## Task 4 - 170514

A distance of $240 \mathrm{~m}$ is divided into four segments such that the following conditions are met:

(1) The first segment is twice as long as the second segment.

(2) The second segment is as long as the sum of the lengths of the third and fourth segments.

(3) The third segment is one-fifth of the total distance.

Determine the lengths of the individual segments!
Let the four segments be $a, b, c, d$. The equations are:

$$
a=2 b ; \quad b=c+d ; \quad 5 c=240 \text{ m} ; \quad a+b+c+d=240 \text{ m}
$$

From the third relation, we immediately get $c=48 \text{ m}$. Substituting $c+d$ (from the 2nd equation) and also (1) into the fourth equation, we get $240 \text{ m}=a+b+c+d=2 b+b+b=4 b$. Thus, $b=60 \text{ m}$ and therefore $a=120 \text{ m}$.

$d$ is obtained from $d=240 \text{ m}-a-b-c=12 \text{ m}$.

The segments have lengths of $120 \text{ m}, 60 \text{ m}, 48 \text{ m}$, and $12 \text{ m}$.

Official solution: According to (2), the sum of the lengths of the second, third, and fourth segments is equal to twice the length of the second, i.e., according to (1), equal to the length of the first segment. Therefore, this is half of the total length $240 \text{ m}$, which is $120 \text{ m}$, and the second segment is $60 \text{ m}$ long.

According to (3) and because $250: 5=48$, the third segment is $48 \text{ m}$ long. From the sum $60 \text{ m}$ of the lengths of the third and fourth segments, $60-48=12$ remains for the length of the fourth segment, which is $12 \text{ m}$.

The required lengths are therefore $120 \text{ m}, 60 \text{ m}, 48 \text{ m}, 12 \text{ m}$.