Answer: $S=\frac{2}{3}$.

Solution. Mark a point $M$ on side $AD$ of square $ABCD$ such that $AM=\frac{1}{3}$. Cut the contour of the square at vertices $B, C$, and $D$, and at point $M$. From the resulting parts, form a right-angled triangle $A_1B_1C_1$ with legs $A_1B_1=1$, $A_1C_1=A_1M_1+M_1C_1=\frac{1}{3}+1=\frac{4}{3}$, and hypotenuse $B_1C_1=B_1F+FC_1=1+\frac{2}{3}=\frac{5}{3}$ (see Fig. 9.3). The existence of this triangle follows from the converse of the Pythagorean theorem. Its area is $S=\frac{1}{2} \cdot 1 \cdot \frac{4}{3}=\frac{2}{3}$.

We will show how one could find this

![](https://cdn.mathpix.com/cropped/2024_05_06_5797a033a3e95e041854g-2.jpg?height=463&width=711&top_left_y=217&top_left_x=1186)
way of cutting (participants in the olympiad are not required to do this). Let $AM=x$. Form a right-angled triangle $A_1B_1C_1$. One of its legs is $A_1B_1=1$, the other is $A_1C_1=A_1M_1+M_1C_1=x+1$. The hypotenuse is $B_1C_1=B_1F+FC_1=1+(1-x)=2-x$. The triangle with these sides is right-angled if $1^2+(x+1)^2=(2-x)^2$, from which $x=\frac{1}{3}$.

It can be proven that the given solution method is unique up to congruence of triangles. This was not required of the olympiad participants. It can also be shown that four is the minimum number of parts into which the contour of the square can be cut to form a triangle.

Grading criteria. “+” A complete and justified solution is provided “士” The correct method of cutting and assembling is shown, including the correct sizes of the parts, but the area of the triangle is not found or is found incorrectly “Ғ” The correct cutting scheme is shown, but the sizes of the parts are not indicated or are found incorrectly

“-” The problem is not solved or is solved incorrectly
9.3. Come up with a way to cut the contour of a square with a side length of 1 into four parts and assemble these parts into the contour of a triangle. Find the area of the resulting triangle. (The contour has no thickness. Bending and unfolding the parts is not allowed)