7. (4 points) Given a right triangle $A B C$, where angle $A$ is 60 degrees, and the hypotenuse $A B$ is $2+2 \sqrt{3}$. Through vertex $B$, a line $p$ parallel to $A C$ is drawn. Points $D$ and $E$ are placed on line $p$ such that $A B=B D, B C=B E$. $F$ is the intersection point of lines $A D$ and $C E$. Find what the perimeter of triangle $D E F$ can be.

Answers:

$$
\begin{gathered}
2 \sqrt{2}+\sqrt{6}+9+5 \sqrt{3} \\
\sqrt{3}+1+\sqrt{2} \\
9+5 \sqrt{3}+2 \sqrt{6}+3 \sqrt{2}
\end{gathered}
$$

$$
1+\sqrt{3}+\sqrt{6}
$$
Solution: First, we calculate the sides of triangle $ABC$: side $AC$ lies opposite the angle $30^{\circ}$, so it is half the hypotenuse, which is $1+\sqrt{3}$. Now, using the Pythagorean theorem, we calculate $BC=\sqrt{(2+2\sqrt{3})^{2}-(1+1\sqrt{3})^{2}}=\sqrt{4+8\sqrt{3}+12-1}$ $\sqrt{9+6\sqrt{3}+3}=6+2\sqrt{3}$.

Triangle $CBE$ is a right isosceles triangle with a right angle at $B$, so line $CE$ intersects $p$ at an angle of $45^{\circ}$. Triangle $ABD$ is isosceles with the angle at vertex $B$ being $120^{\circ}$ or $60^{\circ}$ depending on the position of point $D$. Therefore, line $AD$ intersects $p$ at an angle of $60^{\circ}$ or $30^{\circ}$.

Depending on the positions of points $D$ and $E$ on line $p$, triangle $DEF$ has angles of 30, 45, and $105^{\circ}$; 30, 135, and $15^{\circ}$; 60, 45, and $75^{\circ}$; or 120, 45, and $15^{\circ}$.

Consider the height $FH$ in triangle $DEF$. From the size of angle $E$, it follows that $EH=FH$.

In a right triangle with angles $60^{\circ}$ and $30^{\circ}$, the legs differ by a factor of $\sqrt{3}$, and triangle $DFH$ is precisely such a triangle. Therefore, in the first two cases, $DH=\sqrt{3}FH$, and in the remaining two cases, $DH=\frac{FH}{\sqrt{3}}$. Thus, $DE=|DH \pm EH|$ (the minus sign appears in the obtuse triangle $DEF$ with an obtuse angle at vertex $D$ or $E$), that is, when points $D$ and $E$ lie on the same side of $B$), and therefore, $DE=FH|\sqrt{3} \pm 1|$ or $DE=FH\left|1 \pm \frac{\sqrt{3}}{3}\right|$. Express $FH$ in terms of $DE$. Side $EF=\sqrt{2}$ as the hypotenuse in the right isosceles triangle $EFH$, and in the right triangle $DFH$, the hypotenuse is twice the smaller leg.

Now, let's calculate the desired perimeter in each case separately.

Angles 30, 45, and 105 degrees. Points $D$ and $E$ lie on opposite sides of point $B$ and $DE=BD+BE=AB+BC=3\sqrt{3}+5$. $FH=\frac{DE}{\sqrt{3}+1}=\frac{3\sqrt{3}+5}{\sqrt{3}+1}=\frac{1}{2}(3\sqrt{3}+5)(\sqrt{3}-1)=2+\sqrt{3}$, then $EF=2\sqrt{2}+\sqrt{6}$ and $DF=2FH=4+2\sqrt{3}$. We get a perimeter of $2\sqrt{2}+\sqrt{6}+9+5\sqrt{3}$.

Angles 30, 135, and 15 degrees. Points $D$ and $E$ lie on the same side of point $B$ and $DE=BD-BE=AB-BC=\sqrt{3}-1$. $FH=\frac{DE}{\sqrt{3}-1}=1$, then $EF=\sqrt{2}$ and $DF=2FH=2$. We get a perimeter of $\sqrt{3}+1+\sqrt{2}$.

Angles 60, 45, and 75 degrees. Points $D$ and $E$ lie on opposite sides of point $B$ and $DE=BD+BE=AB+BC=3\sqrt{3}+5$. $FH=\frac{DE}{\frac{\sqrt{3}}{3}+1}=\sqrt{3} \cdot \frac{DE}{\sqrt{3}+1}=2\sqrt{3}+3$, then $EF=2\sqrt{6}+3\sqrt{2}$ and $DF=\frac{2}{\sqrt{3}}FH=4+2\sqrt{3}$. We get a perimeter of $9+5\sqrt{3}+2\sqrt{6}+3\sqrt{2}$.

Angles 120, 45, and 15 degrees. Points $D$ and $E$ lie on the same side of point $B$ and $DE=BD-BE=AB-BC=\sqrt{3}-1$. $FH=\frac{DE}{1-\frac{\sqrt{3}}{3}}=\sqrt{3} \cdot \frac{DE}{\sqrt{3}-1}=\sqrt{3}$, then $EF=\sqrt{6}$, and $DF=\frac{2}{\sqrt{3}}FH=2$. We get a perimeter of $1+\sqrt{3}+\sqrt{6}$.