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So you have a circle with radius 6. Now you have to attach clockwise equilateral triangles with side length 6 to it (from the outside) so that

a) The equilateral triangles do not overlap

b) Of every triangle one corner lies on the circle

c) Two triangles that are put down after each other share exactly one corner and this one lies not on the circle.

How many triangles can be attached to the circles without the triangles overlapping again?

Now prove that with that amount of triangles the first and last triangle share at least one corner.

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Hint: Connect the vertex on the perimeter to the center of the circle. This forms several rhombuses with the edges of the equilateral triangle.

  1. What can you say about these rhombuses?

  2. What can you say about the angles of the rhombuses? Are they the same? If no, what do their angles sum to up? Hint: Use equilateral triangles somewhere.

  3. If you show that the angles at the center sum up to $360^\circ$, then you can conclude that the first and last triangle share a corner.

  4. How many rhombuses do we need, and why?

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  • $\begingroup$ Well the rhombuses alternate, so it is one 'version' then the other one, then the first one again. And the angles at the point where the corners touch, add up to 360°, because the angles at the center of the circle add up to 360° $\endgroup$ – Katy Lin Sep 28 '13 at 22:15
  • $\begingroup$ Sounds like you've got it. $\endgroup$ – Calvin Lin Sep 28 '13 at 22:16
  • $\begingroup$ As a check: 1) If the angle at the center of a circle of a rhombus is $\alpha$, what is the angle at the center for the neighboring rhombus? Hint: Use equilateral triangles somewhere. 2) How many rhombuses do we need, and why? $\endgroup$ – Calvin Lin Sep 28 '13 at 22:22
  • $\begingroup$ @ Calvin Lin 1) 60-α 2) Therefore we need 6 rhombuses to get around. $\endgroup$ – Katy Lin Sep 28 '13 at 22:26
  • $\begingroup$ yes to 1, no to 2. Think about it again. $\endgroup$ – Calvin Lin Sep 28 '13 at 22:28

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