I got a question yesterday, to count the number of triangles in the above figure. I counted them, then thought about a formula to do the counting. I can't do that. Can someone show me how?
[Math] Counting triangles
combinatoricsdiscrete mathematicsgeometrypuzzle
Related Solutions
Say that instead of four triangles along each edge we have $n$. First count the triangles that point up. This is easy to do if you count them by top vertex. Each vertex in the picture is the top of one triangle for every horizontal grid line below it. Thus, the topmost vertex, which has $n$ horizontal gridlines below it, is the top vertex of $n$ triangles; each of the two vertices in the next row down is the top vertex of $n-1$ triangles; and so on. This gives us a total of
$$\begin{align*} \sum_{k=1}^nk(n+1-k)&=\frac12n(n+1)^2-\sum_{k=1}^nk^2\\ &=\frac12n(n+1)^2-\frac16n(n+1)(2n+1)\\ &=\frac16n(n+1)\Big(3(n+1)-(2n+1)\Big)\\ &=\frac16n(n+1)(n+2)\\ &=\binom{n+2}3 \end{align*}$$
upward-pointing triangles.
The downward-pointing triangles can be counted by their by their bottom vertices, but it’s a bit messier. First, each vertex not on the left or right edge of the figure is the bottom vertex of a triangle of height $1$, and there are $$\sum_{k=1}^{n-1}=\binom{n}2$$ of them. Each vertex that is not on the left or right edge or on the slant grid lines adjacent to those edges is the bottom vertex of a triangle of height $2$, and there are
$$\sum_{k=1}^{n-3}k=\binom{n-2}2$$ of them. In general each vertex that is not on the left or right edge or on one of the $h-1$ slant grid lines nearest each of those edges is the bottom vertex of a triangle of height $h$, and there are
$$\sum_{k=1}^{n+1-2h}k=\binom{n+2-2h}2$$ of them.
Algebra beyond this point corrected.
The total number of downward-pointing triangles is therefore
$$\begin{align*} \sum_{h\ge 1}\binom{n+2-2h}2&=\sum_{k=0}^{\lfloor n/2\rfloor-1}\binom{n-2k}2\\ &=\frac12\sum_{k=0}^{\lfloor n/2\rfloor-1}(n-2k)(n-2k-1)\\ &=\frac12\sum_{k=0}^{\lfloor n/2\rfloor-1}\left(n^2-4kn+4k^2-n+2k\right)\\ &=\left\lfloor\frac{n}2\right\rfloor\binom{n}2+2\sum_{k=0}^{\lfloor n/2\rfloor-1}k^2-(2n-1)\sum_{k=0}^{\lfloor n/2\rfloor-1}k\\ &=\left\lfloor\frac{n}2\right\rfloor\binom{n}2+\frac13\left\lfloor\frac{n}2\right\rfloor\left(\left\lfloor\frac{n}2\right\rfloor-1\right)\left(2\left\lfloor\frac{n}2\right\rfloor-1\right)\\ &\qquad\qquad-\frac12(2n-1)\left\lfloor\frac{n}2\right\rfloor\left(\left\lfloor\frac{n}2\right\rfloor-1\right)\;. \end{align*}$$
Set $\displaystyle m=\left\lfloor\frac{n}2\right\rfloor$, and this becomes
$$\begin{align*} &m\binom{n}2+\frac13m(m-1)(2m-1)-\frac12(2n-1)m(m-1)\\ &\qquad\qquad=m\binom{n}2+m(m-1)\left(\frac{2m-1}3-n+\frac12\right)\;. \end{align*}$$
This simplifies to $$\frac1{24}n(n+2)(2n-1)$$ for even $n$ and to
$$\frac1{24}\left(n^2-1\right)(2n+3)$$ for odd $n$.
The final figure, then is
$$\binom{n+2}3+\begin{cases} \frac1{24}n(n+2)(2n-1),&\text{if }n\text{ is even}\\\\ \frac1{24}\left(n^2-1\right)(2n+3),&\text{if }n\text{ is odd}\;. \end{cases}$$
[This is a community wiki answer to resolve the question]
From André Nicolas's comments, we see that if we take a triangle with sides $1, a, b$ and semiperimeter $S$ for which the Heron formula is known to work, and then enlarge it by a scale factor of $\lambda$, then for the enlarged triangle (with sides $\lambda, \lambda a, \lambda b$) we would have
$$\text{semiperimeter} = \frac{\lambda + \lambda a + \lambda b}{2} = \lambda S.$$
So that for the enlarged triangle, the Heron formula would give
$$\sqrt{\lambda S(\lambda S - \lambda)(\lambda S - \lambda a)(\lambda S - \lambda b)} = \sqrt{\lambda^4S(S-1)(S-a)(S-b)} = \lambda^2\sqrt{S(S-1)(S-a)(S- b)}$$
which is the correct expression for the area of the enlarged triangle. Thus, if Heron works for triangles with one side of length $1$, then it works for all triangles, since clearly every possible triangle is an enlargement of such a triangle (any triangle $p, q, r$ being an enlargement of $1, q/p, r/p$).
Best Answer
Its rather easy, once you think about it.
Let $h$ be the number of horizontal lines, and $v$ the number of non-horizontal lines. To form a triangle, you need to choose two non-horizontal lines, and one horizontal as base.
You can choose non-horizontal lines in $^v\text C_2={{v(v-1)}\over{2}}$ ways. Any of the horizontal lines may be selected in $^h\text C_1=h$ ways.
So, the total number of triangles is ${{^v\text C_2}\cdot{^h\text C_1}}$
Which is ${{{v(v-1)}\over{2}}\cdot{h}}$
And thus
$${{hv(v-1)}\over{2}}$$
P.S. In your figure, it is ${{5×6×5}\over2}=75$