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How many triangles can we form if we draw all the diagonals of a hexagon?

I thought that the answer is $\binom{6}{3}=20$ but this is not the right answer, why?

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What kind of hexagon? Convex or not? Regular or not? Before using counting tools, we need to know what we are counting. –  André Nicolas Mar 9 '12 at 17:03
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Must the vertices of the triangles coincide with vertices of the hexagon? –  Matt Mar 9 '12 at 17:07
    
This is interesting, @Andre considering the type of question I guess it should be convex-regular. –  Quixotic Mar 9 '12 at 17:54
    
What makes you say 20 is not the right answer? –  Gerry Myerson Mar 10 '12 at 6:41
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4 Answers

The problem is very unclear (see the comments). Here is one interpretation (which is probably not the one intended, but who knows?):

Drawing all 9 diagonals of a regular hexagon divides it into 24 regions, of which 6 are quadrilaterals, leaving 18 triangles.

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Interesting. I first thought of the 6 triangles you get when drawing the "diagonals" of a regular hexagon, but after thinking about your answer, it is a correct one, provided you are just looking for the number of triangles you can create with the 6 points of a hexagon (or any 6 points for that matter, provided you don't mind "flat triangles"). For a regular hexagon, it gives you 2 equilateral triangles, 6 isoceles (non-equilateral) ones and 12 triangles with a 90 degree angle (which can be put into 2 types by 2D rotation), so 20 in total. –  KIAaze Mar 9 '12 at 17:16
    
(cont) [4 distinct ones by 2D rotation, 3 distinct ones by 3D rotation] To prove there are only 6 triangles, when drawing all the diagonals (lines going through the centre of mass) of a regular hexagon, I am not quite sure how to proceed. edit: It seems I didn't know the actual definition of a diagonal: "a line joining two nonconsecutive vertices of a polygon or polyhedron.". So actually, it's 18 triangles, not 6, as explained by Gerry Myerson. Proof by simple enumeration? Why the $\binom{6}{3}$ doesn't work to get 18 is obvious: you create triangles using intersection points. –  KIAaze Mar 9 '12 at 17:16
    
(cont) And that's not even counting all the triangles consisting of more than 1 of the 24 created regions... So, yes, this problem needs a lot more clarification. (and how can I add comments here instead of only answers? It's frustrating. :/) –  KIAaze Mar 9 '12 at 17:16
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Assuming a regular hexagon:

If you draw all diagonals of a regular hexagon you have $3 \cdot 6 = 18$ possible triangles, but 3 of those are the same (the equilateral triangles) so we have $18 - 3 = 15$ possible triangles.

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let me set of this numbers, where in every number corresponds with a number of sides of every polygon.. ( 3,4,5,6,7,8,9,10 ),,let me answer how many diagonal can be drawn from the fixed vertex?? and how many triangles are formed from this diagonal??

1.) Triangle = 3 sides, 0 diagonal, 1 triangle

2.) quadrilateral = 4 sides, 2 diagonal formed, 8 triangles formed

3.) Pentagon = 5 sides, 5 diagonal formed, 40 triangles formed

4.) hexagon = 6 sides, 9 diagonal formed, ????????? :))

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I can see 35 in a pentagon, by organising my triangles by the quantity of shapes each is constructed of:

10 triangles made of 1 shape. 10 triangles made of 2 shapes. 10 triangles made of 3 shapes. 5 triangles made of 5 shapes. Total of 35 triangles.

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I meant for a pentagon. –  heygale Feb 17 at 17:18
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