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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 makes you say 20 is not the right answer?2012-03-10

4 Answers 4

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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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    (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. :/)2012-03-09
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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.2014-02-17