A problem by Russian mathematician Viktor Prasolov:
A circle is divided by points into 3k arcs such that there are k arcs each of length 1, 2, and 3. Prove that two of the division points are diametrically opposite one another.
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Suppose that this is not the case, that none of the division points are diametrically opposite one another. This means that every arc of length 1 must stand opposite an arc of length 3. Remove one such pair, an arc of length 1 and its opposing arc of length 3. Now we’ve divided the circle into two large arcs.
If one of these large arcs contains m arcs of length 1 and n arcs of length 3, then the other contains n arcs of length 1 and m arcs of length 3. The total number of arcs of length 1 and 3 in the two large arcs is 2(k– 1), so n + m = k – 1.
Apart from arcs of length 1 and 3, each large arc contains only arcs of length 2, which don’t change the parity (odd or even) of its total length. This means that the parity of the length of each large arc will be the same as the parity of k – 1. But the length of each large arc is (6k – 1 – 3)/2 = 3k – 2. That’s a contradiction: k – 1 and 3k – 2 have opposite parities. So our assumption, that none of the division points on the original circle are diametrically opposite one another, cannot be the case.
(From Prasolov’s Problems in Plane and Solid Geometry, 1988.)
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