$ \newcommand{\Set}[1]{\{#1\}} \newcommand{\Floor}[1]{\lfloor{#1}\rfloor} \newcommand{\Ceil}[1]{\lceil{#1}\rceil} \newcommand{\TwoNorm}[1]{||{#1}||} \newcommand{\Modulus}[1]{|{#1}|} \newcommand{\Nat}{\mathbb{N}} \newcommand{\Int}{\mathbb{Z}} \newcommand{\EndDot}{\,.} \newcommand{\EndComma}{\,,} $

I often draw hexagons or rhombi on graph paper. Both are based on equilateral triangles, often as part of a triangular grid.

You can not draw an equilateral triangle with integer coordinates only, because this would solve the angle trisection problem. Hence, you have to use approximations. For graphing by hand, only approximations with small integer coordinates are interesting.

My goto approximation for quick sketching of an equilateral triangle has corners $(0,0)$, $(0,2)$ and $(1,2)$. A more decent one is $(0,0)$, $(4,1)$ and $(1,4)$.

Some day I stumbled upon the question I want to graph an equilateral triangle on graph paper. The answers there are interesting, but there were no concrete values given, either.

# Search space

Consider a triangle with corners $A$, $B$ and $C$ in anticlockwise order. Without loss of generality, translation invariance of $\Int^2$ lets us consider only the case of $A=\vec{0}$. Let $\alpha$ be the angle between the line going through $A$ and $B$ and the $x$-axis.

As $\Int^2$ has a rotational symmetry generated by quarter-rotations, one may take the principal representative of $(\alpha\!\!\mod\frac{1}{2}\pi)$ lying in $[0,\frac{1}{2}\pi]$.

If $\alpha\in]\frac{1}{3}\pi,\frac{1}{2}\pi]$, then flip along the $y$-axis. The angle between the $x$-axis and the image of $C$ is then $0\le\pi-\frac{1}{3}\pi-\alpha\le\frac{1}{3}\pi$.

If $\alpha\in]\frac{1}{6}\pi,\frac{1}{3}\pi]$, then translate by $-C$ and rotate anticlockwise by $\frac{1}{2}\pi$. The angle between the image of $\vec{0}$ and the $x$-axis is $0\le\alpha-\frac{1}{6}\pi\le\frac{1}{6}\pi$.

If $\alpha\in]\frac{1}{12}\pi,\frac{1}{6}\pi]$, then flip along the line $x=y$. The angle between the image of $C$ and the $x$-axis is $0\le\frac{1}{2}\pi-\frac{1}{3}\pi-\alpha = \frac{1}{6}\pi - \alpha \le \frac{1}{12}\pi$.

Thus, wlog $\alpha\le\frac{1}{12}\pi$.

# Enumeration

Let $n\in\mathbb{N}$ with $n\ge{}1$. The enumeration of coordinates for $B$ is the set $$ B_n := \Set{(x,y)\mid{} 0\in\Set{0,\dotsc,n}, y\in\Set{0,\dotsc,\Floor{x\sin(\frac{1}{12}\pi)}}} \EndDot $$ For each $(x,y)\in{}B_n$, let $\alpha :=\arctan(\frac{y}{x})$ and $c:=\sqrt{x^2 + y^2}$. Let $\beta:=\frac{1}{3}\pi-\alpha$. Let $$ C(x):= x - \cos(\beta) c \qquad \text{and} \qquad C(y):= y + \sin(\beta) c \EndDot $$ Let $$ C(x,y):=\Set{\Floor{C(x)},\Ceil{C(x)}}\times\Set{\Floor{C(y)},\Ceil{C(y)}} \EndDot $$ Hence, the full enumeration is $$ E_n := \Set{ (B,C)\mid{} B\in{}B_n, C\in{}C(B) } \EndDot $$

# Approximation error

Let $(B,C)\in{}E_n$. Let $a:=\TwoNorm{B-C}$, $b:=\TwoNorm{C}$ and $c:=\TwoNorm{B}$ be the side-lengths of the triangle with corners $\vec{0}$, $B$ and $C$. Let
$$
m := \min\Set{a,b,c}
\qquad
\text{and}
\qquad
M := \max\Set{a,b,c}
\EndDot
$$
The *relative sidelength error* is $R_S(B,C):=\Modulus{\frac{M}{m}-1}$ and the *absolute sidelength error* ist $A_S(B,C):=M-m$.

