+[b_1],$ considering the area of the four parallelograms. But now notice that two of the edges of $b_2$ are the same as two of the edges of $w_1,$ so the sum of their areas is the length of one of these edges times the sum of their heights. Then, the height of $w_1$ is the component of the sum of all white edges from one of these given edges to another that is perpendicular to that edge, and the height of $b_2$ is the same thing across all black edges. Thus, we see that their sum is the perpendicular component across the path from one given white edge to another, but we must exclude the other swapped edge since it is black before swapping and white after swapping. Thus, the area is just the side length, which is $1,$ times this component. However, the expression we get for $[w_2]+[b_1]$ is symmetric (across the diameter of the polygon connecting the two vertices on two of our swapping edges), so these two are equal and we are done.

Consider a fixed $n$ with all rods being white. WLOG all sides lengths are $1$. Claim: Changing two opposite rods from white to black (or vice versa) changes the area difference between $W$ and $B$ by some constant which only depends on $n.$ Proof: Rotate the $2n$-gon so the rods we are changing are horizontal with the $x$-axis. WLOG we are changing from white to black. We are removing these two white edges. This effectively removes a parallelogram from $W$ with area \[A_w = \sum_{\text{right half of white rods}} y_i = \frac{1}{2}\sum_{\text{all white rods}} |y_i|\]where $y_i$ are the differences in $y$ coordinates of each rod. Note the two sums are the same, just expressed in two ways for easier understanding. Similarly, we've increase the area of $B$ by \[A_w = \sum_{\text{right half of black rods}} y_i = \frac{1}{2}\sum_{\text{all black rods}} |y_i|\]Thus the difference in their area changed by \[\frac{1}{2} \sum_{\text{all rods}} |y_i| = \text{height of $2n$-gon}\]Thus since the height depends on $n$, we're done. $\square$ The difference in areas only depends on how many rods we swap colors for, which only depends on $b$ and $w$. Thus by our claim, since each of these switches only depends on $n = b+w$, the total area difference only depends on $b$ and $w$ as desired.

It suffices to show that swapping pairs of different colors (say $b_1,b_2$ and $w_1,w_2$) preserves $[B]-[W]$. Upon swapping, the effect on $B$ is the deletion of a parallelogram of base $1$ and height (say) $h_{bB}$, followed by the addition of a parallelogram of base $1$ and height $h_{wB}$. Hence, the net effect on $[B]$ is $-h_{bB}+h_{wB}$. Analogously, the net effect on $[W]$ is $-h_{wW}+h_{bW}$, and so it suffices to show \begin{align*} -h_{bB}+h_{wB}&=-h_{wW}+h_{bW}\\ h_{bB}+h_{bW}&=h_{wB}+h_{wW}. \end{align*}But the LHS is merely the distance between $b_1,b_2$ in the initial $2n$-gon barring the components of $w_1,w_2$ in the direction perpendicular to $b_1,b_2$; and the RHS is the distance between $w_1,w_2$ in the initial $2n$-gon barring the components in the $b_1,b_2$ direction perpendicular to $w_1,w_2$. Since $\text{comp}_\textbf{v}\textbf{w}=\text{comp}_\textbf{w}\textbf{v}$ if $\Vert\textbf{v}\Vert=\Vert\textbf{w}\Vert$ (and here that is so), we are done. $\square$

