"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 nn, 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+24)=n(n+2)(n21)24. {n+2 \choose 4} = \frac{n(n+2)(n^2-1)}{24}.

We'll define a new ii equilateral triangle to be an equilateral triangle that can be made on a equilateral triangular lattice of size ii but not in any smaller equilateral triangular lattice. For example, here are all the new 55 equilateral triangles we can make :

Clearly there exist i1i-1 amount of new ii equilateral triangle in an equilateral triangular lattice of size ii for all natural numbers ii.

So how many unique equilateral triangular lattice of size ii can we find in an equilateral triangular lattice of size nn where nin \ge i?

Well we can find n+1in+1 -i in the 1st1^{st} row (the bottom row) and n+1i1n+1 -i -1 in the 2nd2^{nd} row and in general we have n+1ikn+1 -i - k in the kthk^{th} row for knik \le n-i, so the total is given by k=1k=ni+1k=(ni+1)(n+2i)2 \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 ii equilateral triangle ×\times the number of unique equilateral triangular lattice of size ii can we find in an equilateral triangular lattice of size nn which is the same as

i=1i=n(i1)×(ni+1)(n+2i)2and using the substitution $j = i-1$, we have =j=0j=n1j×(nj)(nj+1)2=j=1j=nj×(nj)(nj+1)2=12j=1j=nj3j2(2n+1)+j(n(n+1))=12[n2(n+1)24+n(n+1)(2n+1)26+n2(n+1)22]=n(n+1)24[3n(n+1)+2(2n+1)26n(n+1)]=n(n+1)24[n2+n2]=(n+1)(n)(n1)(n+2)24=(n+24). \begin{aligned} &\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{aligned}

Part 2 : Observations and Questions

Let

  • Tn=T_n = The total number of Equilateral Triangles we can form on an equilateral triangular lattice of size nn = (n+24).{n+2 \choose 4}.

  • Dn=D_n = The total number of Equilateral Triangles we can form on an equilateral triangular lattice of size nn only using lines that are not parallel to the sides of the largest equilateral triangle on the perimeter of the lattice = (n+13){n+1 \choose 3} .

  • Sn=S_n = The total number of Equilateral Triangles we can form on an equilateral triangular lattice of size nn only using lines that are parallel to the sides of the largest equilateral triangle on the perimeter of the lattice =(n+14)= {n+1 \choose 4} .

Note that Dn=(n+13)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...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, Tn=(n+24)T_n = {n+2 \choose 4} and Sn=(n+14)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...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 Tn=(n+24)=((n+1)+14)=Dn+1 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 nn sized lattice is the same as the total number of equilateral triangles we can make in an n1n-1 sized lattice.

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

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

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

Note by Roberto Nicolaides
4 years, 7 months ago

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Nicely explained Roberto :)

A Former Brilliant Member - 4 years, 6 months ago

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Awesome! Hats off!!

Nihar Mahajan - 4 years, 6 months ago

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@Nihar Mahajan , @Azhaghu Roopesh M

Thank you for the kind words :)

Roberto Nicolaides - 4 years, 6 months ago

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