Mathematical misfits two-dimensional solution

Mathematical misfits two-dimensional solution

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.


Go back to the main puzzle page
By submitting this form, you accept the Mollom privacy policy.