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Professor Erasmus informed me today that he has proved another fascinating theorem about chessboards.

Consider a standard $8\times8$ chessboard with 32 black and 32 white squares. Pick an arbitrary (black or white) square $x$. Compute the sum $B$ of the 32 squared distances between the center of $x$ and the centers of the 32 black squares. Compute the sum $W$ of the 32 squared distances between the center of $x$ and the centers of the 32 white squares. Then independently of the choice of $x$, you will always find that $B=W$.

Has the professor once again made one of his well-known mathematical blunders, or does his claimed theorem indeed hold true?

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Place a white 1kg weight at the center of each white square and a black 1kg weight at the center of each black square. Then $B$ is the rotational inertia of the black weights around $x$, and $W$ is the rotational inertia of the white weights around $x$.

By the Parallel Axis Theorem, the rotational inertia of an object depends only on (1) its total mass, (2) its moment about the center of mass, and (3) the distance from its center of mass to the axis of rotation. These quantities are evidently the same for the set of black weights and the set of white weights, by symmetry. So their rotational inertia around any point $x$ is equal for both sets. It follows that $B=W$.

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    $\begingroup$ The rotational inertia also depends on the moment about the center of mass. $\endgroup$ – Julian Rosen Oct 4 '15 at 17:03
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    $\begingroup$ @JulianRosen - correct, this should be added (together with the remark that both moments of inertia are equal). Also, it is worthwhile to note that Lopsy's reasoning generalizes the Erasmus checkerboard theorem to locations non-central to square X. $\endgroup$ – Johannes Oct 4 '15 at 17:16
  • $\begingroup$ Good catch Julian, edited. $\endgroup$ – Lopsy Oct 5 '15 at 2:12
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    $\begingroup$ The first sentence is a bit ambiguous: do you place thirty-two weights of each color or just one? This ambiguity is immediately cleared up, but it might still be worth rewording. $\endgroup$ – Lynn Oct 14 '15 at 6:33
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The professor

is not as dumb as some people may be tempted to think.

Let the intersection cell of row $i$ and column $j$ be $(i,j)$. Let the chosen cell be $(x,y)$.

Therefore,

\begin{align} B &=\sum_{(i,j) \mathrm{ is\ black}} (x-i)^2 + (y-j)^2\\&= \sum_{(i,j) \mathrm{\ is\ black}}(x-i)^2 + \sum_{(i,j) \mathrm{\ is\ black}}(y-j)^2\\&=x_b+y_b \end{align}

Similarly,

\begin{align} W&=\sum_{(i,j) \mathrm{\ is\ white}} (x-i)^2 + \sum_{(i,j) \mathrm{\ is\ white}}(y-j)^2\\&=x_w+y_w \end{align}

Since, there are exactly 4 black and 4 white cells each row,

$$\sum_{(i,j) \mathrm{\ is\ black}} (x-i)^2=4\sum_{i=1}^8(x-i)^2=\sum_{(i,j) \mathrm{\ is\ white}} (x-i)^2$$

So, $x_w=x_b$ Similarly, $y_w=y_b$

Therefore, $B=W$

TM;DR (Too mathy, didn't read):

The first thing to understand is the sum of squares of distances from a cell to all black cells is the sum of squares of distances of the x coordinates $(x_b)$ plus the sum of squares of distances of the y coordinates $(y_b)$. This follows from Pythagoras/distance formula.

But since there are an equal number of black and white cells each row, The set of distances from black cells is actually the same set as the set of distances from white cells. Therefore $x_b$ and $x_w$ are summing the squares of the same set and therefore are equal. Similarly, $y_b$ and $y_w$ are equal.

Therefore B and W are equal.

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Let us take a cell $(x,y)$ and four cells in a block $(a,b),(a+1,b),(a,b+1),(a+1,b+1)$. We know that two of them are black and two are white. We can prove that the sum of squared distances of $(x,y)$ from the black and white cells is equal.

Using Pythagoras theorem, $$B=[d\ to\ (a,b)]+[d\ to\ (a+1,b+1)]$$ $$=[(x-a)^2+(y-b)^2]+[(x-a-1)^2+(y-b-1)^2]$$

and $$W=[d\ to\ (a,b+1)]+[d\ to\ (a+1,b)]$$ $$=[(x-a)^2+(y-b-1)^2]+[(x-a-1)^2+(y-b)^2]$$

On comparinging, we find these two equal. Hence we conclude that for any cell and a block of $4$ other cells, this theorem is true.

We divide the $8\times 8$ checkerboard into $16\ (2\times 2)$ blocks. Since the theorem holds for distances to each of the blocks, it also holds for distances to all of the blocks.

The only block not taken into account is the block where the cell itself resides. It is easy to show that this theorem is true in the cell's own block. ($1^2+1^2=(\sqrt2)^2$)

Hence proved.

P.S.

I did not cheat from @Anachor 's answer. I couldn't understand it because I am yet to learn how to use the $\sum$ symbol properly. I knew Pythagoras theorem had to be used right from the beginning and made up the proof on my own.

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  • $\begingroup$ You don't have to take the cell's own block as a separate case. In that case "d to (.., ..)" equals 0, and that's fine. And by the way, the sigma symbol is very simple, if you just read it as "sum" you'll usually be fine. The subscript and superscript just tell us where the variables come from, like for example, $\sum_{v\in \mathbb{Z}, 1\le v^2\le 20}\, 1/v$ can be read as "sum over all integers v, whose squares are between 1 and 20, of 1/v". Sometimes some constraints on the variable are left out because they can be inferred from context. $\endgroup$ – Ben Frankel Oct 5 '15 at 2:25

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