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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.