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A bit of mathematics out of origami

Dodecahedron Calendar

Last week on Saturday, we went to SMAK Hendrikus, one of senior High-schools in Surabaya. We were there to give a seminar for high-school teachers. The theme of the seminar is “the Amazing Math.”  Personally, I always think that Mathematics is amazing and I really want to convince this to other people.

Apart from the seminar, we were planning to give some souvenir for the participants. After consulting with my colleagues, we decided to give them a rhombic dodecahedron calendar. If there is anyone of you who has never heard of this kind of calendar, you can see it in the picture below, or you can follow this link to download the 2013 calendar along with the instruction on how to fold this model.rhombic_calendarWe were excited to know that this was actually a great gift for them. After reading the instruction we then realized the paper needed to make this calendar has to have a ratio which is normally a ratio of an A4 paper. Unfortunately, we already bought many origami papers which are all square.


The Lichtenberg Ratio

The Lichtenberg ratio is the ratio used in all of A-series paper such that A0, A1, et cetera. This ratio was found by Georg Christoph Lichtenberg. In the A-series paper the ratio between the length and the width of the paper is always \sqrt{2}:1. The unique property of this ratio is the following fact. Suppose the original paper is cut into two equal pieces (smaller pieces) such that the width of the original paper becomes the length of the smaller paper and the width of the smaller paper is half of the length of the original paper. Then the ratio between the length and the width of the original paper is the same as that of the smaller paper.the ratio between the length and the width stays the same.


We go back to our problem above. We do not have such a paper, but we have a bunch of papers whose dimension are all square. We then notice the following theorem that actually solves our problem.

Theorem (Construction a Lichtenberg ratio paper from a square)
Suppose we have a paper whose dimension is a \times a. If we follow the instructions below then we have a paper whose dimension follows the Lichtenberg ratio.


1. Draw a diagonal line as in the above figure.


2. Fold the paper along the diagonal line as the above right figure.


3. Mark the meeting point between the bottom right corner and the diagonal line as O in the above figure. then draw a vertical line. Cut the paper along this line.


4. We have a paper whose dimension is the Lichtenberg ratio as in the above figure.

Proof of the theorem

Consider the following figuresquare4

Let the square be the square ABCD as in the above figure. Assume the square has been folded and unfolded according the above instruction such that E, F, O and G are well defined. Suppose x is equal to AE, c is equal to CO and b is equal to CF. We are going to prove that

a : x = \sqrt{2} : 1.

First notice that AO is equal to a. Using the triangle ABC and the Pythagoras theorem, we get c = a\sqrt{2}-a.

Using the triangle OFC and the fact that the angle OCF is 45 degrees, we get b = a - a/\sqrt{2}.

Using the fact that the triangle OGC is an isosceles triangle we then obtain x=a-2b=a/\sqrt{2}.

This proves the theorem.

Categories: math puzzle, Teachings
  1. July 3, 2013 at 6:40 pm

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