Task B-2.7. Three bicycle paths with a total length of $24 \mathrm{~km}$ form a right-angled triangle. From the vertex of the right angle, two cyclists start simultaneously at constant speeds. One rides along one path, and the other along the other path. While one cyclist covers $5.5 \mathrm{~km}$, the other covers $6.5 \mathrm{~km}$ in the same time. If the cyclists meet exactly at the midpoint of the hypotenuse, what are the lengths of the paths, i.e., the sides of the right-angled triangle?
## Solution.

Let $a$ be the length of the path one cyclist rides, and $b$ be the length of the path the other cyclist rides. The length of the hypotenuse is $c$. By the time they meet, the first cyclist has traveled $a+\frac{c}{2}$, and the second cyclist has traveled $b+\frac{c}{2}$ kilometers. Since the speeds are constant, the ratio of the distances traveled is constant and equals $\frac{5.5}{6.5}$. We write

$$
\begin{array}{rl}
\frac{a+\frac{c}{2}}{b+\frac{c}{2}}=\frac{5.5}{6.5}=\frac{11}{12} \\
a+\frac{c}{2}=11 k & a=11 k-\frac{c}{2} \\
b+\frac{c}{2}=13 k & b=13 k-\frac{c}{2}
\end{array}
$$

From the perimeter $a+b+c=24$, we get $k=1$.

Applying the Pythagorean theorem, we get

$$
\begin{gathered}
\left(11-\frac{c}{2}\right)^{2}+\left(13-\frac{c}{2}\right)^{2}=c^{2} \\
c^{2}+48 c-580=0
\end{gathered}
$$

The positive solution to the equation is $c=10$.

(2 points)

The lengths of the paths are $a=6 \mathrm{~km}, b=8 \mathrm{~km}, c=10 \mathrm{~km}$.

## COUNTY COMPETITION IN MATHEMATICS

## 3rd grade - high school - B variant

## February 15, 2013.

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