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## Mathematical misfits two-dimensional solution

September 2002

## Mathematical Misfits - two-dimensional

Take a square of sidelength 1; it has area . The biggest circle that can fit inside has diameter 1 and area .

The area of this circle divided by the area of the square containing it is .

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

The area of this square is .

The area of this square divided by the area of the circle containing it is

Since , the round peg fits better in the square hole than the square peg fits in the round hole.

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