Mobius Madness

Geometry Level 4

Consider a Mobius strip whose width (the red line) is extended indefinitely in both directions throughout 3D space. Note that the Mobius surface will intersect itself.

Consider all the regions that this surface would divide 3D space into. How many (disconnected) regions are there? And how many of them have an infinite volume?

Enter your answer as a . b a . b , where a a is the number of disconnected regions, and b b is the number of regions which have infinite volume. For example, if you believe there are 4 regions, of which 3 have infinite volume, enter your answer as 4.3 4.3 .


Formally, the Mobius surface as shown in the graphic above is defined as a ruled surface , with a circle as the directrix curve (green), through each point on which (big black point) passes a line (red) perpendicular to the tangent of the circle at that point. As one progresses around the directrix circle, the angle that line makes with the plane of the directrix circle varies by exactly half of the angle around the directrix circle, so that after it has gone around once, the arrow points in the opposite direction, and going around one more time brings the arrow back to the original orientation. We are then extending the red line indefinitely in both directions.


The answer is 2.2.

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1 solution

Michael Mendrin
Mar 25, 2017

The above graphics shows an orthographic view down the z z axis, showing 5 different cross-sections of this mobius surface near where z = 0 z=0 . The cross-section at z = 0 z=0 is shown in purple, and consists of the directrix circle and a straight line through it. The other cross-sections in blue and red are just above and below the purple cross-section.

At any cross-section but the purple one, the plane is divided into 3 3 regions. The purple cross-section is divided into 4 4 . However, each of the semi-circle regions of the purple cross-section is connected with the "loop" region of one color as well as the region of the other color that has a loop in it. But that region of one color that doesn't have a loop in it is connected to the region of the other color that has a loop in it. So for each color, the "loop" region is connected to the region that doesn't contain the loop. Thus, each of the semi-circle regions of the purple cross-section is connected to half of the entire 3D space, and so the answer is 2.2 , which is 2 regions, 2 that have infinite volume.

All other cross-sections, no matter how far above or below the purple cross-section exhibit the same geometry of a curve with a single loop.

Can you explain how one arrives at this result? I am unable to grasp/visualize most of the things being talked about here.

Agnishom Chattopadhyay - 4 years, 2 months ago

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I'm only a sophomore in highschool XD I'm excited to do stuff like this in college though.

Spectate Nate 2 - 4 years, 2 months ago

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Why wait though? Do them now, you won't have time in college.

Julian Poon - 4 years, 2 months ago

This might help. It depicts the the intersection of the mobius strip with the plane z = k z=k . Each of the regions looks sort of like a funnel. The large opening of the funnel takes either the top or bottom half of the space and the smaller lumen extends infinitely

Julian Poon - 4 years, 2 months ago

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Interesting that there is a point where the intesection takes the form of a circle. Why does that happen?

Agnishom Chattopadhyay - 4 years, 2 months ago

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@Agnishom Chattopadhyay That is due to the description of the mobius strip "with a circle as the directrix curve (green)"

Julian Poon - 4 years, 2 months ago

I followed a similar argument but made a mistake and got 4.4 - need to be more careful in my checks!!

Steven Linnell - 4 years, 2 months ago

Let n be the number of half twists. For n = 0 we would have 2.2 as answer too. But what about n = 2, 3,... ? Will we always have the answer as 2.2? I had a headache because of this problem

Mehdi K. - 4 years, 2 months ago

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This might help. Here you can see that for both odd number of half twists and even number of half twists it seems that the number of regions follow an arithmetic sequence, and all the regions are of infinite volume. For 2 n + 1 2n+1 number of half twists, the number of regions appears to be 4 n + 2 4n+2 , while for n n number of full twists, the number of regions appears to be 4 n 4n . I don't have a rigorous proof for this yet though.

Julian Poon - 4 years, 2 months ago

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