4. Two straight railway lines intersect at point $N$ at an angle of $60^{\circ}$. Inside this angle is an airfield (point $A$) located 10 km and 20 km from these railways. Find the locations of points $B$ and $C$ (loading and unloading points) on these railways, so that the cost of cyclic transportation by road from $A$ to $B$, then to $C$, and back to $A$ is minimized. Assuming that the transportation of 1 ton of cargo by road along $A B C A$ costs 100 rubles per km, determine:

a) whether it is possible to transport 10 tons of cargo for a sum of 50 thousand rubles, without considering the costs of loading and unloading the cargo?

b) The same question for a sum of 60 thousand rubles, considering the costs of complete reloading of the cargo (2500 rubles for 10 tons) at 2 out of 4 points? at 3 out of 4 points? at all 4 points?
## Solution.

1) Construct point $A$, and reflect it symmetrically relative to the railways: we get points $A_{1}$ and $A_{2}$ (see fig.)

## (2 points)

2) Connect points $A_{1}$ and $A_{2}$, then the intersection of this line with the railways gives the locations of points $B$ and $C$.

## (+1 point)

3) Then the perimeter of triangle $A B C$ is minimal and equal to the segment $A_{1} A_{2}$. If this is proven, then

![](https://cdn.mathpix.com/cropped/2024_06_04_c790142104ba3f7af959g-18.jpg?height=720&width=597&top_left_y=842&top_left_x=1363)
it is additionally evaluated. (+1

## point)

(The proof is based on the following principle: if at least one of the points $B$ or $C$ moves, then the perimeter of triangle $A B C$ will be equal to a broken line connecting points $A_{1}$ and $A_{2}$.)

4) Measurement with a ruler gives $A_{1} A_{2} \approx 53$ km (Comparison with a compass at a scale of 10 km gives $5060000$.

(+1 point)

6) With a reload at three out of four points, the total will be:

$53000+2500 * 3=60500$.

Here, a more precise estimate is required $2 p_{A B C}=A_{1} A_{2} . \quad \mathbf{( + 1} \quad$ point)(7 points)

## IX Team and Individual Tournament "Mathematical Pentathlon"

October 31 - November 5, 2016, Moscow

## Geometry (solutions)

## Senior League