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Folding paper

Question

Take a standard sized paper. Pick any corner and bring  that corner to the middle of the paper. Fold. Notice that the rectangle that was the original exterior shape is now a Pentagon. Repeat this process, picking corners of your choice for three more folds.

What is the most sides the exterior shape can have?

What is the fewest amount of sides this shape can have?

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Another interesting puzzle! Thanks, BMAD! Here's a question from me: when you say "repeat this process", do you mean "pick any corner and bring that corner to the middle of the paper"? Or do we bring the corner to any place at all that we can choose (not just the absolute center of the paper)?

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My guess for Most sides:

Spoiler

Ten sides

Spoiler

Fold point A to the center
This forms a fold, B is the East end of the fold.
Fold point B to the center.

Symmetrically, fold point C to the center
Fold point D (analogous to B) to the center

10 edges in the outline

Edited by CaptainEd
oops, I got the picture into the spoiler, but I couldn't get it out of the reply itself. Sorry
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6 hours ago, CaptainEd said:

Another interesting puzzle! Thanks, BMAD! Here's a question from me: when you say "repeat this process", do you mean "pick any corner and bring that corner to the middle of the paper"? Or do we bring the corner to any place at all that we can choose (not just the absolute center of the paper)?

It goes to the center every time

3 hours ago, CaptainEd said:

My guess for Most sides:

Hide contents

Ten sides

Hide contents

Fold point A to the center
This forms a fold, B is the East end of the fold.
Fold point B to the center.

Symmetrically, fold point C to the center
Fold point D (analogous to B) to the center

10 edges in the outline

How do you know if this is the most possible?

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How do you know if this is the most possible?

Spoiler

Each fold changes a vertex into an edge and two vertices. That is, each fold adds a vertex and an edge (side.) To be precise, a fold might affect more than one preexisting vertex and side. But a fold cannot increase the vertex (and edge) count by more then unity. Four folds, therefore, will create a shape with no more than 4+4 = 8 sides.

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@BMAD I can't prove this is the largest

@bonanova: in my picture ( representing real folds on real paper ) a "tail" lapped over the Southeast corner, adding an extra two edges beyond the ones identified by your argument. I presume that they become part of the overall envelope of the figure, leading to my answer, greater than yours.

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On 7/1/2016 at 1:17 PM, CaptainEd said:

@BMAD I can't prove this is the largest

@bonanova: in my picture ( representing real folds on real paper ) a "tail" lapped over the Southeast corner, adding an extra two edges beyond the ones identified by your argument. I presume that they become part of the overall envelope of the figure, leading to my answer, greater than yours.

Agree.
When I hit the send button, I realized my thinking was too simple. But instead of deleting my post (moderator privilege) I left it to take its licks.

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You're stronger than I am; I edit the post, removing everything, and then I put in some quiet, gentle, lame stuff.

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14 hours ago, CaptainEd said:

You're stronger than I am; I edit the post, removing everything, and then I put in some quiet, gentle, lame stuff.

LOL, thought that was only me.

On 7/1/2016 at 1:17 PM, CaptainEd said:

@BMAD I can't prove this is the largest

@bonanova: in my picture ( representing real folds on real paper ) a "tail" lapped over the Southeast corner, adding an extra two edges beyond the ones identified by your argument. I presume that they become part of the overall envelope of the figure, leading to my answer, greater than yours.

In (dis)proving this,  consider what events cause the creation of more than one extra vertex. Then I think you will see.

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I have no idea how to prove this but in playing around I found an interesting idea. How many folds using the same folding technique would it take to get back to only having 4 sides?

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6 hours ago, tojo928 said:

I have no idea how to prove this but in playing around I found an interesting idea. How many folds using the same folding technique would it take to get back to only having 4 sides?

4

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@BMAD What sequence of four folds yields a four-sided figure? It appears to me that the least number of sides after four folds is six.

@tojo928 Barring a four-fold solution from BMAD, a six-fold solution is best.

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23 minutes ago, Logophobic said:

@BMAD What sequence of four folds yields a four-sided figure? It appears to me that the least number of sides after four folds is six.

@tojo928 Barring a four-fold solution from BMAD, a six-fold solution is best.

You are right!

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With only four folds, it seems impossible to get back to four sides. However...

Spoiler

It is possible to create a pentagon. I do not have a picture, so I will try my best to explain.

If the long side of the rectangle is vertical, these are the corners. Northeast is marked A, southeast is B, southwest is C, and northwest is D. Fold A to the middle, and then the same for B. Folding B in will create two corners on each end of the fold. Fold both of those corners to the center. This will create 5 sides in the four folds. Two sides of it are part of the beginning sides of the rectangle, while the other three are created using folds.

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"It goes to the center every time ." -- BMAD

What is not made perfectly clear is whether the center is that of the original unfolded paper, or that of the folded paper. For each fold, the center of the external shape will be located at a different point.

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3 hours ago, DejMar said:

"It goes to the center every time ." -- BMAD

What is not made perfectly clear is whether the center is that of the original unfolded paper, or that of the folded paper. For each fold, the center of the external shape will be located at a different point.

Good question, for this problem I meant for the move to mean always going to the original center.

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I'm am curious if maybe I am missing something with my own problem but every time I fold into the center the vertices sum to relatively similar (maybe human error) value. Perhaps this is helpful.

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Spoiler

Buddyboy3000's pentagon solution:

Edited by Logophobic
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1 minute ago, Logophobic said:
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And?

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And that is the solution for the fewest number of sides, as described by Buddyboy3000

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