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What is the best way to count all of the triangles in this shape?

I originally thought to just use casework to go through the ones composed of $1$ triangle, then $2$ triangles, etc., but I still missed several triangles, getting a total of only $75$.

3 Answers 3

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So far, I've found $85$. If we call the three pentagons the major, the minor, and the demi, we have

  • All three points on the major: $10$
  • All three points on the minor: $10$
  • Two adjacent points on the major, one on the minor: $15$
  • Two diagonal points on the major, one on the minor: $5$
  • Two adjacent points on the minor, one on the major: $5$
  • Two diagonal points on the minor, one on the major: $15$
  • Two adjacent points on the minor, one on the demi: $15$
  • Two diagonal points on the minor, one on the demi: $5$
  • Two adjacent points on the demi, one on the minor: $5$
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    You are right ! (+1) I counted in b) $3$ triangles, missing the one between a point on the minor + diagonal points on major. So actually the part b) of my answer should read $20$, and so total coincides with yours. – 2017-02-07
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It would be best to list all of the possible shapes of triangles and count the total amount of each kind.

For example: The triangle that goes from one vertex of the pentagon to the the two vertices opposite of it is repeated 5 times in the big pentagon and 5 times in the little pentagon. You can also multiply triangles found in both pentagons by 2.

Hint: One pentagon has 35 triangles, use this to find the total number of triangles.

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    Perhaps you can expand on the number of "possible shapes"? Or provide a count for one such shape, to illustrate your suggestion? – 2017-02-05
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    I added more description – 2017-02-06
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a) Each vertex of the major pentagon makes a triangle with other two : ${5 \choose 3} = 10$.
b) Each vertex of the minor pentagon makes $3$ triangles with two vertices on the major $=15$.
c) Each vertex of the major pentagon makes a triangle with $4$ couples of vertices of the minor $20$.

Same as above for the minor pentagon, except for point c) where the couple of internal points is only $1$.

So in total that gives $10+15+20+10+15+5=75$, and your answer looks correct, unless I also missed something.
--- amendment ---
And in fact, I missed $1$ triangle in b), as the answer by Briang Tung let me realize: so the total of b) should actually read $20$, leading to $85$ triangles in all.

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    as there are 10 points, can't it be $\dbinom{10}{3}$ – 2017-02-06
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    @Kiran No, because some of the points lie on a straight line (five sets of four points are thus arranged). Using your argument you would expect a set of four points to always produce 4 triangles (4!/3!), but if they were in a straight line you wouldn't get any (unless you counted 'degenerate' triangles with h=0). – 2017-02-06
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    @Penguino, thanks clarifying. – 2017-02-06