Translate the sides of the $2n$-gon to the $2n$th roots of unity. Let $\zeta = e^{\frac{i\pi}{n}}$, and label the root of unity corresponding to $\zeta^k$ as $k$. Furthermore, let $y_k$ be the $y$-coordinate of the $k$th RoU. Note that we only care about the sides $\{0, 1, \dots, n-1 \}$. Claim: If a polygon is made of the sides $A \subset \{0, 1, \dots, n-1 \}$, then the area of $A$ can be expressed as \[ \sum_{i, j \in A \atop i < j} \sin(\frac{\pi}{n} (j-i))\] Proof: We can induct on the number of sides. Base case: 2 sides is trivial by parallelogram area formula. Inductive step: By constructing parallelograms, we can reduce a polygon from $2k$ sides to a polygon with $2k-2$ sides. Let $j$ be the side removed; then, the sum of the areas of the parallelograms is \[\sum_{i \in A \atop i \neq j} \sin(\frac{\pi}{n} |j-i|) \]and we finish by the inductive hypothesis. Claim: If the color of an edge is swapped from black to white, the increase of the difference in area is the same regardless of the position of the edge swapped (and vice versa). WLOG we can rotate to swap edge $0$ (and $n$ by extension). Let $B, W$ be the set of black, white edges, respectively (they can be empty). Then, by again ``parallelogramizing'' the new white polygon into the old white polygon and parallelograms containing edges $0, n$, we have the area increases by \[ \sum_{1 \leq i < n \atop i \in W} |y_i| \] Via a similar process, we can see that the area of the black polygon decreases by \[\sum_{1 \leq i < n \atop i \in B} |y_i|\] Since $W \cap B = \emptyset$ and $W \cup B = \{0, 1, \dots, n-1 \}$, we have that $[W] - [B]$ is the same regardless of which edge we flip. Thus, the statement follows by setting the polygon to all black and flipping edges to white, since the order of which we flip doesn't matter, and we are done.

Hi @above It suffices to show that the difference is the same after swapping two adjacent sides of the same color, so swap two sides $s_1$ and $s_2$ which are different in color. The effect to say $B$ is shown in the following image. Essentially, the before and after polygon, are basically the same: they each consist of the same parts joined by a parallelogram, but the difference is that this parallelogram is different for each one. For the old polygon, it is $\vec{v}$ by $\vec{s_1}$, and the new one is $\vec{v}$ by $\vec{s_2}$, and the other polygon is similar but swapped and using a different vector $\vec{u}$ instead of $\vec v$. Therefore, letting brackets denote area as usual and the parallelograms being defined by their components (and drop arrows for latex reasons), it suffices to show \[[v,s_1]-[v,s_2]=[u,s_2]-[u,s_1] \iff [u,s_1]+[v,s_1]=[u,s_2]+[v,s_2] \iff [u+v,s_1]=[u+v,s_2]\]But now, we consider what $u$ and $v$ actually are. Starting at say the left $s_1$ and going clockwise, one is the vector sum of the black edges and the other is the sum of the white edges. Therefore, $u+v$ is the vector from one endpoint of $s_1$ to the endpoint of $s_2$ on the other side as shown in the image. Therefore we finish by symmetry

