As shown in the figure below, each pair of points on each side of the square trisect that side. How many isosceles triangles can be formed by using the 8 trisection points and the 4 vertices?
How should I start "counting"? Or is there a better way?
My approach:
Let's name the dots: A, B, C, ... L from the top-left vertex.
The top-left vertex, which is point A, can form three triangles
△ABL △ACK △ADJ
The similar applies to the other vertices, which gives s a total of
12 triangles (3 triangles each for the four vertices)
Now for the trisection points. For point B, it can form a triangle △BHJ
Same for the other trisection points, which give us 8 more triangles.
Therefore we have 12+8=20 triangles in total.
So my answer is 20, which was not the correct answer given (36). What did I mis-count?
Best Answer
Some hints:
The peak of the triangle can be a vertex or a trisection point. It is sufficient to analyze these two cases.
For the first case take the upper left vertex $A$ of the square, and write next to each other marked point its distance to $A$. Looking at the resulting numbers you should be able to find the number of isosceles triangles with peak at $A$.
Same for the marked point $B$ at one third of the upper edge of the square.
Now think of how many marked points of type $A$ you have, and how many marked points of type $B$.