Mathematical misfits two-dimensional solution

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Submitted by plusadmin on 1 September, 2002
September 2002

Mathematical Misfits - two-dimensional

Take a square of sidelength 1; it has area $1 \mbox{ unit}^2$. The biggest circle that can fit inside has diameter 1 and area $\pi \times (0.5)^2 = \pi /4 \mbox{ units}^2$.

The area of this circle divided by the area of the square containing it is $\pi /4$.

Now we fit the largest possible square inside that circle of diameter 1. We use Pythagoras’ Theorem to find its sidelength $d$.

  $\displaystyle 2d^2 $ $\displaystyle = $ $\displaystyle 1; $    
  $\displaystyle d $ $\displaystyle = $ $\displaystyle 1/\sqrt{2}. $    

The area of this square is $1/2 \mbox{ units}^2$.

The area of this square divided by the area of the circle containing it is $0.5/\left(\pi /4\right) = 2/\pi .$

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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