Yes,
the beetles will meet! Even if D_{0} were
8 cm (one ladybug on the pole, the other on the
cannonball), the change is thus:
Sec.

Band
Length 
Bugs
Gap 
1) 
D_{1} =
16 [cm]; 
A_{1} =
12 [cm] 
2) 
D_{2} =
24 [cm];

A_{2} =
15 [cm] 
3) 
D_{3} =
32 [cm];

A_{3} =
17 1/3 [cm] 
4) 
D_{4} =
40 [cm];

A_{4} =
19 1/6 [cm] 
5) 
D_{5} =
48 [cm];

A_{5} =
20.6 [cm] 
6) 
D_{6} =
56 [cm];

A_{6} =
21.7 [cm] 
7) 
D_{7} =
64 [cm];

A_{7} =
22 4/7 [cm] 
8) 
D_{8} =
72 [cm];

A_{8} =
23 3/35 [cm] 
9) 
D_{9} =
80 [cm];

A_{9} =
23.422... [cm] 
10) 
D_{10} =
88 [cm];

A_{10} =
23.564... [cm] 
11) 
D_{11} =
96 [cm]; 
A_{11} =
23.524... [cm] 
The
beetles are now closing faster than the distance
between them grows. After 30 seconds,
the rubber band is 248 cm long and the bugs are
0.316... [cm] apart. They will meet just before
the next second.
But...
There is a more mathematical way to solve this
puzzle without ITERATIVE calculations!
Consider
that each time the cannonball stretches the rubber
band, the bug comes with it, which means that the
FRACTION of band left to cross stays the same.
Each
beetle moves specularly to its counterpart toward
the central point of the rubber band. Their relative
position being after each move specular to each
other, to solve this puzzle we just need then to
take into consideration the moves of one ladybug
on a HALF section of the rubber band.
OK.
After one crawl, the beetle has done 1 cm, that
is 1/4 of the rubber band (fig. a.1), while the
cannonball stretches the band to 8/2 = 4 cm (and
the ladybug is now 2 cm along). The beetle crawls
1 cm again, which is now 1/(4 + 4) = 1/8 of the
total band (fig. a.2). His next crawl will only
be 1/(8 + 4) = 1/12 of the band (fig. a.3). And
so on...
The fraction of band that the ladybug has crawled
at 6 steps is:
1/4 + 1/8 + 1/12 + 1/16 + 1/20 + 1/24
or
(1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6)/4
fig.
a)
The
Winners of the Puzzle of the Month are:
Larry Bickford, USA
Fabio Cirigliano, Italy
Congratulations!
Reactions
from our visitors

Posted 06 Dec 2007 by Sam Muir
Dear
Gianni,
I disagree with the solution you posted for the infinite beetle problem
#114. In your solution, you assume that the movements of the beetles and the
growth of the rubber band occur AFTER every second passes as you said in your
solution:
"After each second, A and D change thus:
D_{n+1} = D_{n }+
8 [cm]
A_{n+1} = (A_{n} 
2) x D_{n+1}/D_{n }[cm]"
A
more accurate calculation would be to assume that
the beetles and the rubber band are constantly
moving (not in one second increments, but in infinitely
small increments).
By doing so, and creating a program to calculate the limit as the
increments approach zero (by iterating for increasingly small intervals), I was
able to reach my solution of 53.598 seconds.
Here
is the code for the program used (in C++):
#include <iostream>
#include <cstdlib>
using namespace std;
int main(){
double distance = 4;
double length = 4;
double interval = .0001;
double time = 0;
do{
time = time + interval;
distance = distance  (interval * 1);
distance = distance + ((distance / length)*interval*4);
length = length + interval * 4;
}while (distance > 0);
cout << time;
cout << "seconds";
char line[25];
cin.get( line, 25);
return 0;
}
Sincerely,
Sam Muir

