Finding Equilateral Triangles

Geometry Level 2

Consider a $10 \times 10$ square grid of lattice points.

If we join the dots using straight lines, can we form an equilateral triangle?

No Yes

9 solutions

Jon Haussmann
Oct 25, 2015

Suppose that three lattice points can form the vertices of an equilateral triangle. Let $s$ be the side length. Then $s^2$ is a positive integer, so the area $\frac{\sqrt{3}}{4} s^2$ is irrational. But by the Shoelace Formula , the area is rational, contradiction. Thus, three lattice points cannot form the vertices of an equilateral triangle.

Can you please elaborate about the Above mentioned Shoelace Formula. and its application over here?

Atharva Bagul - 5 years, 7 months ago

Pick's Theorem would do the job as well.

Brian Charlesworth - 5 years, 7 months ago

According to the shoelace formula, the area of the triangle with coordinates $(x_1,y_1)$ , $(x_2,y_2)$ , $(x_3,y_3)$ is $\frac{1}{2} |x_1 y_2 + x_2 y_3 + x_3 y_1 - x_2 y_1 - x_3 y_2 - x_1 y_3|.$ Since $x_1$ , $y_1$ , $x_2$ , $y_2$ , $x_3$ , $y_3$ are all integers, the area must be rational.

Jon Haussmann - 5 years, 7 months ago

How did you say that $s$ is positive integer? It could be any real number, couldn't it?

Manish Mayank - 5 years, 6 months ago

$s$ is not necessarily a positive integer, but $s^2$ is.

Ivan Koswara - 5 years, 6 months ago

Joze Wingedlion
Oct 27, 2015

To construct a equilateral triangle you would need 3 points and the distamce from one side of the triangle to the opposite point (also known as the height) is equal to a times square root of 3 divided by 2 and there is no such number between 1 and 10 that would make the result of this equation an integrer.

How do you know that there is no distance from a point to a line (between two lattice points) that cannot be "a times square root of 3 divided by 2"?

For example, we have distances of a point to a line that is equal to $\sqrt{2}$ .

You seem to be assuming that "the only distances must be integers from 1 to 10".

Calvin Lin Staff - 5 years, 7 months ago

You did not state that no side of the triangle could contain more than 2 lattice points (the vertices). Why not? (if they could, there would be an obvious solution and the correct answer to the problem would be Yes, which is the answer I gave :/)

Migu El - 5 years, 6 months ago

Harish Sasikumar
Oct 26, 2015

How ever you draw two lines, they won't make 60 degrees, which is required to construct an equilateral triangle.

In response to Calvin Lin, the reasoning is

In this case, all the possible acute angles that can be formed will be $tan^{-1}\frac{a}{b}$ where $a,b=0,1,2,3,...9$

Or we can generalize "The possible angles that can be made out of joining three dots from an array of periodic two dimensional dots has to be an inverse tangent of a rational number or infinity. "

And we know that $\pi/3$ = $tan^{-1}\sqrt{3}$ and $\sqrt{3}$ is not a rational number. (Thanks Calvin Lin for correcting the mistake)

Can you explain why?

Calvin Lin Staff - 5 years, 7 months ago

Lim Yu
Dec 5, 2015

When drawing equilateral triangle on this square grid, we try to balance out the length of each sides. Therefore, all we can make is a right triangle. Since all these triangles have a right angle means there is NO way we can make out equilateral triangles. :)

Muhammet Akgül
Dec 5, 2015

tabiki de hayır lo kurdo

Roberto Nicolaides
Dec 5, 2015

We can use pick's theorem and the fact that $tan(\frac{\pi}{n})$ is rational for $n \ge 3$ if and only if $n=4$ to show that a square is the only regular polygon we can create this way.

Michael Long
Dec 5, 2015

I wish I understood all of you better, I'm trying.

Abraham Prieto
Oct 29, 2015

The tangent between two vectors is equal to the modulus of their cross product divided by their scalar product. In the case of integer components, since the cross product only has one component it can't be that its modulus is irrational and the scalar product can't be either. Therefore an angle with irrational tangent can't not be achieved.

Agnishom Chattopadhyay
Oct 25, 2015

Without loss of generality, let us shift the two points of the equilateral triangle to $(0,0)$ and $(m,n)$ . The third point must be $(m/2 - n \sqrt{3} / 2, n/2 + m \sqrt{3}/2)$ . But that cannot be a rational point, and not a lattice point either.

Not quite. There are 2 possibilities for the third point. But otherwise, you've got the correct idea.

Do you know what is a necessary and sufficient condition for a triangle to be expressible on a square lattice grid?

Calvin Lin Staff - 5 years, 7 months ago

Something relating to constructibility of the T-Ratios of the angles, I think. Not sure, though.

Agnishom Chattopadhyay - 5 years, 7 months ago

Yup. The tangents of the angles have to be rational. Do you know why?

Calvin Lin Staff - 5 years, 7 months ago

@Calvin Lin – Something like this?

WLOG, assume that the first two points lie on (0,0) and (a,b) and the tangents of the angles this segment makes with the third point are m and n.

Then, the third point is at $(b+na,m(b+na))$ .

If m,n were rational, we could scale up the triangle by a factor such that the resulting points lie on lattice points.

Agnishom Chattopadhyay - 5 years, 7 months ago

@Agnishom Chattopadhyay – I've no idea what you're trying to say here. What do you mean by "tangent of the angle this segment makes with the third point are m, n"? Do you mean that $\tan \alpha = \frac{m}{n}$ ?

Furthermore, it is not clear to me why the third line must lie on $y = m x$ (and in particular is independent of (a,b) ).

There are 2 parts to be shown

Given 3 integer points, the angles between them have a rational tangent.
Given a triangle with 3 rational tangent angles, there is a similar triangle whose vertices are lattice points.

Calvin Lin Staff - 5 years, 7 months ago

Finding Equilateral Triangles

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