Counting in geometry is challenging!

I have several methods of approaching this problem and I have posted them here: (Hope you understand with the easiest one)

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M1: This is the same as finding the number of ways to choose two points out of 14 on the slanting side which is $N = \dbinom{14}{2} = \dfrac{14!}{(14 - 2)! \times 2!} = \boxed{91}$

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M2: Let $n$ be the number of points between the 2 vertices of the hypotenuse, $n>0$ then the number of triangles is given by:

$f(n) = \dfrac{n^2 + 3n + 2}{2}.$

Substituting $n=12$ will give us the desired answer of $\boxed{91}$

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M3:

In the following diagram, a triangle can be formed from any 3 of the 15 vertices. So it is the number of ways of choosing 3 vertices from 15 vertices. But since 14 points lie on the same line (a triangle can't be formed by 3 points which lie on the same line), we should subtract the number of ways of choosing 3 vertices from 14 of those vertices.

So the answer is: $\dbinom{15}{3} - \dbinom{14}{3} = \dbinom{14}{2} = \dfrac{14!}{(14 - 2)! \times 2!} = \boxed{91}$

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M4:

It is clearly noticeable that this follows the triangular number pattern. Let $k$ be the number of small triangles of the diagram, then $k^{th}$ triangular number will give us the answer; in our case, $k = 13$ $\implies \dfrac{k(k+1)}{2} = \dfrac{13(13 + 1)}{2} = \boxed{91}$

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Notes:

In M3: it is generally true $\dbinom{n}{k} - \dbinom{n-1}{k} = \dbinom{n-1}{k -1}$ which can be proved by simple algebra:

Tip: Here I will prove ${ n \choose k } + { n \choose k-1 } = { n+1 \choose k }$

$\begin{aligned} { n \choose k } + { n \choose k-1} & = \frac{n!}{k! (n - k)!} + \frac{n!}{(k - 1)!(n - k + 1)!} \\ & = n! \left[ \frac{n -k + 1}{k!(n -k + 1)!} + \frac{k}{k!(n -k + 1)!}\right] \\ & = \frac{n!(n+1)}{k!(n -k + 1)!} = \binom{n+1}{k} \end{aligned}$

In M4: We note that there is a pattern of triangular numbers. If anyone can elaborate why this pattern follows up it would be great. Personally, I am not sure but I have a few ideas such as the fact of summing up triangular numbers: Let $S_n$ be the sum, then:

$S_n = \displaystyle \sum_{n=1}^{n} \dfrac{n(n+1)}{2} = \dfrac{1(1+1)}{2} + \dfrac{2(2+1)}{2} + \cdots + \dfrac{n(n+1)}{2}$

Which is also the same as:

$S_n = \displaystyle \sum_{n=2}^{n} \dbinom{n}{2} = \dbinom{2}{2} + \dbinom{3}{2} + \cdots + \dbinom{n}{2}$

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Do you have any other ways? :D

1. Count all of the individual triangles: that gives us 13.

2. Then, count all of the triangles made by combining two adjacent triangles: 12

3. Next, do the same but for 3 triangles: 11

4. You probably see where this is going. If not, do the same with 4 triangles and get 10

The pattern goes on, until you get to 13 triangles which is the entire triangle (or 1).

This gives us the sum: $\displaystyle\sum_{n=1}^{13} n=\frac{13(13+1)}{2}=13\cdot 7 =91$ .

In the most general form, if any 2-dimensional figure is split into $n$ polygons such that all of the polygons share a vertex, then there are $\frac{n(n+1)}{2}$ polygons in the entire figure. $\beta_{\lceil \mid \rceil}$

To avoid double-counting, we will consider only triangles below each triangular wedge. * The orange triangle is part of one triangle (itself). * The pink triangle above it is part of two, itself and pink+orange. * The brown triangle up next is part of three: itself, itself+pink, and itself+pink+orange.

Continuing in this fashion, we see that the $n$ th triangle up from the bottom is a piece in $n$ different triangles (again, only considering triangles below it). Thus the total number of triangles is $1+2+\cdots+13 = \frac{13\cdot14}{2} = \boxed{91}$