Day 12: Rotating Tables!

A rectangle is small if its diagonal is less than or equal to $b+2$ . So we need to find the pairs that satisfy this inequality where a and b are positive integers and $a \geq b$ : $\sqrt{a^2+b^2} \leq b+2$ As both sides are positive we can square both sides to give: $a^2+b^2 \leq b^2+4b+4$ Therefore: $b^2 \leq a^2 \leq 4b+4$ We can solve $b^2 \leq 4b+4$ to show that $b \leq 2+2\sqrt{2}$ so that the possible values of $b$ are $1, 2, 3$ and $4$ . Analysing all cases gives the pairs for $(a,b)$ : $(1,1),(2,1),(2,2),(3,2),(3,3),(4,3),(4,4)$ and therefore the answer is $\boxed{7}$ .