Odd "How many triangles are there in the picture?" Question From the First Stage of Israel's International Mathematical Olympiad Selection Process So stage 1 of of Israel's International Mathematical Olympiad selection process took place on Monday, and this was one of the questions (6 or 7 out of 10).
Q:
(The way the shape is produced wasn't inscribed, but its a pretty easy guess):
Draw an interval from each vertex of a square to the 2 midpoints of the sides it is not on.
How many triangles are there in the shape?

Is there a (viable) way to solve questions similar to the one I showed?
 A: You can follow this picture, the number of triangles are written inside of each drawing:

A: A triangle is determined by three lines. There are $12$ lines in the picture, so as a first approximation we have $\binom{12}3=220$ possible triangles. But not every triple of lines makes a triangle, so we have to count the bad triples and subtract. There are three types of bad triples.
Type I. Three lines meeting at one point. This happens $\boxed{20}$ times; the point of concurrency can only be a corner of the big square or the midpoint of a side of the big square.
Type II. Two of the three lines are parallel. There are $6$ pairs of parallel lines, so we get a bad triple of this kind by taking two parallel lines and a random third line, in $6\cdot10=\boxed{60}$ ways.
Type III. None of the three lines are parallel, but at least two of them fail to meet within the picture. In this case one of the lines, call it $S$, must be a side of the big square, and another is one of the two lines, call them $A$ and $B$, which are not parallel to $S$ but do not meet $S$ within the picture. (If $S$ is the top side, then $A$ is the line from the lower left corner of the big square to the middle of the right side, and $B$ is its mirror image.) There are $8$ type III triples containing $S$ and $A$, as the third line can be any of the $8$ lines not parallel to $S$ or $A$. Likewise there are $8$ type III triples containing $S$ and $B$. Since the triple $\{S,A,B\}$ has been counted twice, the total number of type III triples containing $S$ and $A$ or $B$ is $8+8-1=15$. Since there are $4$ choices for $S$ (and no overlaps between different choices for $S$), the total number of bad triples of type III is $4\cdot15=\boxed{60}$.
The number of triangles in the picture is $$220-20-60-60=\boxed{80}.$$
A: There are several types of triangles. Two types are symmetric (the first two in the following figure), they occur $4$ times each. All other types occur $8$ times each. I found $9$ of them, so that we have $80$ triangles in all.

