You may also like

problem icon

Golden Thoughts


problem icon

From All Corners


problem icon

Star Gazing


Circuit Training

Stage: 4 Challenge Level: Challenge Level:1

A neat solution by Nazia, and similar ones from a number of other students from Nottingham Girls' High School, showed that an organised approach could be fruitful; but where else did they meet?

When Mike has finished the first circuit, Monisha will be $\frac{2}{3}$ of the way round.

When Mike has finished the second circuit, Monisha will be $1\frac{1}{3}$ of the way round.

When Mike has finished the third cicuit, Monisha will have just finished her second circuit.

This is where they will cross for the first time, and when they both will meet at the start again.

The answer is that they will meet at the start again after Mike has run $3$ circuits and Monisha has run $2$ circuits.

Fred and Matt of Albion Heights took a slightly different approach and were able to identify an intermediate passing place:

The first question asks if they will ever meet at the start if Mike runs a lap in $2/3$ the time it takes Monisha. Since speed is directly related to time, Mike's speed is $3/2$ greater than Monisha's speed.

They will meet on Mike's third lap and Monisha's second lap.

The second question asks where will they meet on the first lap.

Well, if Mike can travel $3$ units in one minute and and Monisha can travel $2$ units in one minute, then the total distance travelled will be $5$ units.

Therefore, if they were on opposite ends of a $400$ m track, Mike will be $\frac{3}{5}$ of the way round the track and Monisha $\frac{2}{5}$ of the way around when they meet. If you make the denominator of $\frac{3}{5}$ $400$, you get $\frac{240}{400}$.

Therefore, they will meet $240 \; \text{m}$ down the track starting from Mikes end of the track.

But will they meet again part of the way around this circular track?
Last but not least Andrei of the Tudor Vianu National College suggests the following:, which starts from exactly the same argument as the one used by Fred and Matt:

Let Mike be runner No. $1$, and Monisha runner number $2$. All physical quantities (period: time for each to run around the circuit, velocity, distance travelled) for Mike will be indexed with $1$, while for Monisha with $2$.

From the data of the problem, I see that the relation between their periods is: $$ T_1 = \frac{2}{3}T_2$$ This implies that their velocities are in an inverse relation: $$v_1 = \frac{3}{2}v_2$$ These relations could be read as: in the same time, runner $1$ completes $3$ units, runner $2$ only completes $2$ circuits.

So, if the circuit is divided into $5$ parts, runner $1$ completes $3$ and runner $2$ only $2$ parts. They meet first at $\frac{3}{5}$ of the circuit in the direction Mike runs, and at $\frac{2}{5}$ of circuit in the direction Monisha runs.

For the first meeting they must complete a circuit ($400 \; \text{m}$).

The second meeting corresponds to completing $2$ circuits, with $\frac{6}{5}\times 400 = 480$ m travelled by Mike, and $\frac{4}{5} \times 400 = 320 \; \text{m}$ by Monisha. This point is at $80 \; \text{m}$ from the start in the direction Mike runs.

The third meeting is at a multiple of three of the first (as distances travelled) so at $320 \; \text{m}$ from the start in the direction Mike runs.

The fourth is at $160 \; \text{m}$ from the start, in the same direction.

The fifth is at the start, because the circuit was divided into $5$ parts, as dictated by the relation between the periods.

More generally, if the relation between the periods would be $$\frac{T_1}{T_2} = \frac{p}{q}$$ with $p$ and $q$ coprime, then the runners meet again at the start after completing in total $(p+q)$ circuits, this means they meet the $(p+q)$th time.