# "Too Many Triangles?!":General Solution and Thoughts

This note is for the question Too may Triangles?!

Consider an equilateral triangular lattice like the one below :

We'll say that this specific lattice is of size 5 since it has 5 dots on its outer perimeter.

Here we'll think about the question :

On an equilateral triangular lattice of size $$n$$, by joining dots with straight lines, how many different equilateral triangles can we make?

Part 1 :The Answer and a Solution

Here we'll show that the answer is ${n+2 \choose 4} = \frac{n(n+2)(n^2-1)}{24}.$

We'll define a new $$i$$ equilateral triangle to be an equilateral triangle that can be made on a equilateral triangular lattice of size $$i$$ but not in any smaller equilateral triangular lattice. For example, here are all the new $$5$$ equilateral triangles we can make :

Clearly there exist $$i-1$$ amount of new $$i$$ equilateral triangle in an equilateral triangular lattice of size $$i$$ for all natural numbers $$i$$.

So how many unique equilateral triangular lattice of size $$i$$ can we find in an equilateral triangular lattice of size $$n$$ where $$n \ge i$$?

Well we can find $$n+1 -i$$ in the $$1^{st}$$ row (the bottom row) and $$n+1 -i -1$$ in the $$2^{nd}$$ row and in general we have $$n+1 -i - k$$ in the $$k^{th}$$ row for $$k \le n-i$$, so the total is given by $$\sum_{k=1}^{k=n-i+1} k = \frac{(n-i+1)(n+2-i)}{2}$$ by the sum of consecutive integers formula.

So the total amount of equilateral triangles is given by

the number of new $$i$$ equilateral triangle $$\times$$ the number of unique equilateral triangular lattice of size $$i$$ can we find in an equilateral triangular lattice of size $$n$$ which is the same as

\begin{align} &\sum_{i=1}^{i = n} (i-1)\times \frac{(n-i+1)(n+2-i)}{2} \\ & \text{and using the substitution j = i-1, we have } \\=& \sum_{j=0}^{j = n-1} j\times \frac{(n-j)(n-j+1)}{2} = \sum_{j=1}^{j = n} j\times \frac{(n-j)(n-j+1)}{2} \\ =&\frac{1}{2} \sum_{j=1}^{j = n}j^3 -j^2(2n+1) + j(n(n+1)) \\ =& \frac{1}{2}[\frac{n^2(n+1)^2}{4} + \frac{n(n+1)(2n+1)^2}{6} + \frac{n^2(n+1)^2}{2}] \\=&\frac{n(n+1)}{24}[3n(n+1) + 2(2n+1)^2 - 6n(n+1)] \\=&\frac{n(n+1)}{24}[n^2+n-2] \\=&\frac{(n+1)(n)(n-1)(n+2)}{24} = {n+2 \choose 4}. \end{align}

Part 2 : Observations and Questions

Let

• $$T_n =$$ The total number of Equilateral Triangles we can form on an equilateral triangular lattice of size $$n$$ = $${n+2 \choose 4}.$$

• $$D_n =$$ The total number of Equilateral Triangles we can form on an equilateral triangular lattice of size $$n$$ only using lines that are not parallel to the sides of the largest equilateral triangle on the perimeter of the lattice = $${n+1 \choose 3}$$ .

• $$S_n =$$ The total number of Equilateral Triangles we can form on an equilateral triangular lattice of size $$n$$ only using lines that are parallel to the sides of the largest equilateral triangle on the perimeter of the lattice $$= {n+1 \choose 4}$$ .

Note that $$D_n = {n+1 \choose 3}$$ correspond to the fourth diagonal line of Pascal's Triangle containing the entries of $$1,4,10,20,35,56...$$. We can interpret these as the number of points needed to describe a tetrahedron of "size n+1".

Similarly, $$T_n = {n+2 \choose 4}$$ and $$S_n = {n+1 \choose 4}$$ corresponds to the fifth diagonal line of Pascal's Triangle containing the entries $$1,5,15,35,70,126,210...$$. We can interpret this row as "The number of points we need to describe a fourth dimensional triangle (aka a pentatope) of "size" n+2 and n+1 respectively" .

It is true that $T_n = {n+2 \choose 4} = {(n+1) + 1 \choose 4} = D_{n+1}$ which means that the total number of "non parallel equilateral triangles" in an $$n$$ sized lattice is the same as the total number of equilateral triangles we can make in an $$n-1$$ sized lattice.

Is there a more intuitive reason why? This also true for the analogue square question.

Note that as $$n \rightarrow \infty$$ the area enclosed by the new n equilateral triangles seems to tend to a Releaux Triangle. In this case $$n = 200.$$

and seems to have similar results for square analogue. $$\square$$

Note by Roberto Nicolaides
3 years, 3 months ago

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Thank you for the kind words :)

- 3 years, 3 months ago

Awesome! Hats off!!

- 3 years, 3 months ago

Nicely explained Roberto :)

- 3 years, 3 months ago