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Life's greatest treasures are simple ones such as erasers, for example. These small pieces of molded rubber are underappreciated but indispensable tools for home and office.
In 1752, the proceedings of the French Academy of Sciences recorded that caoutchouc obtained from the Hevea brasiliensis rubber tree could be used as pencil mark eraser. It is Edward Naime, however, an English engineer, who is usually credited with the invention of the eraser (1770). The story goes that he picked up a piece of rubber rather than the usual wad of bread (some artists still use bread to lighten charcoal or graphite marks) and discovered its properties. Naime began selling rubber erasers, the first practical application of the substance.
Rubber, like bread, was perishable and would go bad over time. Charles Goodyear's invention of the process of vulcanization (1839) led to widespread use of rubber. Erasers became commonplace.
In 1858, Hymen Lipman of Philadelphia patented the pencil with an eraser at the end.  # Previous Puzzles of the Month + Solutions

October-November 2007, Puzzle nr 114 Back to Puzzle-of-the-Month page | Home Puzzle # 114  Italiano Français Difficulty level:    , basic geometry knowledge.

 Infinite beetle path    A high quality rubber band is fastened and hung from a horizontal pole with a cannonball at its end. Two facing ladybugs are crawling along this rubber band toward each other.    From their respective starting positions (8 cm apart -- see image), each small beetle crawls toward the other at a speed of 1 cm per second. However, in the length of time each beetle crawls 1 cm, the cannonball, thanks to the force of gravity, stretches the rubber band an additional 8 cm. Will the poor ladybugs ever meet? And, if yes, when? If not, why?! (the drawing is only representative)
[enlarge] • The rubber band stretches evenly 8 [cm/s.] along its entire length. • The more it stretches, the less the distance between the ladybugs grows. • The distance between the ladybugs is originally A0 = 8 [cm] and the rubber band is D0 centimeters long. • After each second, A and D change thus: Dn+1 = Dn + 8 [cm] An+1 = (An - 2) x Dn+1/Dn [cm] [In words: every second the distance A is reduced by 2 from them moving, but expands by a factor of (new band length Dn+1)/(old band length Dn)]

Yes, the beetles will meet! Even if D0 were 8 cm (one ladybug on the pole, the other on the cannonball), the change is thus:

 Sec. Band Length Bugs Gap 1) D1 = 16 [cm]; A1 = 12 [cm] 2) D2 = 24 [cm]; A2 = 15 [cm] 3) D3 = 32 [cm]; A3 = 17 1/3 [cm] 4) D4 = 40 [cm]; A4 = 19 1/6 [cm] 5) D5 = 48 [cm]; A5 = 20.6 [cm] 6) D6 = 56 [cm]; A6 = 21.7 [cm] 7) D7 = 64 [cm]; A7 = 22 4/7 [cm] 8) D8 = 72 [cm]; A8 = 23 3/35 [cm] 9) D9 = 80 [cm]; A9 = 23.422... [cm] 10) D10 = 88 [cm]; A10 = 23.564... [cm] 11) D11 = 96 [cm]; A11 = 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) This is more evident when we analyze the beetle's route until step 6. As shown in fig. b) below, the ladybug has covered m + m/2 + m/3 + m/4 + m/5 + m/6, but the total route it must crawl to meet its counterpart is 4m long. Then the fractions of band that both ladybugs have to crawl respectively until the meeting are:
m + m/2 + m/3 + m/4 + m/5 + m/6 + ... + m/n = 4m
or
1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + ... + 1/n = 4

fig.b) The sum 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + ... + 1/n represents the first n terms of consecutive reciprocals of natural numbers called "harmonic series".

There is a relatively 'simple' formula to estimate the sum H of n terms of an harmonic series:
Hn ≈ ln(n) + 0.5772156649 + 1/2n (it gets closer, the larger n is)
Since in our case Hn = 4, therefore
ln(n) + 0.5772156649 + 1/2n ≈ 4
ln(n) + 1/2n ≈ 3.422784336
We can omit here 1/2n, then the equation becomes:
ln(n) ≈ 3.422784336
n ≈ e3.422784336 ≈ 30.65464915...

Approximatively one half second after 30 seconds, the ladybugs will meet. 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:
Dn+1 = Dn + 8 [cm]
An+1 = (An - 2) x Dn+1/Dn [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;
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 time-step 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 rubber-band 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 time-steps that are sufficiently small to estimate continuous functions.
Using your equations, but adapted to include a time-step that is not necessarily equal to 1 sec.:

Dn+1 = Dn + 8 [cm/s] x dt
An+1 = (An - 2 [cm/s] x dt) x Dn+1/Dn

Performing the solution using iteration in a spreadsheet (see attached) yields:

dt time-to-meet
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:

Dn+1 = Dn + 8 [cm/s] x dt
An+1 = An x Dn+1/Dn - 2 [cm/s] x dt

Performing this solution using iteration in a spreadsheet yields:

dt time-to-meet
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 time-to-meet 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 time-to-meet 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 time-to-meet value, which is: e4 - 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 Marie-Jo 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,
Marie-Jo and Gianni

Do you agree with the comments above? Send 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.

© 2004 G. Sarcone, www.archimedes-lab.org
You can re-use content from Archimedes’ Lab on the ONLY condition that you provide credit to the authors (© G. Sarcone and/or M.-J. Waeber) and a link back to our site. You CANNOT reproduce the content of this page for commercial purposes.

You're encouraged to expand and/or improve this article. Send your comments, feedback or suggestions to Gianni A. Sarcone. Thanks!

More Math Facts behind the puzzle

Curiously enough, the sum of harmonic series diverges. That is, if the last term n is big enough the sum is as large as we want. This surprising result was first proved by a mediaeval French mathematician, Nicole Oresme, also known as Nicholas Oresme, who lived over 600 years ago.

Oresme noted that if you replace the series:
S = 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + 1/9 + 1/10 + 1/11 + ...
by the series of lesser terms and bracket the terms as shown
T = 1 + (1/2) + (1/4 + 1/4) + (1/8 + 1/8 + 1/8 + 1/8) + (1/16 + 1/16 + ... + 1/16) + ...
then the latter series is just
T = 1 + 1/2 + 1/2 + 1/2 + 1/2 + ... which is infinite for infinitely many terms.

Conclusion:
Since T < S for any finite number of terms, but T = infinite for infinitely many terms, then S must also be infinite!

The math symbol for the sum of harmonic series is:
 ∞ ∑ k = 1 1 k

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