why is my writeup so complicated Since $\Lambda^2(\mathbb{R}^{\oplus2})\cong\mathbb{R}$, we identify the elements of $\Lambda^2(\mathbb{R}^{\oplus2})$ as real numbers. Call a unit vector in $\mathbb{R}^{\oplus2}$ xooks if the angle it makes with the $x$-axis is in $[0,\pi)$. For a set $V\subseteq\mathbb{R}^{\oplus2}$ of xooks vectors, let $A(V)$ be the area of the convex polygon formed with $2|V|$ rods in the directions of the vectors in $V$, and each direction of a vector in $V$ is present in exactly $2$ rods. Lemma. Let $V\subseteq\mathbb{R}^{\oplus2}$ be a set of xooks vectors. Let $v\in V$ be the vector forming the smallest angle with the $x$-axis. Then we have \[ A(V)=A(V\setminus\{v\})+\left|v\wedge\sum_{u\in V\setminus\{v\}}u\right|=A(V\setminus\{v\})+\sum_{u\in V\setminus\{v\}}|v\wedge u|. \] Proof. Drawing the vectors and their opposites (vectors in $V\cup(-V)$) in place of the rods, it follows that \[ \sum_{u\in V\setminus\{v\}}u \]is the vector from the head of $v$ to the the tail of $-v$. Since the wedge product is just the area of the parallelogram, the conclusion follows. $\square$ By the lemma and a simple induction, we have \[ A(V)=\sum_{\{v,u\}\in[V]^2}|v\wedge u|. \]Note that $|v\wedge u|$ is equal the sine of the angle between $v$ and $u$. Consider the set \[ X=\left\{\left(\cos\frac{\pi x}{n},\sin\frac{\pi x}{n}\right)^\top\mid x\in\{0,\ldots,n-1\}\right\} \]of xooks vectors. $X$ represents a regular $2n$-gon, and each vector in $X$ is represented by a unique element of $\{0,\ldots,n-1\}$. Note that the angle between $\left(\cos\frac{\pi x}{n},\sin\frac{\pi x}{n}\right)^\top$ and $\left(\cos\frac{\pi y}{n},\sin\frac{\pi y}{n}\right)^\top$ is $\left|\frac{\pi}{n}(y-x)\right|$. Let $\sim$ be an equivalence relation on $\mathbb{Z}$ where $x\sim y$ if and only if $\left|\sin\frac{\pi x}{n}\right|=\left|\sin\frac{\pi y}{n}\right|$. Note that $\sim$ is given by $x\sim x+n$ and $x\sim -x$. Claim. Let $(B,W)$ be a partition of $S:=\{0,\ldots,n-1\}$. Then the $(n-1)$-tuple \[ (r_1(B,W),\ldots,r_{n-1}(B,W)) \]depends only on $|B|$ and $|W|$, where \[ r_i(B,W):=|\{\{b_1,b_2\}\subseteq B\mid b_2-b_1\sim i\}|-|\{\{w_1,w_2\}\subseteq W\mid w_2-w_1\sim i\}|. \] Proof. Consider two elements $x,y\in\mathbb{Z}_n$ and a partition $(B',W')$ of $S\setminus\{x,y\}$. We have \begin{align*} |\{z\in B'\mid z-x\sim i\}|+|\{z\in W'\mid z-x\sim i\}|&=|\{z\in S\setminus\{x,y\}\mid z-x\in\{i,-i,i-n,n-i\}\}|\\ &= \begin{cases} 1, & \text{if $i=\frac{n}{2}$}\\ 2, & \text{otherwise} \end{cases} \\ &=|\{z\in S\setminus\{x,y\}\mid z-y\in\{i,-i,i-n,n-i\}\}|\\ &=|\{z\in B'\mid z-y\sim i\}|+|\{z\in W'\mid z-y\sim i\}| \end{align*}so \[ |\{\{b_1,b_2\}\subseteq B'\cup\{x\}\mid b_2-b_1\sim i\}|+|\{\{w_1,w_2\}\subseteq W'\cup\{x\}\mid w_2-w_1\sim i\}|=|\{\{b_1,b_2\}\subseteq B'\cup\{y\}\mid b_2-b_1\sim i\}|+|\{\{w_1,w_2\}\subseteq W'\cup\{x\}\mid w_2-w_1\sim i\}|. \]Thus \[ r_i(B'\cup\{x\},W'\cup\{y\})=r_i(B'\cup\{y\},W'\cup\{x\}) \]for all $i\in\{1,\ldots,n-1\}$. Since switching an element of $B$ with an element of $W$ keeps $(r_1,\ldots,r_{n-1})$ constant, the conclusion follows. $\square$ Let $(\mathcal{B},\mathcal{W})$ be a partition of $X$ corresponding to the black and white rods, and $(B,W)$ be a partition of $\{0,\ldots,n-1\}$ corresponding to $\mathcal{B}$ and $\mathcal{W}$. It is easy to see that \[ A(\mathcal{B})-A(\mathcal{W})=\sum_{i=1}^{n-1}r_i(B,W)\sin\frac{\pi i}{n}, \]which depends only on $b$ and $w$. $\square$

