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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.

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    $\begingroup$ Have you plotted the points? You need to look at the points and figure out what groups of three make isosceles triangles. There are a few different classes of triangles of interest. $\endgroup$ – Ross Millikan Aug 16 '15 at 21:05
  • $\begingroup$ So I know two sides will have equal distance if I go about this do I have to worry about using the points more then once $\endgroup$ – HighSchool15 Aug 16 '15 at 21:09
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    $\begingroup$ You know that the triangle will either have 2 vertices in the top row and 1 in the bottom row, or vice versa. If you have 2 vertices in the top row, try to see how many ways you can choose a vertex in the bottom row so the triangle is isosceles; this will depend on how far apart the 2 vertices are. $\endgroup$ – user84413 Aug 16 '15 at 21:30
  • $\begingroup$ In this problem (unlike others), the issue of double-counting is unlikely to arise. Divide into cases, $\endgroup$ – André Nicolas Aug 16 '15 at 21:46
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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 $A_i$.

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  • $\begingroup$ Note that my $A_i$ are different from the $A_i$ of the problem statement. $\endgroup$ – André Nicolas Aug 17 '15 at 0:29

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