66th Belarusian Mathematical Olympiad

Answer: 3.

We show that at most three circumscribed quadrilaterals can be obtained. Suppose, contrary to our claim, that there are at least four circumscribed quadrilaterals. Then some two of them are obtained when one diagonal (say, $AD$) is constructed.



Fig. 1



Fig. 2

Consider one of the circumscribed quadrilaterals remained. Without loss of generality we assume that this is the quadrilateral $BCDE$. For this quadrilateral we have $BC+DE=CD+BE$. For the quadrilateral $ABCD$ we have $AB+CD=BC+AD$ since, by our assumption, this quadrilateral is circumscribed. Summing these two equalities we obtain $AB+DE=AD+BE$. The last equality is impossible because $AD+BE>AB+DE$ for the convex quadrilateral $ABDE$. Indeed, if $M$ is the intersection point of $AD$ and $BE$, then, by the triangle inequality,

$$\begin{gathered} AD+BE=(AM+MD)+(BM+ME)=(AM+BM)+(DM+EM)> \\ >[AM+BM>AB,DM+EM>DE]>AB+DE. \end{gathered}$$

Thus the number of the circumscribed quadrilaterals is less than or equal to 3.

There are many ways to construct the convex hexagon with three circumscribed quadrilaterals. One of such hexagons is shown in Fig. 2, where $A_1{B}_1{A}_2{B}_3$, $A_2{B}_2{A}_3{B}_1$, $A_3{B}_3{A}_1{B}_2$ are the circumscribed quadrilaterals.