Find the number of all isosceles triangles, where all three vertices belong to the set $\{A_1,A_2, \cdots,A_{30}\}$ 
In the coordinate plane let $A_i=(i,1)$ for $l\leq i\leq15$, and let $A_i=(i-15,4)$ for $16\leq i \leq 30$. Find the number of all isosceles triangles, where all three vertices belong to the set $\{A_1,A_2, \cdots,A_{30}\}$

I don't have any background in combinatorics so I'm having trouble on how to reword this so I can look up information to solve it. 
 A: Outline: Draw the $30$ points. The counting will be completely picture-based.  An isosceles triangle has either (i) $1$ vertex on the upper line and $2$ vertices on the lower line, or (ii) $1$ vertex on the lower line, and $2$ vertices on the upper line. By symmetry there are just as many of Type (i) as of Type (ii), so we count the Type (i) isosceles triangles and double. From now on we only consider Type (i) triangles.
Call the vertex on the upper line $A$, and the two vertices on the lower line $B$ and $C$.  If $A$ is a point on the upper line, it will be convenient to call the point on the lower line immediately below it by the name $A'$.
We first count the obvious isosceles triangles, the ones in which $AB=AC$. Let the $A$'s, from left to right, be $A_1$, $A_2$, and so on up to $A_{15}$.
It is clear that there is no isosceles triangle with $AB=AC$ and $A=A_1$. There is $1$ with $A=A_2$, there are $2$ with $A=A_3$, there are $3$ with $A=A_4$, and so on for a while. Continue. I think finding the total will not be difficult.
Now comes the trickier part, counting the isosceles Type (i) triangles in which $BC$ is equal to one of $AB$ or $AC$. Note that $BC$ is an integer, so $AB$ or $AC$ has to be an integer. 
Under what circumstances can $AB$ (or $AC$) be an integer? There are two possibilities, $P_1$ where $B$ (or $C$) is just below $A$, and $P_2$, where neither $B$ nor $C$ is directly below $A$.
Counting the $P_1$ kind should not be hard, for our triangle is then right-angled isosceles with two sides equal to $3$. There is $1$ such triangle with $A=A_1$, there is $1$ with $A=A_2$, $1$ with $A=A_3$, but there are $2$ with $A=A_4$, $2$ with $A_1=A_5$. Continue.
Finally, we look at Type (i) isosceles triangles of the $P_2$ kind, where $BC$ is equal to one of $AB$ or $AC$ but neither $B$ nor $C$ is directly below $A$. L If $AB$ is an integer, then $AA'B$ is an integer-sided right triangle, with $AA'=3$. Thus $A'B=4$ and $AB=5$. Similarly, if $AC$ is an integer then $AC=5$. 
Again we consider the various possibilities for $A$. If $A=A_1$, then there is $1$ such triangle, with $B=A_5'$ and $C=A_{10}'$. Similarly, there is $1$ such triangle with $A=A_2$. Continue, noting the left-right symmetry of the $A_i$.
