# Problem 3.

Two swimmers are training in a rectangular quarry. The first swimmer finds it more convenient to exit at a corner of the quarry, so he swims along the diagonal to the opposite corner and back. The second swimmer finds it more convenient to start from a point that divides one of the quarry's shores in the ratio $2018: 2019$. He swims along a quadrilateral, visiting one point on each shore, and returns to the starting point. Can the second swimmer choose points on the other three shores such that his path is shorter than the first swimmer's? What is the minimum value that the ratio of the length of the longer path to the shorter can have?
# Solution.

Let's draw a rectangular quarry $A D C D$ and the inscribed quadrilateral route of the second swimmer $N L M K$.

Reflect the drawing symmetrically first with respect to side $C D$, then with respect to side $C B^{\prime}$, and finally with respect to side $A^{\prime} B^{\prime}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_e4646b86c42e5aa31aa7g-2.jpg?height=905&width=1202&top_left_y=1558&top_left_x=313)

By construction, segments $N D$ and $N^{\prime} D^{\prime}$ are equal and parallel to each other. Therefore, $A A^{\prime} N^{\prime} N$ is a parallelogram, and its side $N N^{\prime}$ is equal to twice the diagonal of the original rectangle $A C$ (i.e., the length of the first swimmer's path).

Olympiad for schoolchildren "Nadezhda Energetiki". Final stage. Solutions

It is clear that the sum of the lengths of segments $N K, K M^{\prime}, M^{\prime} L^{\prime}$, and $L^{\prime} N^{\prime}$ is equal to the perimeter of the quadrilateral $N L M K$ (i.e., the length of the second swimmer's path).

But the length of the broken line $N K M^{\prime} L^{\prime} N^{\prime}$ cannot be less than the length of the segment $N N^{\prime}$, which is the shortest distance between points $N$ and $N^{\prime}$. From this, it follows that the second swimmer's path cannot be shorter than the first swimmer's path.

The paths can be made equal if we transfer (using the symmetry of the construction) the points of intersection of the segment $N N^{\prime}$ with the sides $D C, C B^{\prime}$, and $B^{\prime} A^{\prime}$ to the original rectangle. (It can be shown that the resulting figure will be a parallelogram, but this is not part of the problem statement.)

With such a construction, the swimmers' paths will be equal, and their ratio will be 1.

Note that nowhere in the reasoning is it used that the initial point $N$ divides the corresponding side of the rectangle $A D$ in a certain proportion.

## Answer:

the second swimmer's path cannot be shorter than the first swimmer's path; the minimum ratio of the length of the longer path to the shorter one is 1.