Here's a fun little problem I had to do on an application for a tutoring position.

A 6-foot-diameter disk and an 18-foot-diameter disk sit side by side as shown in the figure below. Determine the shortest length of rope that will go around both disks.


Good luck!

Place one disk on top of the other disk, and wrap the rope around the two of them for a total length of 18 feet + 18 feet = 36 feet.

This is my sixth and final answer. Did I cheat or just see the forest through the trees . . . finally?

Do I have to add 1 inch to tie the rope, since they did not say rubber band?

Did they supply multiple choice answers?

Very resourceful, and definitely the simplest way to wrap the rope "around" both disks. Mathematically, though, it's a much more interesting problem to wrap the rope around the perimeters of the disks when they are adjacent in the same plane, as shown in the picture.

By the way, the questions weren't multiple choice, so there's no way of knowing what the "right" answer might be except to work it out.

Nice problem. It's great to get the math juices flowing again! Here's what I get for the length L of the rope:

L = 12√3 + 24(pi) - 36(tan^-1(1/√3)) - 12(tan^-1(√3))

The first term gives the length of straight parts; the other terms are for the circular arcs.

√3 ≈ 1.732
tan^-1(1/√3) = pi/6
tan^-1(√3) = pi/3

Thus:
L = 12√3 + 24(pi) - 6(pi) - 4(pi)
L = 12√3 + 14(pi)
L ≈ 20.78 + 43.98
L ≈ 64.8 feet.

This message has been edited. Last edited by: Professor,

silenteuphony: Thank you for the encouragement.

Professor: Unless you explain a bit as to how you arrived at your first line, you know "L =", some dummies around here will accuse you of a mix of imagination and visual hallucinations. I recently saw an interview with a scientist who was in favor of a movement to make things clear to nonscientists.

Anyone: Is the following an alternative solution "somehow justifiably"?

The circumference C1 of the big circle is 18pi, and the circumference C2 of the little circle is 6pi. Subtract 1/3 of C1, or 6pi, from the 18pi to get 12pi for the cord around the big circle, and subtract 2/3 of C2, or 4pi, from the 6pi to get 2pi for the cord around the little circle so that the total cord around the two circles (but not between them) is 12pi + 2pi = 14pi = 14(3.14) = 43.96 feet.

Let L1 = length between the "vital points"--along the horizontal diameters--in the two circles = (1/3 of 18) + (2/3 of 6) = (6 + 4) = 10 feet. (Where the circles are tangent, the 6 feet are to the right along the big circle's diameter, and the 4 feet are to the left along the little circle's diameter.)

The 10 feet is the height of a right triangle (also the horizontal line of 6 feet + 4 feet) whose other leg is 1/3 of 9 (the big circle's radius) = 3 feet so that the right triangle's hypoteneuse is the square root of (10)(10) + (3)(3) = 100 + 9 = 109. The square root is 10.44 feet. We are interested in twice this hypoteneuse as length L2 = the two straight cords between the circles = 2(10.44) = 20.88 feet.

43.96 feet + 20.88 feet = 64.84 feet

So is my eighth attempt pretty good, or did I just luck out?

This message has been edited. Last edited by: tsaeb,

tsaeb, I think we're all in agreement on the curved part of the rope. Your answer for the straight part, while very close numerically, differs from the result the Professor (and I) got. Here is my diagram from the original problem, showing 6√3 for the length of the upper straight piece, which agrees with the Professor's result. (Sorry the text is so faint; Microsloth Equation Editor didn't translate to a picture very well.)

Sorry I didn't explain my analysis -- it's hard to do geometry with text only. I don't know how to post a small diagram I made, but I can tell you it used a bunch of similar and congruent right triangles but is not as elegant as silenteuphony's diagram.

My use of the arctangent function was a brute force approach before I realized I was dealing with 30-60-90 triangles. tsaeb is correct that the rope wraps exactly 1/3 of the way around the little circle and 2/3 of the way around the big circle.

How did you guys post those diagrams? I tried in MS Word and the image wouldn't copy.

I wish I could illustrate this answer but...

The line connecting the two centres is the 12 ft hypotenuse of a 30-60-90 triangle with sides 6, 6xsqroot3, 12.

Since 60 degrees = (pi)/3, the string wraps around the small disk for [2(pi)/3]x3ft.

It wraps around the large disk for 4(pi)/3]x9ft.

The two tangent lengths between the disk edges are 6x(sqroot3) each.

The total is 14(pi) + 12(sqroot3) =64.77 ft. (Which rounds to yours, 64.8)

I'm just posting this because I want to find out how to do the images, and I don't want to go off-topic (again) just to ask you that. I'm not math-y at all.

My diagram was also in a Word document originally. I couldn't just save it as a picture from the Word document. I had to copy it into another Word document, save the new doc as a web page, then copy the picture from the folder for the web page. Then I uploaded it to my own website, and put the URL for the picture in the image tool on the UBB tool bar in the Post a Reply window. Easy!

Thank you! I'll give it a try.

To clarify, I posted a response as to how someone who cannot otherwise know how to solve the problem might get an approximation by eye. I wanted to encourage math dummies and to make the solution's discussion more interesting to our math brains here. (I could have figured out how to solve the problem, which is not a particularly difficult one, more rigorously.) When I have more time, I will look over the work of silenteuphony and babthrower and also try to understand how silenteupony got the fascinating picture to appear. So this particular problem has stirred up quite a few interesting challenges, making me happy.