Fitting an Equilateral Triangle Through 3 Points

The slope of the line passing through $A$ is $m_A = \tan^{-1} 45° = 1$ . Since an equilateral triangle has 60° angles, the slope $m_B$ through $B$ must satisfy the equation $\tan (-60°) = \frac{m_B - 1}{1 + m_B}$ , which solves to $m_B = -2 + \sqrt{3}$ , and the slope $m_C$ through $C$ must satisfy the equation $\tan 60° = \frac{m_C - 1}{1 + m_C}$ , which solves to $m_C = -2 - \sqrt{3}$ .

The equation of the line through $A(5, 0)$ with a slope of $m_A = 1$ is $y = x - 5$ . The equation of the line through $B(1.8, 2.4)$ with a slope of $m_B = -2 + \sqrt{3}$ is $y = (-2 + \sqrt{3})x + \frac{30 - 9\sqrt{3}}{5}$ . And the equation of the line through $C(0, 0)$ with a slope of $m_C = -2 - \sqrt{3}$ is $y = (-2 - \sqrt{3})x$ .

The intersection of $y = x - 5$ and $y = (-2 + \sqrt{3})x + \frac{30 - 9\sqrt{3}}{5}$ is $(\frac{69 + 14\sqrt{3}}{15}, \frac{-6 + 14\sqrt{3}}{15})$ , and the intersection of $y = x - 5$ and $y = (-2 - \sqrt{3})x$ is $(\frac{15 - 5\sqrt{3}}{6}, \frac{-15 - 5\sqrt{3}}{6})$ , and the distance between these two points (and also the side length of the equilateral triangle) is $\frac{63\sqrt{2} + 53\sqrt{6}}{30} \approx \boxed{7.29728}$ .

With a little bit calculation, we know that $AB=4, BC=3, AC=5$ also $AC$ will form an angle $45^\circ$ with a side of the equilateral triangle. Construct an ex-triangle with vertices $D, E, F$ (see the figure below)

$\dfrac{EC}{\sin45^\circ}=\dfrac{5}{\sin60^\circ}$ gets $EC=\dfrac{5\sqrt6}3$ ;

$EA=EC\cos60\circ+5\cos45^\circ=\dfrac{15\sqrt2+5\sqrt6}6$ .

Assume the angles $\angle BAC=\theta, \angle BAF=\alpha,$ we then have $\tan\theta=\dfrac34$ and $\tan(\theta+\alpha)=\tan135^\circ=-1$ , so

$\begin{aligned}\tan\alpha&=\tan(\theta+\alpha-\theta)\\&=\frac{\tan(\theta+\alpha)-\tan\theta}{1+\tan(\theta+\alpha)\tan\alpha}\\&=\frac{-1-\dfrac34}{1-\dfrac34}\\&=-7\end{aligned}$

So other trigonometric of $\alpha$ will be $\sin\alpha=\dfrac{7}{5\sqrt2}$ and $\cos\alpha=-\dfrac1{5\sqrt2}$ .

Finding $BF$ :

$\dfrac{BF}{\sin\alpha}=\dfrac4{\sin60^\circ}$ to get $BF=\dfrac{28\sqrt6}{15}$ , then we find $AF$ by using same method:

$AF=4\cos\alpha+BF\cos60^\circ=\frac{-12\sqrt2+28\sqrt6}{30}$ .

Adding up $EA$ and $AF$ will give us the side length of the equilateral triangle:

$\begin{aligned}EF&=EA+AF\\&=\frac{15\sqrt2+5\sqrt6}{6}+\dfrac{-12\sqrt2+28\sqrt6}{30}\\&=\boxed{\dfrac{63\sqrt2+53\sqrt6}{30}}\end{aligned}$

$As~ seen~from~the~sketch,~the ~Y_G,~Y_B,~Y_{Br}~the~equations~of~the~\Delta~sides~are:-\\ Y_G=X-5.~~~~~~~~~~Y_B=-Tan15(X-1.8)+2.4.~~~~~~~~~~Y_{Br}=-Tan75.X\\ \therefore~Y_G~and~Y_B~~intersect~at~vertex~(X_{G-B},Y_{G-B})=\left (\dfrac{7.4+1.8Tan15}{1+Tan15}~,~\dfrac{7.4+1.8Tan15}{1+Tan15}~-~5 \right ) \\ And~~Y_G~and~Y_{Br}~~intersect~at~vertex~(X_{G-Br},Y_{G-Br})=\left (\dfrac 5 {1+Tan75}~,~\dfrac 5 {1+Tan75}~-~5 \right ) \\ \therefore~the ~side~length\\$ $=\sqrt{\left ( \dfrac{7.4+1.8Tan15}{1+Tan15}-\dfrac 5 {1+Tan75} \right )^2~ ,\left (\dfrac{7.4+1.8Tan15}{1+Tan15}-5 ~~-\dfrac 5 {1+Tan75}~+5 \right )^2 } \\ =\sqrt2*\left( \dfrac{7.4+1.8Tan15}{1+Tan15}-\dfrac 5 {1+Tan75} \right )=\dfrac{\sqrt2}{1+Tan15}*(7.4-3.2Tan15)............................Since~~Tan75=\dfrac 1 {Tan15} \\ =\Huge \color{#D61F06}{7.29728}$