1972 IMO Problems/Problem 2
Prove that if , every quadrilateral that can be inscribed in a circle can be dissected into quadrilaterals each of which is inscribable in a circle.
Actually, should work. Our initial quadrilateral will be .
For , we do this:
Take with sufficiently close to respectively. Take such that is an isosceles trapezoid, with close enough to (or close enough to ) that we can find a circle passing through (or ) which cuts the segments in . Our four cyclic quadrilaterals are .
For we do the exact same thing as above, but now, since we have an isosceles trapezoid, we can add as many trapezoids as we want by dissecting the one trapezoid with lines parallel to its bases.
The above solution was posted and copyrighted by grobber. The original thread for this problem can be found here: 
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