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 2 of 3: Close the loop when the difference is 1

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\left(\sum_{a\in A}a\right)\sqrt2+\frac{(n-3)+(n-1)}{2}\sqrt2=\left(\sum_{b\in B}b\right)\sqrt2+\frac{n+(n-2)}{2}\sqrt2

Detailed analysis

String the squares of $A$ along one $45^\circ$ diagonal (opposite corners touching) and the squares of $B$ along a parallel $45^\circ$ diagonal. Place the pair $\{n,n-3\}$ on one perpendicular end-diagonal and $\{n-1,n-2\}$ on the other. When $\sum_{a\in A}a-\sum_{b\in B}b=1$ , attach the squares of sides $n-3$ and $n-1$ to the $A$ -diagonal and those of sides $n$ and $n-2$ to the $B$ -diagonal; since $(n+(n-2))-((n-3)+(n-1))=2=2\cdot1$ , the two span lengths are equal by the identity above.