## Task 4 - 140624

A cyclist rode with constant speed on a road from $A$ to $B$. He started in $A$ at 6:00 AM and covered $14 \mathrm{~km}$ every hour. A second cyclist rode on the same road with constant speed from $B$ to $A$. He started on the same day at 8:00 AM in $B$ and covered $21 \mathrm{~km}$ every hour.

Both cyclists met exactly at the midpoint of the road from $A$ to $B$.

At what time did they meet? How long is the road from $A$ to $B$?
The first cyclist had been cycling for exactly two hours by 8:00 AM, covering a distance of 28 km.

From this point on, the second cyclist covered exactly 7 km more per hour than the first.

Since they met exactly at the midpoint of the route from A to B, and because $28 \div 7 = 4$, this happened exactly 4 hours after the second cyclist started, which was at 12:00 PM.

By this time, because $6 \times 14 = 84$ and $4 \times 21 = 84$, each of them had covered exactly 84 km.

The length of the route from A to B is therefore $2 \times 84 = 168$ km, because $2 \times 84 = 168$.

Solutions of the second round 1974 taken from [5]

### 3.16 XV. Olympiad 1975

### 3.16.1 First Round 1975, Class 6