Fitting an equilateral triangle to three points

The equations for $(P_1, P_2, P_3)$ are as follows:

$P_{1x} = x_c + R \cos(90^\circ + \theta) \\ P_{1y} = y_c + R \sin(90^\circ + \theta) \\ P_{2x} = x_c + R \cos(-30^\circ + \theta) \\ P_{2y} = y_c + R \sin(-30^\circ + \theta) \\ P_{3x} = x_c + R \cos(210^\circ + \theta) \\ P_{3y} = y_c + R \sin(210^\circ + \theta) \\ R = \frac{8}{\sqrt{3}}$

This link has a nice formula for the distance from a point to a line segment, which I used. Find $(x_c, y_c, \theta)$ such that the following equations are satisfied. Since we want the distance to be zero, the denominators of the distance equation can be disregarded.

Segment with $(P_1, P_2, C)$

$(P_{2x} - P_{1x})(P_{1y} - C_y) - (P_{1x} - C_x)(P_{2y} - P_{1y}) = 0$

Segment with $(P_2, P_3, B)$

$(P_{3x} - P_{2x})(P_{2y} - B_y) - (P_{2x} - B_x)(P_{3y} - P_{2y}) = 0$

Segment with $(P_3, P_1, A)$

$(P_{1x} - P_{3x})(P_{3y} - A_y) - (P_{3x} - A_x)(P_{1y} - P_{3y}) = 0$

A search algorithm finds the following values for the parameters:

$x_c \approx 2.452 \\ y_c \approx 4.754 \\ \theta \approx 43.863^\circ$

Draw $AB$ , $BC$ , and $AC$ , and let $D$ be the vertex of the equilateral triangle nearest to $A$ , $E$ be the vertex of the equilateral triangle nearest to $B$ , and $F$ be the vertex of the equilateral triangle nearest to $C$ . Also 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 these three equations solve to $a \approx 0.7029$ , $b \approx 2.6857$ , $c \approx 3.9021$ for $a < b < c$ .

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 $y_D \approx 0.3176)$ for $y_D < 1$ . 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 $(x_E - 3)^2 + (y_E - 7)^2 = 4.0979^2$ , which solves to $x_E \approx 6.9364$ and $y_E \approx 5.8610)$ for $x_E > 4$ .

Because $\triangle DEF$ has a side of $8$ , its center $(x_C, y_C)$ is $\frac{8\sqrt{3}}{3}$ away from both $D$ and $E$ , so $(x_C - 1.1684)^2 + (y_C - 0.3175)^2 = (\frac{8\sqrt{3}}{3})^2$ and $(x_C - 6.9364)^2 + (y_C - 5.8610)^2 = (\frac{8\sqrt{3}}{3})^2$ , which solves to $x_C \approx 2.4522$ and $y_C \approx 4.7544$ for $x < 3$ .

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 $\theta \approx \tan^{-1} m \approx \tan^{-1} 0.9611 \approx 43.8625°$ .

Therefore, $\theta + x_C + y_C \approx 43.8625 + 2.4522 + 4.7544 \approx \boxed{51.07}$ .