I want to graph an equilateral triangle. It would be ideal if I had a set of three points: $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ with $x_i, y_i \in \mathbb{Z}$ as the vertices. However, this is impossible. I am willing to "settle" for a triangle that is off by the width of a pencil lead. (That is, if we draw circles of radius $\delta$ around all of the lattice points, the true location of the vertices of the equilateral triangle are inside of these error circles.)

For a given $\delta$ what is the smallest such triangle?

My instinct is to let the computer guess, is there a more elegant way?

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    $\begingroup$ Well, you can start by looking at the continued fraction approximants to $\sqrt{3}$... $\endgroup$ Oct 23, 2010 at 22:54

3 Answers 3


Great question! Here is an upper bound: let $\frac{p_n}{q_n}$ be the sequence of convergents to the continued fraction of $\sqrt{3}$. It is known that $| \sqrt{3} - \frac{p_n}{q_n} | < \frac{1}{q_n q_{n+1}}$. Then the vertices

$$(0, 0), (2q_n, 0), (q_n, p_n)$$

are within $|\sqrt{3} q_n - p_n| < \frac{1}{q_{n+1}}$ of an equilateral triangle. Asymptotically I believe that $q_n \sim C \cdot \sqrt{3}^n$ for some constant $C$, which means that for small $\delta$ we can always find a triangle with side length something like $\frac{2}{\delta}$.

Here is a way to get a lower bound: suppose $(x_i, y_i)$ is a collection of points which are $\delta$-close to an equilateral triangle, and suppose WLOG that $(x_1, y_1) = (0, 0)$. Then $\frac{x_3 + iy_3}{x_2 + iy_2}$ is a rational approximation to $\frac{1}{2} + \frac{\sqrt{3}}{2} i$, hence taking imaginary parts, $2 \frac{x_2 y_3 - x_3 y_2}{x_2^2 + y_2^2}$ is a rational approximation to $\sqrt{3}$. It is known that the convergents $\frac{p_n}{q_n}$ give the best rational approximations to $\sqrt{3}$ (in the sense that better ones must have larger denominators), so one should be able to write down how good this approximation is in terms of $\delta$ and use this to give a lower bound on how large $x_2^2 + y_2^2$ must be. But the details look messy. In any case, I would be very surprised if one didn't end up with a lower bound of the form $\frac{C}{\delta}$.

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    $\begingroup$ N.B. The lower bound argument is really based on the fact that it is not possible to approximate algebraic numbers too well; see en.wikipedia.org/wiki/Thue%E2%80%93Siegel%E2%80%93Roth_theorem . $\endgroup$ Oct 23, 2010 at 23:31
  • $\begingroup$ Oh thank you! You are using $\sqrt{3} = 1 + \frac{2}{2+ frac{2}{2+ frac{2}{2+ ...}}}$ right? $\endgroup$
    – futurebird
    Oct 23, 2010 at 23:34
  • $\begingroup$ @a little don: I think the c.f expansion is $\sqrt{3}=1+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\cdots }}}$ $\endgroup$ Oct 24, 2010 at 16:08

If you want one of the sides to be along the axis, you can use (0,0),(2x,0),and (x,y). Then you want $\sqrt{y^2+x^2}=2x$ or $y^2=3x^2$. Pell's equation will help you find solutions in the integers to $y^2=3x^2\pm1$. $x=30, y=26$ is within $.017$ in $y$. Once you have one solution (say (a,b)=(1,0)), you can get another one as (2a+3b,a+2b). These get very close to equilateral. I don't know if you can do better if you don't put one side along an axis.


If you are interested in fast sketching of nearly equilateral triangles by hand, based on $5$mm graph paper and assuming a $0.5$mm pencil, then the triangle with corners on the grid coordinates $(0,0)$, $(1,4)$ and $(4,1)$ has an error of less than $0.7$mm (based on an upper bounds of two times the difference of the circumscribed circle based on the shortest and largest edge of this triangle). Even if you sketch without a ruler, the result is nearly indistinguishable from an equilateral triangle.


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