Marquis De L'Hopital puzzle

Jeremy Desmond · 2022-08-22 21:31:47

Hello,

I have been studying some of the Maths Is Fun tutorial pages to brush up on my math knowledge and skills. (I studied math at university 40 years ago but haven’t looked at it since. Now that I am retired from work I want to delve back in to the world of math.)

I came across an interesting puzzle here: Maths Is Fun puzzles and I have been wrestling with it but getting nowhere. I would be very grateful for a hint of how to begin to solve it please. But I don’t want to be spoon-fed the answer.

Thanks very much in advance.

Last edited by Jeremy Desmond (2022-08-24 00:11:51)

Bob · 2022-08-23 00:34:22

hi Jeremy,

Welcome to the forum.

I haven't tried this yet but it looks like you have to make equations for the forces, and also for the geometry.

The tension in the string section BC will equal the tension in section CD, which means it will equal the weight. Horizontal and vertical forces must balance. And you can calculate BC using trigonometry.

The three forces (tension, force in the strut AC , weight) will make a 'triangle of forces' so this may be a short cut.

Bob

Jeremy Desmond · 2022-08-23 20:35:07

Thank you, Bob, for your welcome and for the hints about the puzzle. I will follow your guidance and see if I can make any progress.

The Marquis de L’Hopital was a brilliant French mathematician of the 17th century. I find the lives and work of ancient mathematicians fascinating. This puzzle was published in his famous textbook on calculus.

Thanks again for your help.

Bob · 2022-08-24 00:40:30

I've been working on my own solution. I have a vector geometry program called Sketchpad that I find very helpful for making decent diagrams. The measurements it gives are not perfectly accurate but what I've done shows a pssible way to an answer.

I started with AB = 10cm, and with A as centre, made a circle radius 4 cm.

I chose an arbitary point on the circle, labelled C. Then with B as centre, made a second cicrle radius BC.

I constructed a line CD perpendicular to AB. At this stage D is an arbirary point on that line.

So A, B, C, D represents the geometry of the situation.

There are three forces acting at point C and these must be in equilibrium if C is static. These are the force in the strut AC; the tension in the string acting towards B; and the weight at D. If the pulley is frictionless then the tension acting towards B will be the same as the tension acting towards D and, as tension at D must equal the weight, this means the tension and weight are equal.

I wanted to construct a vector force diagram for equilibrium at C.

So I extended the line AC to cut the circle centred on B at E. The size of the forces (in Newtons, poundals, or any other force unit of your choice is irrelevant) , so I can choose the force scale so that BC also represents the tension in the string. EC is the force in the strut and BE the weight, equal in magnitude to BC.

Not surprisingly, BE was not vertical. I'd have been very shocked if my arbitary choice for C had caused this to happen. But the software allows me to select C and move it. It is locked to the circle, but can travel around the circumference. I juggled it until BE became vertical.

Finally I made some measurements including CD and used this to calculate the total length of the string, BC + CD. I moved D until this total was 10cm.

That's the screen shot you see above.

You can see that there is a unique solution, but the measurements are not accurate enough and anyway a diagram is hardly a 'proper' way to get an answer.

Somewhere in that diagram must be the means to calculate BC but it eludes me at the moment.

Bob

phrontister · 2022-08-24 04:38:16

Hi Jeremy & Bob,

The maths is beyond me, but I wanted to try graphically, drew this up in Geogebra, and got ED ≈ 51.90438 to 5 decimal places.

I moved C (radius 100 - BC) along circle A (radius 40) to get the largest ED (approx).

Is that anywhere near the mark?

Bob · 2022-08-24 06:40:40

hi Phro,

I have just measured that distance on my diagram and got that value too.

I'm still trying to find a way (presumably trig) to get these distances.

Bob

Jeremy Desmond · 2022-08-24 20:28:47

Hi Phro and Bob,

Very interesting to read your two approaches. And that is impressive software!

I have given up trying to solve this myself. It is far above my ability. I had thought that because it was included in the Maths Is Fun puzzle pages, it would be fairly easy to solve with a bit of basic knowledge and effort. But googling the problem has revealed to me that it has taxed some very advanced mathematicians over the centuries. There seem to be numerous ways of solving it and I don’t understand any of the detailed explanations of the solutions! Guillaume François Antoine de l'Hôpital himself solved it using calculus but other methods exist.

Before I started this thread I searched past posts on this forum to see if this problem had been discussed before. I did not find anything but this morning I came across this thread in the course of my google search: previous discussion.

