1972 IMO SL #10

I got several emails about how I haven't posted in a week, while I usually post during the first few days of the week.

So don't worry, I'm still alive :D 

Now let's get to the math.

Given $n>4$, prove that every cyclic quadrilateral can be dissected into $n$ cyclic quadrilaterals.

We prove a stronger statement, namely $n \ge 4$. 

Let the cyclic quadrilateral be $ABCD$. 

First, we prove for $n=4$. Take $E\in AB,F\in CD,\ EF\|AD$ with $E,F$ sufficiently close to $A,D$ respectively. Take $U\in AD,V\in EF$ so that $AEVU$ is an isosceles trapezoid, with $V$ close enough to $F$ (or $U$ close enough to $D$) so 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 then $BCFE,\ AEVU,\ VFYX,\ YXUD$.

For $n \ge 5$ we can do the exact same thing as $n=4$, but this time, since we have an isosceles trapezoid, we can add as many isosceles trapezoids we need (depending on the value of $n$) by dissecting the one trapezoid with lines parallel to its bases. 

And we're done. $\square$


However, if you don't want to get into proving the extra stuff (the "stronger statement" part), you can simply do the $n \ge 5$ part. 


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