Uzair from Wilson's School, and Katherine
from St Joseph's School both correctly said that at the same speed,
boats would always meet in the middle, and if one was going twice
as fast as the other they would meet $\frac{2}{3}$ of the way
across.
Philip from Wilson's School continued this
line of thinking to explain in general where the boats will
meet:
When both boats start at opposite ends and travel at constant
speed, their first meeting will be halfway down the lake and every
meeting thereafter will occur halfway down the lake.
If the ratio of the boats is $2:1$, then boat A is travelling at
twice the speed as boat B. They would first meet $\frac{2}{3}$ of
the way down the lake. Their second meeting would be at the end of
the lake, after boat A has gone two lengths and boat B one.
If the ratio is $3:1$, with boat A going three times as fast as
boat B, they would first meet $\frac{3}{4}$ of the way down the
lake. Their next meeting would be $\frac{1}{4}$ of the way down the
lake, after boat A has gone 2.25 lengths and boat B 0.75
lengths.
If the ratio is $4:1$, with boat A going four times as fast as
boat B, they would first meet $\frac{4}{5}$ of the way down the
lake.
There is a consistent pattern here: say the amount of times boat A
is going faster than boat B is $n$. They would then meet
$\frac{n}{n+1}$ of the way down the lake.
Robert clearly explained that boat B must
travel further than boat A:
No matter what speed the boats are travelling at, the lake cannot
be $600$m or any smaller than that distance. If the lake was
smaller than $600$m, the two boats wouldn'??t be able to get $600$m
from side A. If the lake was $600$m long, that suggests that boat B
would not be moving, however this would be impossible as boat B
needs to get $400$m from his starting point for the second
crossing.
Boat B must travel further than Boat A. For example, if the lake
was $1000$m long, Boat A travels $600$m for the first crossing, and
Boat B travels $400$m for the first crossing. Then Boat A has to
travel $400$m to the end of the of the lake, and Boat B $600$m to
the end of the lake, at this point both boats have travelled
$1000$m. Now Boat A has to travel $400$m to the second crossing
point $400$m from side B, and Boat B has to travel $600$m to the
second crossing point.
So altogether, Boat A has travelled $600 + 400 + 400 = 1400$m and
Boat B has travelled $400 + 600 + 600 = 1600$m.
If the lake is $1500$m long, Boat A travels $600$m to the first
crossing point, and Boat B travels $900$m to the first crossing
point. Next, Boat A has to travel $900$m to the other side of the
lake, and Boat B only has to travel $600$m to the other side of the
lake. At this point both boats have travelled $1500$m. For the
final leg Boat A has to travel $400$m to the second crossing point,
and Boat B has to travel $1100$m to the second crossing
point.
So altogether, Boat A has travelled $600 + 900 + 400 = 1900$m and
Boat B has travelled $900 + 600 + 1100 = 2600$m.
In general, Boat A's distance travelled is Length of lake +
$400$m
Boat B's distance travelled is Length of lake $+$ (Length of lake +
$400$m).
Extending Robert's ideas:
We know boat B travels further than boat A, so the lake must be
more than $1200$m long, as Boat A has travelled $600$m when they
meet for the first time.
Let's say the first meeting happens after $1$ unit of time. Then
the second meeting is more than $1$ unit of time later as A has to
complete the length and turn the corner.
After $3$ units of time, A has travelled $1800$m. If this was the
second meeting point, A has travelled Length of lake + $400$m, so
the lake must be $1400$m. This would suggest that Boat B travels
$800$m in $1$ unit of time, so travels $2400$m in three units of
time.
The second meeting point is when Boat B has travelled $1400 + (1400
- 400) = 2400$m, so the solution is that the lake is $1400$m long
and Boat A travels at $\frac{3}{4}$ of the speed of Boat B.