Math CentralQuandaries & Queries


Question from Alexander:

Have you ever solved a problem, in which you have a rectangle, from which you need to cut the largest two circles of equivalent diameter? I bisected a rectangle diagonally, but the circles, while tangent to two of the sides, are not tangent to eachother. Can you devise a method for two equivalent circles, that are tangent to two sides, are also to eachother?

Take for example a piece of paper, Each if the two largest circles has a diameter that is greater than the distance to the midpoint of the diagonal bisector of the rectangle.

Hi Alexander,

The answer depends on the dimensions of your rectangle. Start with a long, thin rectangle (whose length is in the horizontal direction and is at least twice its width); the two LARGEST congruent circles are those tangent to both long sides; if you want, you can take them tangent to one another at the center of the rectangle. Their common tangent is the vertical line through the center of the rectangle. They will be tangent also to the short sides, one on the right, the other on the left, when the rectangle's length is exactly twice the width. Now keep the width fixed but shrink the length. One circle will stay tangent at the top and left, and the congruent mate will be tangent on the bottom and the right; they will be tangent to one another at the center of the rectangle, and the common tangent will be a slanted line. Eventually the rectangle becomes a square (with length equal to width), and the common tangent will be the diagonal that runs from the bottom left corner to the top right corner.


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