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

I have not made much progress yet.

When I tried to do it for $n=4$, I noticed that it may not be possible to move from $4$ dissections to $5$ or more while maintaining the given conditions.

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marked as duplicate by Jack D'Aurizio, G. Sassatelli, C. Falcon, Joel Reyes Noche, Jennifer Jun 28 at 5:06

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up vote 1 down vote accepted

Clearly, it is sufficient to show that any cyclic quadrilateral can be dissected into 5,6 or 7 cyclic quadrilaterals.

Dissection in 5: let $ABCD$ be our initial ciclic quadrilateral, with $X=AC\cap BD$. Let $A',B',C',D'$ the images of $A,B,C,D$ through a circular inversion with respect to a circle centered in X with a small radius. The quadrilateral $A'B'C'D'$ is cyclic since the circumcircle of $ABCD$ is mapped into a circle containing $A',B',C',D'$; $AA'B'B$ is cyclic since $XA\cdot XA'=XB'\cdot XB$.

Dissection in 7: cut from ABCD two small cyclic quadrilateral $ABC'D'$,$A'B'CD$, such that $A'B'$ is parallel to $AB$ and $C'D'$ is parallel to $CD$, then dissect $A'B'C'D'$ into 5 cyclic quadrilaterals.

Dissection in 6: cut from ABCD two small cyclic quadrilateral $ABC'D'$,$A'B'CD$, such that $A'B'$ is parallel to $AB$ and $C'D'$ is parallel to $CD$, then connect the circumcenter of $A'B'C'D'$ with the midpoints of the sides, splitting $A'B'C'D'$ into 4 cyclic quadrilaterals.

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