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Two circles (Posted on 2004-10-21) Difficulty: 3 of 5
In a 8½x11 sheet of paper I drew two equal non-overlapping circles -- both completely inside the paper, of course.

What's the largest portion of the paper I could cover with the circles?

What would be the answer if I drew THREE equal circles?

No Solution Yet Submitted by Federico Kereki    
Rating: 3.2500 (4 votes)

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Solution three circle case | Comment 4 of 11 |

We can assume that two of the circles will lie vertically on the paper tangent to one side (say the left side) of the paper and the third tangent to the opposite (right) side.  The lone circle will be tangent to each of the other two. As the circles expand in size, the two that are tangent to the left side of the paper will separate, and a maximum size will be reached when they become tangent to the top and bottom.

Consider just half the paper: 5.5 inches high and 8.5 inches wide. On this, consider the line joining the centers of the top-left and the right-hand (middle) circles.  The top-left center is r units down and r units from the left.  The right-hand center is on the bottom edge of the half-page and r units from the right.

By the Pythagorean Theorem,

(2r)^2 = (5.5-r)^2 + (8.5-2r)^2

which works out to

r^2 - 45r + 5.5^2 + 8.5^2 = 0

which comes out to r = 2.4064687..., or a solution outside the bounds of the paper.

The only thing to watch out for would be if this is too large a radius to fit four of them into the 11-inch height of the paper. Indeed it fits, so there is no overlap of the circles.

  Posted by Charlie on 2004-10-21 17:32:50
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