Number of triangles in a shape 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 $75$.
 A: 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: $5$

A: 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.
A: 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.
