Tokaeva I.

Let $F_{1}, F_{2}, F_{3}, \ldots$ be a sequence of convex quadrilaterals, where $F_{k+1}$ (for $k=1,2,3, \ldots$) is obtained by cutting $F_{k}$ along a diagonal, flipping one of the parts, and gluing it back along the cut line to the other part. What is the maximum number of different quadrilaterals that this sequence can contain? (Different polygons are those that cannot be made to coincide by a movement.)
Let $ABCD$ be the original quadrilateral $F_{1}$. We can assume that each time the half of the quadrilateral containing side $CD$ is flipped, while side $AB$ remains stationary. In this process, the sum of angle $A$ and the opposite angle does not change. Additionally, the set of side lengths does not change. However, the three lengths (sides other than $AB$) can be ordered in only six ways.

We will prove that if two resulting quadrilaterals $ABKL$ and $ABMN$ have the same ordered side lengths ($BK = BM$, $KL = MN$, $LA = NA$), then the quadrilaterals are equal. It is sufficient to prove the equality of the diagonals $AK$ and $AM$. Assuming, for example, that $AK > AM$, we get $\angle ABK > \angle ABM$, $\angle ALK > \angle ANM$, which contradicts the equality of the sums $\angle ABK + \angle ALK$ and $\angle ABM + \angle ANM$.

Six different quadrilaterals will result from any quadrilateral where all sides are different, the sums of opposite angles are also different, and the quadrilateral remains convex and does not degenerate into a triangle during transformations. Only quadrilaterals with the reverse order of sides might be suspected of being equal, but they are not equal either, since flipping one of them will not align the sums of opposite angles. The highlighted condition is satisfied, for example, for quadrilaterals whose area is greater than half the product of the two longest sides: this inequality is clearly not true for non-convex and degenerate quadrilaterals. A specific example: a right trapezoid with bases 3 and 6 and height 4.

## Answer

6 quadrilaterals.