## Task A-1.7.

In the figure, square $ABCD$ has a side length of 1. If $ABEF$ and $KLMN$ are congruent rectangles, determine the length of $|BE|$.

![](https://cdn.mathpix.com/cropped/2024_05_30_4cdaff2107ef283b1d94g-05.jpg?height=505&width=522&top_left_y=1478&top_left_x=1098)
## Solution.

Triangles $K E L$ and $M D N$ have the same angles and congruent hypotenuses, so $|D N|=|E L|$. 1 point Let $N^{\prime}$ and $L^{\prime}$ be the feet of the perpendiculars from point $N$ to side $\overline{B C}$ and from point $L$ to side $\overline{A D}$, respectively.

Rectangles $L N^{\prime} N L^{\prime}$ and $K L M N$ share a common diagonal, so they are congruent. 2 points

From this, it follows that $\left|L N^{\prime}\right|=|K L|=|B E|$.

![](https://cdn.mathpix.com/cropped/2024_05_30_4cdaff2107ef283b1d94g-06.jpg?height=714&width=708&top_left_y=622&top_left_x=603)

Now we conclude that $|B L|=|B E|+|E L|=\left|L N^{\prime}\right|+|N D|=\left|L N^{\prime}\right|+\left|N^{\prime} C\right|=|L C|$. Therefore, $L$ is the midpoint of side $\overline{B C}$, i.e., $|L C|=\frac{1}{2}$.

The right triangle $M L C$ has a hypotenuse of length 1 and one leg of length $\frac{1}{2}$, so $\varangle C L M=60^{\circ}$ and $|C M|=\frac{\sqrt{3}}{2}$.

It also holds that $\varangle D M N=60^{\circ}$, so

$$
|B E|=|M N|=2|D M|=2 \cdot\left(1-\frac{\sqrt{3}}{2}\right)=2-\sqrt{3}
$$

## SCHOOL/CITY COMPETITION IN MATHEMATICS

## 2nd grade - high school - A variant

January 28, 2019.

## IF A STUDENT HAS A DIFFERENT APPROACH TO SOLVING THE PROBLEM, THE COMMITTEE IS OBLIGED TO GRADE AND EVALUATE THAT APPROACH APPROPRIATELY.