Problem 6. (Variant 2).

$C A K D$ is a square with a side length of 6. On side $C D$, a point $B(B D=2)$ is chosen, and on line $A D$, a point $E$ is chosen such that the perimeter of triangle $B E C$ is the smallest possible. Then, on line $D C$, a point $F$ is marked such that the perimeter of triangle $F E A$ is the smallest possible. Find $E F$.

## Construction and Proof:

Mark point $\mathrm{B}_{1}$ on side DK $\left(\mathrm{B}_{1} \mathrm{D}=\mathrm{BD} \Rightarrow B_{1} B \perp A D\right)$. Draw line $\mathrm{B}_{1} \mathrm{C}$, which intersects AD at point E. The perimeter of triangle CBE is the smallest because, among all possible points $\mathrm{E}_{1}$ on line $\mathrm{AD}$, the sum of the lengths of segments $\mathrm{B}_{1} \mathrm{E}+\mathrm{EC}$ is the smallest ($\mathrm{B}_{1} \mathrm{E}+\mathrm{EC}<\mathrm{B}_{1} \mathrm{E}_{1}+\mathrm{E}_{1} \mathrm{C}$ - triangle inequality) and $\mathrm{B}_{1} \mathrm{E}=\mathrm{EB}$.

Similarly, mark point $\mathrm{A}_{1}$ on side $\mathrm{AC}$ $\left(\mathrm{A}_{1} \mathrm{C}=\mathrm{AC}\right)$. Draw line $\mathrm{A}_{1} \mathrm{E}$, which intersects CD at point F. The perimeter of triangle AFE is the smallest because, among all possible points $F_{1}$ on line $\mathrm{AD}$, the sum of the lengths of segments $\mathrm{A}_{1} \mathrm{~F}+\mathrm{EF}$ is the smallest $\left(\mathrm{A}_{1} \mathrm{~F}+\mathrm{EF}<\mathrm{A}_{1} \mathrm{~F}_{1}+\mathrm{F}_{1} \mathrm{E}\right.$ - triangle inequality) and $\mathrm{A}_{1} \mathrm{~F}=\mathrm{FA}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_3fdcc7ef9dfdd3dab8ecg-08.jpg?height=839&width=554&top_left_y=203&top_left_x=1362)
# Solution:

$C D=A C=6, \quad B D=B_{1} D=2$. Since triangles $A C E$ and $\mathrm{B}_{1} \mathrm{ED}$ are similar ( $\angle A E C=\angle B_{1} E D$ - vertical, $\angle E A C=\angle B_{1} D E=45^{0}$ ),

then $E C: B_{1} E=A C: D B_{1}=A E: E D=6: 2=3: 1$.

Let $E H \perp D C \Rightarrow$ right triangles ACD and EHD are similar ( $\angle A C D-$ common), then $C H: H D=A C: E H=A D: E D=4: 1 \Rightarrow E H=H D=1.5 \Rightarrow C H=4.5$.

Right triangles $\mathrm{FHE}$ and $\mathrm{FCA}_{1}$ are similar ( $\angle E F H=\angle A_{1} F C-$ vertical), then:

$C F: H F=A_{1} F: E F=A_{1} C: E H=6: 1.5=4: 1 ; C H=4.5 \Rightarrow H F=0.2 \cdot 4.5=0.9 \Rightarrow$ $F E^{2}=H F^{2}+E H^{2}$

$$
F E=\sqrt{0.81+2.25}=\sqrt{3.06}=0.3 \sqrt{34}
$$

Answer: $0.3 \sqrt{34}$.

## Grading Criteria for Tasks

| Task | 1 | 2 | 3 | 4 | 5 | 6 | Total |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| Points | 15 | 15 | 15 | 15 | 20 | 20 | 100 |