Posted 22 Dec 2007 by Greg Dunn
Dear
Sirs:
Your solution to puzzle #114 is incorrect. If you perform your iterative
method but with a smaller timestep that 1 sec., you will see a significant change
in the time it takes for the lady bugs to meet. Convergence must always be checked
when performing iterative solutions (such as in this problem) or finite element
analysis (FEA)  just because an analysis produces a result, it does not mean
that result is correct.
The problem does not state, and one cannot assume, that the rubberband
stretch and the lady bug movement is quantized (or in other words, in complete
packets). Rather, since it mentions 'speed' one must treat it as a continuum,
and to do this, one must use timesteps that are sufficiently small to estimate
continuous functions.
Using your equations, but adapted to include a timestep that is
not necessarily equal to 1 sec.:
D_{n+1} = D_{n }+
8 [cm/s] x dt
A_{n+1} = (A_{n} 
2 [cm/s] x dt) x D_{n+1}/D_{n}
Performing the solution using iteration in a spreadsheet (see attached)
yields:
dt timetomeet
1.00 s 31.00 s
0.50 s 41.00 s
0.25 s 47.25 s
0.10 s 51.00 s
0.05 s 52.30 s
0.02 s 53.08 s
0.01 s 53.34 s
0.005 s 53.47 s
0.002 s 53.55 s
The
above analysis is constructed assuming the bugs
move PRIOR to the rubber band stretching, but due
to the continuous nature of the problem, it will
work equally well to construct the equations such
that the bugs move AFTER the rubber band stretches.
Therefore, the equations can be modified to:
D_{n+1} = D_{n }+
8 [cm/s] x dt
A_{n+1} = A_{n} x D_{n+1}/D_{n } 
2 [cm/s] x dt
Performing this solution using iteration in a spreadsheet
yields:
dt timetomeet
1.00 s 82.00 s
0.50 s 67.50 s
0.25 s 60.50 s
0.10 s 56.30 s
0.05 s 54.95 s
0.02 s 54.14 s
0.01 s 53.87 s
0.005 s 53.74 s
0.002 s 53.65 s
Since
the bugs move before the rubber band in the first
method, it predicts a timetomeet that is less
than the actual value, so it provides a lower bound
to the true meeting time. Since the second move
after the rubber band in the second method, it
predicts a timetomeet that that is greater than
the actual value, so it provides an upper bound.
As the two answers get closer, the potential error
decreases until they converge at which time the
error range is zero.
dt range
1.00 s 51.0 s
0.50 s 26.5 s
0.25 s 12.1 s
0.10 s 5.30 s
0.05 s 2.65 s
0.02 s 1.06 s
0.01 s 0.53 s
0.005 s 0.27 s
0.002 s 0.10 s
The
two methods converge on the same timetomeet value,
which is: e^{4}  1 = 53.598...
sec. I sent you one proof using differential calculus
demonstrating that this. Attached is a different
proof written by my Mark Dunn which comes to the
same answer.
Regards,
Greg Dunn

Posted 29 Dec 2007 by MarieJo Waeber
Dear
Sam, Dear Greg,
I
asked a math expert, Alexander
Bogomolny, to help me to settle this
problem and here is his answer:
"I
understand from your site that there are two approaches
that are judged different: discrete and continuous.
When the latter is presented there is a remark
that the discretization differentiates between
two sequences of events: stretching/crawling and
crawling/stretching, and that at the limit there
is no difference. It also argued that the answer
provided by a discretization depends on the chosen
unit interval. I think that the two arguments should
settle the problem.
Neither the gravity nor the bugs are not concerned
with the fact that they became a part of a puzzle.
They act independently of us puzzle fans, and the
bugs meet when they meet: at the unique moment in
time. Since the discrete solution depends on the
two aforementioned factors (sequence and interval)
it may only provide a numeric approximation to the
real solution."
To
conclude, I will just add that since I didn't mention
if the ladybugs move uniformly or not, I think
both approaches determine approximately the limits
of the time t the beetles will meet: 30
sec. < t < 54 sec.
Best
wishes,
MarieJo and Gianni
Do
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us your reactions!
Answers
that made us smile
Those unlucky ones below didn't win but we find their
answer hilariously funny! So we decided to post them
anyway.
A.S.
from India wrote
[...] So it is concluded that under the given conditions the
poor ladybugs are never going to meet each other
as long as they dont try to imitate Jackie Chan and launch
at each other...
M.
S. from USA wrote
[...] Yes, they will eventually meet. The bugs will
eventually die and the bug at the top of the rubber
band will fall down and hit the dead bug below.