Copyright © University of Cambridge. All rights reserved.
Well done to Daniel from Kings School, New
Zealand, Josh from Chatham Grammar School for Boys, and Anna
from the Sandon School, who all correctly worked out the dimensions
of the running track. Here is Josh's solution:
Radius of the curved sections:
As the straight sections of the track are both 85m long, 170m of
the track is straight. That means that the remaining 230m of the
track is comprised of the two semicircular sections. As there are
two semicircular sections, each of equal radii, the sum of their
inside edges (of the inside lanes) is equal to the circumference of
a circle that would fit neatly into the inside curves of the
track.
The circumference of a circle, $c$, is given by the equation: $c =
2 \times \pi \times r$, where $r$ is the radius.
Therefore, as we have deduced that the circumference of this
imaginary circle is equal to the sum of the inside curves of the
track, we can write: $230 = 2 \times \pi \times r$, so $r =
\frac{230}{2\pi} = 36.606$ metres (to 3 d.p.)
This measurement is only the measurement of the radii of the
semicircles which comprise the inside edge of the inside lane on
the curved section. We want to find a measurement for the radii of
the semicircles which comprise the whole of both curved sections;
there are 8 lanes that we must consider in order to do this. As
each lane is a constant 1.25 metres, the sum of the thickness of
the lanes is equal to 8 x 1.25 = 10 metres. Therefore, the radius
of the curved sections, in order to produce a scale drawing, is
equal to: 36.606 + 10 = 46.606 metres (to 3 d.p.)
The 200m Staggered Start:
As it is mentioned in the problem for the 400m start that the
measurement for the length of a given lane is that of its inside
edge, I will assume that it is the inside edges of each lane that
must be made constant in the 200m race in order to make the race
fair. The runner in lane 1 starts at the curved section on the
bottom right of the track, so I will use that lane for comparison
with the others. The straights do not need to be compensated for in
the staggered start, as the runners would all run the same distance
here. Therefore, we only need to consider the 115m of track that
the runners will run on before they hit the straight; in this time,
all runners will be running on the curved section of track.
The runner in lane 2 is running on a track with an inside edge that
is 1.25m further outwards than that of lane 1. Therefore, the
radius of runner 2's track is equal to $36.606 + 1.25 = 37.856$
metres. Thus, the circumference of the semicircle that makes up the
curved section is equal to: $\frac{2 \pi r}{2} =
\pi r = 118.927$ metres This is 3.927 metres longer than the
inside edge of lane 1, so the rune in lane 2 will start 3.927
metres in front of the runner in lane 1.
We can also generate an nth term sequence for the lanes. The nth
term for this sequence is: $(36.606 + 1.25(n-1)) \times \pi$ I
arrived at this conclusion because the formula for the arc of a
semicircle is pi x r, hence the reason for multiplying the sequence
by pi, and the section in brackets is the way to determine the
radius of the inside edge of any given track. Each runner, as we
move from lane 1 to lane 8, will start 3.927 metres (3 d.p.) in
front of the previous runner.
The 400m Staggered Start:
The obvious conclusion for this problem would be to say that each
runner starts twice as far behind the runner in the next inside
lane as they did in the 200m race, but I will investigate this
mathematically.
Each runner is running 400 metres, but 170 metres of the track
they will run (85 x 2) is made up of straight sections, where the
runners will run the same distance regardless of the lane they are
in, so this does not need to be compensated for. Thus, 230m of the
track they will run on is comprised of a curved section, which must
be compensated for. When the runner in lane 2 reaches a curved
section, the radius of the inner edge of his/her lane is equal to
$36.606 + 1.25 = 37.856$ metres. Thus, the arc lengths of both
semicircles together is equal to: $2 \times \pi \times 37.856
= 237.854$ metres (3 d.p.) This is 7.854 metres longer than the
length of curved track in lane 1, so the runner in lane 2 will
start 7.854 metres in front of the runner in lane 2.
We can also generate an nth term sequence for the lanes. The nth
term for this sequence is: $2 \times ((36.606 + 1.25(n-1))
\times pi)$. I arrived at this conclusion using the nth term
formula I generated before to calculate the arc length of one of
the semicircular pieces of track, then multiplied it by 2 (for the
two semicircles involved in the 400m. As we move from lane 1 to
lane 8, each runner starts 7.854 metres in front of the runner on
the next inside lane to them.
Rajeev, from Haberdashers' Aske's Boys'
School, sent us his calculations for the staggers presented clearly
in a table. You can see it here.