1972 IMO Problems/Problem 2

Problem

Prove that if $n > 3$, every quadrilateral that can be inscribed in a circle can be dissected into $n$ quadrilaterals each of which is inscribable in a circle.

Solution

Actually, $n\ge 4$ should work. Our initial quadrilateral will be $ABCD$.

For $n=4$, we do this:

Take $E\in AB,F\in CD,\ EF\|AD$ with $E,F$ sufficiently close to $A,D$ respectively. Take $U\in AD,V\in EF$ such that $AEVU$ is an isosceles trapezoid, with $V$ close enough to $F$ (or $U$ close enough to $D$) that we can find a circle passing through $U,D$ (or $F,V$) which cuts the segments $UV,DF$ in $X,Y$. Our four cyclic quadrilaterals are $BCFE,\ AEVU,\ VFYX,\ YXUD$.

For $n\ge 5$ 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: [1]

See Also

1972 IMO (Problems) • Resources
Preceded by
Problem 1
1 2 3 4 5 6 Followed by
Problem 3
All IMO Problems and Solutions
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