4. A square piece of paper $A B C D$ with side 1 is folded along a line so that vertex $A$ coincides with the midpoint of side $C D$. What is the area of the resulting hexagon?
Answer: $61 / 96$.

Solution. The diagram is as follows:

![](https://cdn.mathpix.com/cropped/2024_05_06_e307cd835ab5e38efb5cg-1.jpg?height=488&width=559&top_left_y=2162&top_left_x=748)

Let $E$ be the midpoint of side $CD$, $Q$ the midpoint of $AE$, $FQ \perp AE$, $R$ the intersection point of line $FQ$ with side $BC$, $P$ the intersection point with line $AB$, $ET \perp FE$, $RH \perp TH$. Folding the paper along a line is performing an axial symmetry with respect to that line. Since point $A$ maps to point $E$, the line along which we fold is the axis of symmetry, i.e., the perpendicular bisector of segment $AE$; in our diagram, this is line $FQ$. We need to find the area of the hexagon $RHTCDF$. By the Pythagorean theorem for triangle $ADE$, $AE = \sqrt{5} / 2$. Right triangles $ADE$ and $AQF$ are similar, so

$$
AF = AE \cdot AQ / AD = 5 / 8
$$

Triangle $FAP$ is also similar to them, so $AP = 2AF = 5 / 4$, from which

$$
S_{FAP} = AF \cdot AP / 2 = 25 / 64
$$

Next, triangle $PBR$ is also similar to all of them, so $RB = PB / 2 = 1 / 8$, since

$$
PB = PA - AB = 5 / 4 - 1 = 1 / 4
$$

from which $S_{PBR} = 1 / 64$. Finally, $HR = BR = 1 / 8$, triangle $RHT$ is similar to triangle $FDE$, so $TH = RH \cdot ED / FD = 1 / 6$ (since $FD = 3 / 8$), and thus $S_{THR} = 1 / 96$. Ultimately,

$$
S_{RHTCDF} = S_{ABCD} - S_{FAP} + S_{PBR} + S_{THR} = 61 / 96.
$$