Problem 1

Let

n\ge5

be an integer. Consider

n

squares with side lengths

1,2,\ldots,n

, respectively. The squares are arranged in the plane with their sides parallel to the

x

and

y

axes. Suppose that no two squares touch, except possibly at their vertices. Show that it is possible to arrange these squares in a way such that every square touches exactly two other squares.

Step 3 of 3: Close the loop when the difference is 2 and check non-overlap

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\left(\sum_{a\in A}a\right)\sqrt2+\frac{(n-3)+(n-2)}{2}\sqrt2=\left(\sum_{b\in B}b\right)\sqrt2+\frac{n+(n-1)}{2}\sqrt2,\qquad \frac{a_i+b_j}{2}\sqrt2\le(n-4)\sqrt2<\frac{2n-3}{2}\sqrt2

Detailed analysis

When $\sum_{a\in A}a-\sum_{b\in B}b=2$ , attach $n-3$ and $n-2$ to the $A$ -diagonal and $n$ and $n-1$ to the $B$ -diagonal; since $(n+(n-1))-((n-3)+(n-2))=4=2\cdot2$ , the first identity above holds and the two chains again close up at the four corner squares. In both cases the perpendicular distance between the $A$ - and $B$ -diagonals is $\tfrac{(n-3)+n}{2}\sqrt2=\tfrac{2n-3}{2}\sqrt2$ , whereas any $a_i\in A$ and $b_j\in B$ satisfy $a_i,b_j\le n-4$ , so $\tfrac{a_i+b_j}{2}\sqrt2\le(n-4)\sqrt2<\tfrac{2n-3}{2}\sqrt2$ . Hence the two parallel chains never touch each other, and every square in the loop touches only its two neighbours along the cycle.