We will show that swapping an adjacent pair of black and white edges (and their opposite sides) will not affect the difference in the area. Let the two sides we are swapping be $s_1$ and $s_2$ and suppose that they are of different colors. By doing this, only a parallelogram is changed in the respective polygons. In $B$, let the vector that is not $s_1$ or $s_2$ that defines the respective parallelograms be $\vec{u}$. Similarly, let the vector in $W$ be $\vec{v}$. In order to grasp the areas of the parallelograms, we will denote $[s,\vec{a}]$ as the area of the parallelogram with base $s$ shifted by the vector $\vec{a}$. We wish to prove that \[[s_1,\vec{u}]-[s_2,\vec{u}] = [s_2, \vec{v}] - [s_1,\vec{v}].\] This can be manipulated to \[[s_1,\vec{(u+v)}] = [s_2,\vec{(u+v)}].\] Vector $\vec{u}$ is the vector sum of the black edges and $\vec{v}$ is the vector sum of the white edges. Hence, the vector $\vec{(u+v)}$ goes from one endpoint of $s_1$ to the opposite endpoint of $s_2$ on the other side of the polygon. We finish by symmetry. $\blacksquare$

I hope this is right (also I'm sorry for the bad writeup) Let the vectors of the sides of $B$ be $v_1$, $v_2$, $\dots$, $v_b$, $-v_1$, $\dots$, $-v_n$. We will use the following formula \[[B]=\sum_{1\le i<j\le b}v_i\times v_j\] To see why this is true, translate the vectors $v_2$, $v_3$, $\dots$, $v_b$ by vector $v_1$. The sum \[v_1\times\left(\sum_{i\ge2}v_i\right)\]is equal to the area lost when translating and the area remaining is \[\sum_{2\le i<j\le b}v_i\times v_j\]by induction. Hence the formula follows. Now consider what happens when we swap a white edge and its pair to black. Label the vectors in the regular $2n$-gon by $t_1$, $t_2$, $\dots$, $t_n$, $-t_1$, $\dots$, $-t_n$. Suppose WLOG that we change $t_1$ and $-t_1$ from white to black. Also say that $t_{x_1}$, $t_{x_2}$, $\dots$, $t_{x_b}$ are the vectors of the sides of $B$ before adding $t_1$ and define $t_{y_i}$ similarly. Notice that by the same argument as above, $[B]$ increases by \[t_1\times\left(\sum_{i=1}^bt_{x_i}\right)\]and $[W]$ decreases by \[t_1\times\left(\sum_{i\ge2}^wt_{y_i}\right)\]Hence the difference changes by \[t_1\times\left(\sum_{i\ge 2}t_i\right)\]and this is equal to the area formed by the parallelogram with opposite sides $t_1$ and $-t_1$ in the $2n$-gon. This is clearly constant so we are done. $\blacksquare$

It's not spiral, but it's still great. Wonderful. It suffices to show that swapping an adjacent pair of black and white rods (and their parallel opposite pair as well) preserves $[B]-[W]$. Consider how this swapping only affects the position of four segments, two white and two black. Each half of black on the two sides of this swapping can be pieced into two halves of $B$ in the exact same way as before, albeit missing two pieces. The only change in area will be the parallelogram that connects these two halves, made from the two missing pieces. This parallelogram has a single base of length $1$ and a variable height. Call the heights of $W$ and $B$ to this base in the original (pre-swapping) configuration $h_{W1}$ and $h_{B1}$, respectively, and call them $h_{W2}$ and $b_{B2}$ in the new configuration. By parallelogram area, we seek to prove that \[h_{W1} - h_{W2} = h_{B1} - h_{B2}.\]Now consider dropping the perpendicular from one of the swapped sides in the original configuration as so (red is white): The sum of the lengths of the black segments is $h_{B1}$, while the sum of the lengths of the red (white) segments is $h_{W2}$. We then have the relation \[h_{B1}+h_{W2} = L,\]where $L$ is the length of the entire height. Dropping the perpendicular from the other swapped side gives \[h_{B2}+h_{W1} = L.\]Now subtracting these two equations yields the result. $\blacksquare$