# Is it possible to put an equilateral triangle onto a square grid so that all the vertices are in corners?

In the following collection of problems - arXiv:1110.1556v2 [math.HO] - the following question is posed:

Is it possible to put an equilateral triangle onto a square grid so that all the vertices are in corners?

The first approach that springs to mind is to use Pick's Theorem (e.g. http://www.geometer.org/mathcircles/pick.pdf) assuming that all vertices are on lattice points. It turns out that it is not possible (by Pythagoras, the area of an equilateral triangle with two vertices on lattice points is a rational multiple of $\sqrt3$).

My question then is - how can one establish the impossibility of such a placement without resorting to Pick's Theorem?

No, because that would imply an infinite sequence of smaller and smaller triangles with the same property:

$\hspace{90pt}$

The key to the proof below is this property:

For any point $(x,y) \in \mathbb{Z}^2$ we have that $(-y,x)$ and $(y,-x)$ are its ccw and cw rotations around $(0,0)$ by $\frac{\pi}{2}$.

This implies that we can rotate points of $\mathbb{Z}^2$ around other points of $\mathbb{Z}^2$ by $\frac{\pi}{2}$ and we will still end up in $\mathbb{Z}^2$.

The proof:

Consider an equilateral triangle $\triangle ABC$ with vertices in $\mathbb{Z}^2$ and perform a rotation of $A$ around $C$ to get $A''$ which also has integer coordinates:

$\hspace{70pt}$

Then translate $B$ along $\vec{A''C}$ to get $B'$, again with integer coordinates:

$\hspace{70pt}$

Do this two more times to get also $A'$ and $C'$, all with integer coordinates:

$\hspace{80pt}$

However, observe that $\triangle A'B'C'$ is also an equilateral triangle with vertices in $\mathbb{Z}^2$, but it is strictly smaller than $\triangle ABC$ (for convenience marked with the gray shaded area). This implies an infinite descending chain of equilateral triangles with coordinates in $\mathbb{Z}^2$ which is clearly impossible.

Edit:

For those of you who would like a construction in which the relative sizes are more apparent, observe that existence of an equilateral triangle with vertices in $\mathbb{Z}^2$ implies existence of a regular hexagon with the same property, and that in turn implies an infinite sequence of smaller and smaller regular hexagons:

$\hspace{30pt}$

I once saw this method applied just for hexagons, but was unable to find the source (if someone knows it, I would be grateful for the reference).

I hope this helps $\ddot\smile$

• Is there an easy way to show that the new triangle is smaller, and not larger ? For people with no eyes ? Commented Jul 14, 2015 at 12:18
• @mercio You could calculate the size of the new triangle, but there are other ways. With slight complication to the construction you could make the new triangle be contained in the old one. You could also say that the existence of equilateral triangle implies the existence of a hexagon, and then it's very easy to construct a smaller hexagon inside the larger one. Commented Jul 14, 2015 at 12:38
• That's a beautiful proof. Infinite descent in geometry - I like it! +1 Commented Jul 14, 2015 at 12:53
• woah, it's very clear with the hexagons ! Commented Jul 14, 2015 at 14:15

Theorem: A triangle is embeddable in $\mathbb{Z}^2$ if and only if all its angles have rational tangents.

But $\tan(\pi/3)=\sqrt{3},$ so an equilateral triangle is not embeddable.

Reference: Triangles with Vertices on Lattice Points Michael J. Beeson, The American Mathematical Monthly, Vol. 99, No. 3 (Mar., 1992), pp. 243-252

• +1 Thanks - that looks like an interesting article. I want to see what results they get in $\mathbb{R}^{n>2}$. Commented Jul 13, 2015 at 21:04
• @Marconius You're welcome. I just learned this theorem last month, so it was fresh in my mind.
– user940
Commented Jul 13, 2015 at 21:15

Let $P=(a,b),Q=(c,d)$ be two of the three points on the grid (assumed to be rational). The midpoint between these is $M=(\frac{a+c}{2},\frac{b+d}{2})$. Let $w=\sqrt{(d-b)^2+(c-a)^2}$ the length of a side. A vector in the direction of this side is given by $\vec{u}=\langle c-a,d-b\rangle$ and $\|\vec{u}\|=w$. The vector perpendicular to $\vec{u}$ is given by $\vec{v}=\langle d-b,a-c\rangle$ also of length $w$. The third vertex is then found at $M+h\hat{v}=M+(\frac{\sqrt{3}}{2}w)\frac{\vec{v}}{w}=M+\frac{\sqrt{3}}{2}\vec{v}$, which cannot be rational because $M,\vec{v}$ are rational.

Imagine the grid is $\mathbb{Z}[i]$ and one of the vertices is $0$. Let the others be $z$ and $w$. Then without loss $z$ is equal to $w$ after a $2\pi/3$ rotation, so $z=e^{2i \pi/3} w$. This is impossible (considering real/imaginary parts) as $\sqrt{3}$ is irrational.

Without loss of generality, let the angle $A$ be on the origin:

$\hspace{5cm}$

Slope of the line $AB$: $$\tan \alpha=\frac{n_2}{m_2}.$$ Slope of the line $AC$: $$\tan (\alpha+60^\circ)=\frac{\frac{n_2}{m_2}+\sqrt{3}}{1-\frac{\sqrt{3}n_2}{m_2}}=\frac{n_1}{m_1} \iff \frac{\sqrt{3}m_2+n_2}{m_2-\sqrt{3}n_2}=\frac{n_1}{m_1} \iff \\ \frac{\sqrt{3}(m^2+n^2)+4m_2n_2}{m^2-3n^2}=\frac{n_1}{m_1}.$$ Given $m_1,m_2,n_1,n_2$ are integers, the left hand side is never rational.

WLOG, let one of the vertices lie at the origin. We then have that the square of the three lengths are given by:

\begin{align}a^2+b^2=c^2+d^2&=(a-c)^2+(d-b)^2\\&=a^2+b^2+c^2+d^2-2(ac+bd)\end{align}

We can assume that not all $$a,b,c,d$$ are even, otherwise there is a smaller equilateral triangle. Now let's consider some cases:

• $$a^2+b^2\equiv_4 1\iff c^2+d^2\equiv_4 1$$. This yields $$a^2+b^2+c^2+d^2-2(ac+bd)\equiv_4 2(1-ac-bd)$$ This is impossible.

• $$a^2+b^2\equiv_4 2\iff c^2+d^2\equiv_4 2$$. This yields $$a^2+b^2+c^2+d^2-2(ac+bd)\equiv_4 2(ac+bd)\equiv_4 0$$ This is also impossible.