Fitting an equilateral triangle to three points (corrected)

Draw $AB$ , $BC$ , and $AC$ , and label the two equilateral triangles $\triangle DEF$ and $\triangle D'E'F'$ , as shown below. Also let $a = AD$ , $b = BE$ , and $c = CF$ , so that $8 - a = AF$ , $8 - b = BD$ , and $8 - c = CE$ , and similarly let $a' = AD'$ , $b' = BE'$ , and $c' = CF'$ , so that $8 - a' = AF'$ , $8 - b' = BD'$ , and $8 - c' = CE'$ .

By the distance formula, $AB^2 = (5 - 1)^2 + (4 - 1)^2 = 25$ , $BC^2 = (3 - 5)^2 + (7 - 4)^2 = 13$ , and $AC^2 = (3 - 1)^2 + (7 - 1)^2 = 40$ .

Since $\triangle DEF$ is an equilateral triangle, $\angle ADB = 60°$ , so by the law of cosines on $\triangle ABD$ , $AB^2 = AD^2 + BD^2 - 2 \cdot AD \cdot BD \cdot \cos 60°$ or $25 = a^2 + (8 - b)^2 - a(8 - b)$ . Similarly on $\triangle BCE$ and $\triangle ACF$ , $13 = b^2 + (8 - c)^2 - b(8 - c)$ and $40 = c^2 + (8 - a)^2 - c(8 - a)$ , and to match the diagram, these three equations solve to:

$a \approx 0.7029$ , $b \approx 2.6857$ , and $c \approx 3.9021$

From $AD \approx 0.70290$ we have $(x_D - 1)^2 + (y_D - 1)^2 = 0.7029^2$ and from $BD \approx 8 - 2.6857 \approx 5.3143$ we have $(x_D - 5)^2 + (y_D - 4)^2 = 5.3143^2$ , which solves to $x_D \approx 1.1684$ and $y_D \approx 0.3176)$ (to match the diagram). Similarly, from $BE \approx 2.6857$ we have $(x_E - 5)^2 + (y_E - 4)^2 = 2.6857^2$ and from $CE \approx 8 - 3.9021 \approx 4.0979$ we have $(x_E - 3)^2 + (y_E - 7)^2 = 4.0979^2$ , and to match the diagram, these two equations solve to:

$x_E \approx 6.9364$ and $y_E \approx 5.8610$

Because $\triangle DEF$ has a side of $8$ , its center $(x_O, y_O)$ is $\frac{8\sqrt{3}}{3}$ away from both $D$ and $E$ , so $(x_O - 1.1684)^2 + (y_O - 0.3175)^2 = (\frac{8\sqrt{3}}{3})^2$ and $(x_O - 6.9364)^2 + (y_O - 5.8610)^2 = (\frac{8\sqrt{3}}{3})^2$ , and to match the diagram, these two equations solve to:

$x_O \approx 2.4522$ and $y_O \approx 4.7544$

Finally, since the slope $m$ of $DE$ is $m = \frac{y_E - y_D}{x_E - x_D} \approx \frac{5.8610 - 0.3176}{6.9364 - 1.1684} \approx 0.9611$ , the counter-clockwise angle $\theta$ is $\theta \approx \tan^{-1} m \approx \tan^{-1} 0.9611$ or:

$\theta \approx 43.8625°$

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By the same steps on equilateral triangle $\triangle D'E'F'$ :

$a' \approx 5.5871$ , $b' \approx 3.9463$ , and $c' \approx 7.1758$

$x_{D'} \approx 6.5364$ and $y_{D'} \approx 0.2487$

$x_{E'} \approx 3.5043$ and $y_{E'} \approx 7.6519$

$x_{O'} \approx 2.8832$ and $y_{O'} \approx 3.0750$

$\theta' \approx 112.2723°$

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Therefore, $\theta + x_O + y_O + \theta' + x_{O'} + y_{O'} \approx 43.8625 + 2.4522 + 4.7544 + 112.2723 + 2.8832 + 3.0750 \approx \boxed{169.3}$ .

If the fitted triangle is centered at $r_0$ and tilted by an angle $\theta$ from the standard orientation, then using simple geometry on the equilateral triangle, we can write the following equation iterated for the three points.

$(P_i - r_0)^T R(\theta) v_i = \dfrac{1}{2 \sqrt{3} } L , \hspace{12pt} i = 1,2,3$

where $\{ P_i \}$ are the three given points, and $R(\theta)$ is the standard 2D rotation matrix, and is given by

$R(\theta) = \begin{bmatrix} \cos \theta && -\sin \theta \\ \sin \theta && \cos \theta \end{bmatrix}$

And the $\{ v_i \}$ are three unit vectors pointing from the center of the equilateral in standard orientation (without rotation) to the center of the three sides of the triangle, and they are given by,

$v_1 = [\cos \frac{\pi}{6} , \sin \frac{\pi}{6} ]^T$

$v_2 = [\cos \frac{5 \pi}{6} , \sin \frac{5 \pi}{6} ]^T$

$v_3 = [\cos \frac{3 \pi}{2} , \sin \frac{3 \pi}{2} ]^T$

Now the key trick here is to eliminate $r_0$ from the three equations, by simply adding them up. This is possible because $v_1 + v_2 + v_3 = 0$ .

Hence, the result of addition is a single equation,

$P_1^T R(\theta) v_1 + P_2^T R(\theta) v_2 + P_3^T R(\theta) v_3 = \dfrac{\sqrt{3}}{2} L$

The left hand side of the above equation is the sum of three quadratic forms, and can be expressed as a certain linear combination of $\cos \theta$ and $\sin \theta$ , so that the equation becomes,

$A \cos \theta + B \sin \theta = C$

where

$A = \displaystyle \sum_{i = 1 }^{3} P_{ix} v_{ix} + P_{iy} v_{iy}$

$B = \displaystyle \sum_{i=1}^{3} P_{iy} v_{ix} - P_{ix} v_{iy}$

$C = \dfrac{\sqrt{3}}{2} L$

Evaluating $A, B, C$ , we obtain,

$A = -2.598076211$

$B = -7.964101615$

$C = 6.92820323$

The trigonometric equation can be solved using standard methods, and its two solutions are:

$\theta_1 = 2.8599456 = 163.86^{\circ}$

$\theta_2 = 4.0539157 = 232.27^{\circ}$

Next, we want to find the center of the triangle for each value of $\theta$ , so for each of the above two values, we build a linear system of $2$ equations in the $2$ unknowns which are $x_0$ and $y_0$ , and these can be solved readily resulting in,

$(x_{01}, y_{01} ) = (2.452, 4.754)$ corresponding to $\theta_1$ and $(x_{02}, y_{02} ) = (2.883, 3.075)$ corresponding to $\theta_2$

Since rotating an equilateral triangle by $120^{\circ}$ about its center does not change it, we'll take the angles we obtained mod $120$ .

The final answer is $43.86 + 2.45 + 4.75 + 112.27 + 2.88 + 3.08 = 169.3$