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This question already has an answer here:

I am trying to fold paper so that it looks like 3 x 4 grid of 12 rectangles of equal size.

Like this https://www.wolframalpha.com/input/?i=3+x+4

Its easy to get 4 rectangles. Just fold twice. But how to get 3?

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marked as duplicate by user63181, colormegone, Claude Leibovici, AlexR, Magdiragdag Apr 5 '14 at 9:30

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ The Pythagoras theorem may help $3^2+4^2=5^2$ is the relationship b/w sides of a right triangle. $\endgroup$ – DVD May 16 '13 at 6:55
  • $\begingroup$ math.stackexchange.com/questions/736346/… This would help you divide it into 3 parts $\endgroup$ – Harshal Gajjar Apr 5 '14 at 7:43
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    $\begingroup$ This question was asked before that question so that one is duplicate of this one :D lol $\endgroup$ – Pratik Deoghare Apr 5 '14 at 18:18
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    $\begingroup$ This question is 11 months older $\endgroup$ – Henry Apr 7 '14 at 7:19
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Let the paper be $[0,a]\times[0,b]$ By halving you find $(\frac a2,0)$, make a crease to find the line through $(\frac a2,0)$ and $(0,b)$. This intersects the diagonal at $(\frac a3,\frac b3)$.

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http://upload.wikimedia.org/wikipedia/commons/8/8d/Haga_theorem_1.svg

If your piece of paper is a square ABCD of side length 1, then from the diagram, if AP=1/2, then QC=1/3.

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(Stealing from Hagen von Eitzen's answer and Ross Millikan diagram at How can a piece of A4 paper be folded in exactly three equal parts?)

Graphical representation of the folds

  1. Fold the paper into four equal horizontal strips with two folds

  2. Fold a diagonal across the top three horizontal strips

  3. Fold vertical strips at the two points where the diagonal crosses the horizontal strips.

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Fold the side in such a way that folded sides overlap each other then you will get three equal folds.For this you need to fold them at same rate. Thanks julian fernandez

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  • $\begingroup$ I apologize, by my comment (and original answer) was wrong! (same as yours!) $\endgroup$ – Wolphram jonny May 16 '13 at 6:28

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