Mathematical misfits two-dimensional solution
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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