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# Answer to the July puzzle: [Preview]

For a circle with a radius of 10 centimeters or more, any pattern of red and blue must satisfy the 10-centimeter bicoloration condition. Here¿s the proof: Consider two points in the circle, R (which is red) and B (which is blue), which are between 10 and 20 centimeters apart (for the proof that two such points must exist, see the following paragraphs). Draw circles with a radius of 10 centimeters around each point. These circles must intersect at one or two points, and at least one of these points must lie within the original circle. Label this point Q. If Q is red, then the line segment B-Q is 10 centimeters long and bicolored. If Q is blue, then the segment R-Q is 10 centimeters long and bicolored. Either way, the pattern satisfies the 10-centimeter rule.

How do we know that a circle C with a radius of at least 10 centimeters must contain a red point R and a blue point B that are between 10 and 20 centimeters apart? Suppose the circle¿s center point P is red. Draw a smaller circle CP with a radius of 10 centimeters around P. If any point on the circle¿s perimeter is blue, we know the circle satisfies the 10-centimeter condition. So assume the entire perimeter is red. Now consider these two cases:

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