Let $\alpha$, $\beta$ and $\gamma$ be the angles opposite of $A$, $B$ and $C$, respectively. Let
$$
\delta := \min\Set{\alpha,\beta,\gamma}
\qquad
\text{and}
\qquad
\Delta := \max\Set{\alpha,\beta,\gamma}
\EndDot
$$
The *relative angle error* is $R_W(B,C):=\Modulus{\frac{\Delta}{\delta}-1}$ and the *absolute angle error* ist $A_W(B,C):=\Delta-\delta$. Here $W$ stands for “Winkel”.

Let $I$ and $i$ be the radii of the circumscribed and inscribed circles respectively. Then the *relative radii error* is $R_C(B,C):=\Modulus{\frac{I}{i}-1}$ and the *absolute radii error* is $A_C(B,C):=I-i$.

With $s:=\frac{a+b+c}{2}$ and $$ A := \sqrt{s(s-a)(s-b)(s-c)} $$ the area of the triangle by Heron’s formula, the radius of the inscribed circle is wikipedia $$ r = \frac{A}{s} $$ and the radius of the circumscribed circle is diameter formula $$ R = \frac{abc}{4A} ,. $$

# Best solutions for small n

See the F# script.

## Usual candidates

$B$ | $C$ | $L(B,C)$ | $W(B,C)$ |
---|---|---|---|

$(0,2)$ | $(1,2)$ | $0.105572809$ | $0.1624474625$ |

$(4,1)$ | $(1,4)$ | $0.028174684$ | $0.0466879655$ |

## Ordered by angles quality

The best solutions for $n\le 10$.

$B$ | $C$ | $L(B,C)$ | $W(B,C)$ |
---|---|---|---|

$(8, 0)$ | $(4, 7)$ | $0.00772212328$ | $0.01270192683$ |

$(10, 1)$ | $(4, 9)$ | $0.02000202061$ | $0.03274063767$ |

$(10, 0)$ | $(5, 9)$ | $0.02871413764$ | $0.04653653752$ |

$(4, 1)$ | $(1, 4)$ | $0.02817468419$ | $0.04668796554$ |

$(6, 0)$ | $(3, 5)$ | $0.02817468419$ | $0.04668796554$ |

$(8, 2)$ | $(2, 8)$ | $0.02817468419$ | $0.04668796554$ |

$(9, 2)$ | $(3, 9)$ | $0.02817468419$ | $0.04668796554$ |

## Ordered by sidelength quality

The best solutions for $n\le 10$.

$B$ | $C$ | $L(B,C)$ | $W(B,C)$ |
---|---|---|---|

$(8, 0)$ | $(4, 7)$ | $0.00772212328$ | $0.01270192683$ |

$(10, 1)$ | $(4, 9)$ | $0.02000202061$ | $0.03274063767$ |

$(4, 1)$ | $(1, 4)$ | $0.02817468419$ | $0.04668796554$ |

$(6, 0)$ | $(3, 5)$ | $0.02817468419$ | $0.04668796554$ |

$(8, 2)$ | $(2, 8)$ | $0.02817468419$ | $0.04668796554$ |

$(9, 2)$ | $(3, 9)$ | $0.02817468419$ | $0.04668796554$ |

$(10, 0)$ | $(5, 9)$ | $0.02871413764$ | $0.04653653752$ |

## Larger solutions

$B$ | $C$ | $L(B,C)$ | $W(B,C)$ |
---|---|---|---|

$(30, 0)$ | $(15, 26)$ | $0.0005550930208$ | $0.0009177507162$ |

$(45, 4)$ | $(19, 41)$ | $0.0009784738155$ | $0.001618174436$ |

$(49, 11)$ | $(15, 48)$ | $0.001384905205$ | $0.002289754747$ |

# Summary

So $((4,1),(1,4))$ is a nice symmetric (along $x=y$) solution usable for fast sketching. The large solutions $((45, 4),(19, 41))$ and $((49, 11),(15, 48))$ are interesting because they are not symmetric along $x=y$.

Returning to the original question. If one assumes standard $5$mm grid paper and a $0.5$mm pencil, then the wanted error bound is $0.1$. The solution $((4,1),(1,4))$ has a difference between a circumscribed circle based on the smallest and biggest side of less than $0.07$. Hence it fulfils the error bound $0.14$, which is just slightly more and visually nearly indistinguishable.