3. On the sides $B C$ and $C D$ of the square $A B C D$, points $E$ and $F$ are chosen such that the angle $E A F$ is $45^{\circ}$. The length of the side of the square is 1. Find the perimeter of triangle $C E F$.
Solution: If point $D$ is reflected relative to line $A F$, and then relative to line $A E$, it will transition to point $B$. Indeed, the composition of two axial symmetries relative to intersecting lines is a rotation by twice the angle between the lines. In our case, these two symmetries are equivalent to a $90^{\circ}$ rotation around point $A$. This means that the image of point $D$ under symmetry relative to $A F$ and the image of point $B$ under symmetry relative to $A E$ are

![](https://cdn.mathpix.com/cropped/2024_05_06_ad676a4db1fdd5e6ad2eg-1.jpg?height=420&width=384&top_left_y=2303&top_left_x=1547)

## Interregional School Olympiad in Mathematics for Students of Departmental Educational Organizations

the same point; on the diagram, it is denoted by $K$. From point $K$, segments $A E$ and $A F$ are seen at an angle of $90^{0}$ (angles are preserved under symmetry, so, for example, angles $A B E$ and $A K E$ are equal). Therefore, point $K$ is the foot of the perpendicular dropped from point $A$ to line $E F$. Finally, since $B E = E K$ and $D F = F K$ (lengths of segments are preserved under symmetry), we see that the perimeter of triangle $C E F$ is equal to the sum of the lengths of sides $B C$ and $C D$ of the square.

Answer: 2.