## Task 4 - 120614

A distance of $168 \mathrm{~m}$ in length is to be divided into three segments, whose lengths are denoted in sequence by $a, b, c$. The second segment is to be three times as long as the first, and the third segment is to be four times as long as the first.

Determine all possibilities for specifying the lengths $a, b, c$ of the segments so that a division with these properties is achieved!
Assuming the lengths $a, b, c$ have the required properties. Then:

$a+b+c=168 \mathrm{~m}$, further $b=3 a$ and $c=4 a$.

This leads to: $a+3 a+4 a=168 \mathrm{~m}$, thus $8 a=168 \mathrm{~m}$, from which we get $a=21 \mathrm{~m}$, so $b=63 \mathrm{~m}$ and $c=84 \mathrm{~m}$.

Therefore, only these lengths can satisfy the given conditions. Since $63 \mathrm{~m}=3 \cdot 21 \mathrm{~m}$, $84 \mathrm{~m}=4 \cdot 21 \mathrm{~m}$, and $21 \mathrm{~m}+63 \mathrm{~m}+84 \mathrm{~m}=168 \mathrm{~m}$, they indeed have the required properties.

Solutions of the I. Round 1972 taken from [5]

### 3.13.2 II. Round 1972, Class 6