Mathematical Misfits - two-dimensional

Take a square of sidelength 1; it has area $1{\text{ unit}}^2$. The biggest circle that can fit inside has diameter 1 and area $\pi \times (0.5{)}^2=\pi /4{\text{ units}}^2$. \par The area of this circle divided by the area of the square containing it is $\pi /4$. \par Now we fit the largest possible square inside that circle of diameter 1. We use Pythagoras' Theorem to find its sidelength $d$. $$\begin{array}{rcl}2d^2& =& 1;\\ d& =& 1/\sqrt{2}.\end{array}$$ The area of this square is $1/2{\text{ units}}^2$. \par

The area of this square divided by the area of the circle containing it is $0.5/\left(\pi /4\right)=2/\pi .$ \par Since $\pi /4>2/\pi$, the round peg fits better in the square hole than the square peg fits in the round hole.

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