Minimal Area

Consider the linear transformation $T:\mathbb{R}^2 \to \mathbb{R}^2$ such that $T\binom{4}{1} = \binom{6}{0}$ and $T\binom{2}{5} = \binom{3}{3\sqrt{3}}$ , so that $T\binom{x}{y} \; = \; \left(\begin{array}{cc} \frac32 & 0 \\ -\frac16\sqrt{3} & \frac23\sqrt{3}\end{array}\right)\binom{x}{y} \; = \; A\binom{x}{y}$ Then $T$ maps $(0,0)$ , $(4,1)$ , $(2,5)$ to the points $(0,0)$ , $(6,0)$ , $(3,3\sqrt{3})$ respectively, which are the vertices of an equilateral triangle. If $\mathcal{E}$ is an ellipse passing through the vertices $(0,0)$ , $(4,1)$ , $(2,5)$ , then $T(\mathcal{E})$ is an ellipse passing through the vertices $(0,0)$ , $(6,0)$ , $(3,3\sqrt{3})$ . If $\mathcal{E}$ is an ellipse with area $|\mathcal{E}|$ ,then $T(\mathcal{E})$ is an ellipse with area $\sqrt{3}|\mathcal{E}|$ ,since $A$ has determinant $\sqrt{3}$ .

Thus the minimum area ellipse $\mathcal{E}$ passing through $(0,0)$ , $(6,0)$ , $(3,3\sqrt{3})$ is such that $T(\mathcal{E})$ is the minimum area ellipse passing through $(0,0)$ , $(6,0)$ , $(3,3\sqrt{3})$ . Now is it pretty easy to show that the minimum area ellipse passing through the vertices of an equilateral triangle is the circumcircle of the triangle. Thus we want to find $\mathcal{E}$ such that $T(\mathcal{E})$ is the circle $\begin{aligned} (x - 3)^2 + (y - \sqrt{3})^2 & = \; 12 \\ x^2 + y^2 - 6x - 2\sqrt{3}y & = \; 0 \\ \mathbf{x}^T\mathbf{x} - (6\;\;\;\;2\sqrt{3})\mathbf{x} & = \; 0 \end{aligned}$ where $\mathbf{x} = \binom{x}{y}$ . Thus $\mathcal{E}$ has equation $\begin{aligned} (A\mathbf{x})^TA\mathbf{x} - (6\;\;\;\;2\sqrt{3})A\mathbf{x} & = \; 0 \\ \mathbf{x}^TA^TA\mathbf{x} - (6\;\;\;\;2\sqrt{3})A\mathbf{x} & = \; 0 \\ \tfrac13\mathbf{x}^T\left(\begin{array}{cc} 7 & -1 \\ -1 & 4 \end{array}\right)\mathbf{x} - (8\;\;\;\;4)\mathbf{x} & = \; 0 \\ 7x^2 - 2xy + 4y^2 - 24x - 12y & = \; 0 \end{aligned}$ Now the matrix $3A^TA \; = \; \left(\begin{array}{cc} 7 & -1 \\ -1 & 4 \end{array}\right)$ has positive eigenvalues $\tfrac12(11 \pm \sqrt{13})$ with corresponding eigenvectors $\binom{3 \pm \sqrt{13}}{-2}$ . The major axis of the ellipse $\mathcal{E}$ points along the direction of the eigenvector associated with the smaller eigenvalue, namely $\binom{3-\sqrt{13}}{-2}$ . Thus $m \; = \; \frac{-2}{3 - \sqrt{13}} \; = \; \frac{-2(3+\sqrt{13})}{9 - 13} \; = \; \tfrac12(3 + \sqrt{13})$ so that $2m-3=\sqrt{13}$ , and hence $(2m-3)^2 = \boxed{13}$ .