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Can you make 10 triangles with just 10 sticks?

The sticks don't have to have the same length.

Hint 1:

Try making 4 triangles with 6 sticks of equal length. Now put more shorter sticks inside this shape.

Hint 2:

If the world would have four or more dimensions, you could have made the shape with sticks that have exactly the same length.

There are several other possibilities than the one I gave hints for.

I recently found one with at least 40 triangles. However, for that, overlapping sticks are needed. Why not try searching for the greatest amount of triangles?

If you disallow overlapping sticks, 13 triangles is possible.

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Using non-overlapping sticks I get 36.

I'll give measurements and positions to make this easy to picture. Imagine we're drawing this on graph paper. Make one horizontal stick of width eight units. Pick a point somewhere above it, and attach sticks from it to each of 9 points on the line, all one unit apart.

Now how many triangles are there?

say the horizontal stick is 0, and the others are 1-9 in order.

There are eight triangles with no sticks between them (012, 023, 034, etc...) where when I say 0 it is understood that I mean the segment of 0 necessary to complete a triangle.

There are 7 triangles with 1 interior stick (013, 024, 035, etc...)

etc... 6 with 2 interior sticks (014, 025, ...)

etc... until we get 1 with 7 interior sticks.

Then 8+7+6+5+4+3+2+1 = 36.

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Two 5-point stars should do the trick! This solution uses 10 'sticks' (each stick spans vertex-to-vertex) of exactly the same length in 2D.

As someone commented briefly (sadly, it was deleted too quickly for me to reference), we get 10 triangles even with a single 5-point star: 5 large, 5 small, and the large triangles overlap the small triangles.

Note: This answer addresses the original question. As @DrXorile commented, a single 10-point star can produce many more triangles.

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  • $\begingroup$ You could even make a 10 sided star, which would, I think, have more than 10 triangles. $\endgroup$ – Dr Xorile Dec 5 '15 at 17:23
  • $\begingroup$ @DrXorile It has at least 40, which is the 40 triangles I mentioned in the question. $\endgroup$ – wythagoras Dec 5 '15 at 17:34
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From the hints I suspect the OP was also looking for a solution based on a tetrahedron. The simplest (non overlapping) solution is to have each face divided into two through one of the vertices. This gives 12 triangles (3 per face) and 10 sticks.

If you wanted to do this exactly, you could make a smaller triangle parallel to one of the faces and with edges on the 3 remaining faces. This gives 8 triangles with 9 sticks, and the two extras can be made by dividing one of the remaining triangle in two.

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  • $\begingroup$ Actually I was looking for this, hence the second hint. $\endgroup$ – wythagoras Dec 5 '15 at 18:29

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