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Let's say you have 6 equal business cards.
How can you place them on a table to create these shapes:

  • regular hexagon
  • regular dodecagon

Edit. Feel free to propose an answer for options:
a) you have 50-by-90 units rectangles,
b) the sizes of rectangles doesn't have matter.

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    $\begingroup$ What are the dimensions of these cards? $\endgroup$ – Deusovi Mar 20 at 1:00
  • $\begingroup$ @deusovi, it does not matter, for instance, 90 by 50 mm. $\endgroup$ – Nick Mar 20 at 1:55
  • $\begingroup$ Do we need to cover the whole interior of the polygon? Or is it enough to create a boundary that has the shape? $\endgroup$ – sedrick Mar 20 at 2:29
  • $\begingroup$ @sedrick, boudary for 6-gon, and vertecies for 12-gon $\endgroup$ – Nick Mar 20 at 2:37
  • $\begingroup$ @dooper, could you please add answer with a sketch? $\endgroup$ – Nick Mar 20 at 3:13
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With the right sized rectangles, a hexagon can be made with just three cards, otherwise with 6 cards a hexagon is quite easy.
The dodecagon requires 12 points, so each card has to provide two points to the shape. So half of the cards for the hexagon can simply be 'rotated' 30°.
The ratio for the 'perfect rectangle' for making a hexagon with three cards is 1:√3 enter image description here
With the right sized rectangles, a dodecagon can be filled perfectly too.
enter image description here

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  • $\begingroup$ Your answer is very good. But how one can define 30°? Also I see the solution in bottom figure includes rectangles which size are not business cards. Could you please give some extensions? $\endgroup$ – Nick Mar 20 at 7:38
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    $\begingroup$ @Nick You said in a comment on the question that the size/dimensions doesn't matter. $\endgroup$ – Rand al'Thor Mar 20 at 20:49
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    $\begingroup$ The four initial diagrams use cards 90 * 50. The last diagram was more of an extra thing, since it's not business card size. (I think it's about half business card size) $\endgroup$ – Matthew Jensen Mar 20 at 23:03
  • $\begingroup$ @randal'thor, you are right. In the comment I wrote that the size does not matter because I have the solution for a regular 6-gon and doesn't have the solution for a regular 12-gon. I am omitting the size's requirement. But after the Matthew Jensen's answer I hope that someone can solve the puzzle for business cards too. $\endgroup$ – Nick Mar 21 at 0:24
  • $\begingroup$ @matthewjensen, if a=50, b=90, then tan(alpha)=a/b. And alpha approximately equals 30 degrees. $\endgroup$ – Nick Mar 21 at 1:05
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I hopped onto google drawings to make the sketch you asked for. If all you want is the boundary than making a hexagon is very simple. I feel like I am missing some aspect of this...

enter image description here

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    $\begingroup$ @dooper, how to proof that 6-gon is regular? $\endgroup$ – Nick Mar 20 at 7:11
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    $\begingroup$ All sides are equal. This is given because all the cards have the exact same dimensions. To be regular, all that is left is to make all the angles equal. This cannot be proven, since it is possible to construct a hexagon with equal sides but different angles. Thus, it is up to the person constructing this to actually make the angles equal. $\endgroup$ – Falc Mar 20 at 10:46

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