By the way, I confirm your answer to the problem. Relying on trigonometry I used my calculator to work out the various altitudes of the weight depending on different angles of BAC, and I repeated the laborious calculation to gradually home in on the maximum result. I got the answer to be 51.90438 to 5 decimal places. This method gives the right answer but in the most boring and prosaic manner possible. It misses out on the beauty and elegance inherent in math.

If you have any more insights on this problem I will be interested to hear your thoughts.
Jeremy

Bob · 2022-08-24 20:30:26

Got it!

In my method I combine two triangles and it is necessary to slide C until the constraints of the forces diagram 'fits' with the geometry diagram. Since a single diagram emerges that fits all the constraints, I reasoned that there must be a way to calculate the lengths/angles using what is known. What follows is my answer.

Let angle BAC = alpha.

Triangle BCE is isosceles so let F be the midpoint of CE.

angle BEA = 90 - alpha so angle EBF = alpha.

also in triangle ABE

Substituting to eliminate alpha gives

but also by pythag

Subtracting to eliminate x^2

This has one positive root, AE = 122.82856 and this gives x = 71.32221

Bob

ps. I tried to find the original problem. The nearest I got was http://mrsk.ca/AP/Chapter4.pdf questio 21. The diagram given there is L/R flipped and the rod becomes a fixed length rope. And in this question a different distance is required so I did some further calculation using my values in the expression given there and compared with my Sketchpad values and got agreement. Although L'Hospital uses potential energy his calculation is based on equilibrium so I think this validates my approach.

pps. Sketchpad is not free but Geogebra is and can be downloaded from their site if you google it.

phrontister · 2022-08-25 03:24:37

Here are some interesting links about L'Hôpital's Pulley Problem:
Simple Geometric Solutions to De L'Hôpital's Pulley Problem
Optimization – L'Hôpital's Pulley Problem
Proof of length of line segment x

Bob wrote:

Triangle BCE is isosceles...

Shouldn't that read CBE, Bob?

But maybe placing the letter of the vertex angle of an isosceles triangle in the middle is just a habit of mine...similar idea to the order in naming angles.

Last edited by phrontister (2022-08-25 04:26:22)

Bob · 2022-08-25 04:40:32

I've never bothered with the order of letters in this situation. As two sides are radii to a circle, it shouldn't matter I hope.

B

phrontister · 2022-08-25 15:02:18

Here's a link to a video I created from my Geogebra file.

Marquis De L'Hôpital's pulley puzzle

The animation is from the Slider function that I linked to BC values, and is set to oscillate between BC=70 & BC=72, with 0.00001 increments.

I didn't capture all the increments, of course(!), although it can be done.

Changing BC values moves Point C along Circle A's circumference, impacting on other objects' measurements...including the length of ED.

The oscillation shows ED approaching and reaching its max of ≈51.90438 at BC≈71.322, continuing past it until BC=72, then returning past it until BC=70, and finally back again to ED's max of ≈51.90438.

Jeremy Desmond wrote:

Relying on trigonometry I used my calculator to work out the various altitudes of the weight depending on different angles of BAC, and I repeated the laborious calculation to gradually home in on the maximum result. I got the answer to be 51.90438 to 5 decimal places. This method gives the right answer but in the most boring and prosaic manner possible. It misses out on the beauty and elegance inherent in math.

Oh well...I'm sure I'll get over it!

Last edited by phrontister (2022-08-25 17:28:15)

Math Is Fun Forum

#1 2022-08-22 21:31:47

Marquis De L'Hopital puzzle

#2 2022-08-23 00:34:22

Re: Marquis De L'Hopital puzzle

#3 2022-08-23 20:35:07

Re: Marquis De L'Hopital puzzle

#4 2022-08-24 00:40:30

Re: Marquis De L'Hopital puzzle

#5 2022-08-24 04:38:16

Re: Marquis De L'Hopital puzzle

#6 2022-08-24 06:40:40

Re: Marquis De L'Hopital puzzle

#7 2022-08-24 20:28:47

Re: Marquis De L'Hopital puzzle

#8 2022-08-24 20:30:26

Re: Marquis De L'Hopital puzzle

#9 2022-08-25 03:24:37

Re: Marquis De L'Hopital puzzle

#10 2022-08-25 04:40:32

Re: Marquis De L'Hopital puzzle

#11 2022-08-25 15:02:18

Re: Marquis De L'Hopital puzzle

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