4
$\begingroup$

I know that it's not possible to construct an equilateral triangle on lattice square grid if its vertices must be lattice points.

But, suppose you only have a lattice square grid and a straightedge (and a tool like pencil or pen to draw). Is it possible to construct several lines that will then, in fact, create an equilateral triangle?

$\endgroup$
2
  • 2
    $\begingroup$ The height will always be a multiple of $\sqrt{3}$ so I don't think you can do it without a compass. $\endgroup$
    – Paul
    Commented Jun 8 at 13:19
  • $\begingroup$ @Paul, this fact concerned me as well, but I couldn't come up with proof using this so far. But I feel like I have something on my mind that might work as a proof of impossibility... $\endgroup$
    – LLL
    Commented Jun 8 at 14:29

2 Answers 2

1
$\begingroup$

By the claim below, no equilateral triangle formed with lattice lines can exist.

Let $L$ be a square lattice in the Euclidean plane. A line will be called a lattice line if it passes through (at least) two lattice points.

Claim. There are no two lattice lines spanning an angle of $60$ degrees.

To justify the claim, let us suppose that there are two such lines. Using translations, we can assume that they intersect in a lattice point, say $O$ (the origin of our frame in the sequel). Let these lines be $(OA)$ and $(OB)$, with $A = (a_1, a_2)$ and $B=(b_1, b_2)$ two lattice points. Since $$ \sigma([OAB]) = \sqrt{3} \langle \vec{OA}, \vec{OB} \rangle, $$ multiplying by $2$ and taking the square, we arrive in coordinates at $$ (a_1b_2 -a_2b_1)^2 = 3(a_1b_1 +a_2b_2). $$ But this is impossible; inspect the multiplicity of $3$ on both sides of the equality.

$\endgroup$
1
  • $\begingroup$ It's still not clear to me how you obtained a square on the left side without a square on the right side of equality. Please, could you provide more steps of required mathematical manipulations? $\endgroup$
    – LLL
    Commented 5 hours ago
1
$\begingroup$

The only way you can construct a new point is as the intersection of two lines joining two pairs of pre-existing points (either the original lattice points or ones you've already constructed).

Since this is equivalent to solving a pair of simultaneous equations, it's easy to see that you can never construct a point with an irrational coordinate (you start with a subset of $\mathbb{Q}^2$; all coefficients involved will therefore always be rational).

Say there existed an equilateral triangle whose vertices were all in $\mathbb{Q}^2$; then scaling by an appropriate integer factor would result in a triangle in $\mathbb{Z}^2$; but you already know it's impossible to have an equilateral triangle whose vertices are all lattice points.

$\endgroup$
1
  • $\begingroup$ That's what I was looking for! Thanks. I knew I needed to use the fact of irrationality somehow, and that it should contradict everything I construct using rational numbers like that, but couldn't come up with a way to do it. $\endgroup$
    – LLL
    Commented 4 hours ago

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .