Consider an equilateral triangle ABC as given in the following diagram: [Equilateral triangle with A at the top, C at bottom-left, B at bottom-right; arrows indicate clockwise movement along the sides.] Two people start at the same time from points A and B with speeds 30 km per hour and 20 km per hour respectively, and move on the sides of the triangle in the clockwise direction. They meet each other for the first time at
- Apoint C
- Ba point between C and A
- Ca point between A and B
- Dpoint A
Explanation
The faster person (30 km/h) gains on the slower (20 km/h) at a relative speed of 10 km/h. Starting one side-length s behind along the clockwise track, he closes the gap in s/10 hours, during which he travels 30 × (s/10) = 3s — exactly one full lap of the triangle. He therefore catches the slower person back at the starting point A. Hence they meet at point A.