In square units, what is the area of this ellipse?

The Steiner ellipse is an ellipse with the least area that touches a triangle at its three vertices. And its area is $A_S = \dfrac {4\pi}{3\sqrt 3}A_\triangle$ , where $A_\triangle$ is the area of the triangle.

To find $A_\triangle$ from its three vertices $A(3.15674, 3.49665)$ , $B (2.19157, 3.34378)$ , and $C(2.80654, 2.58435)$ , we use the formula below:

$\begin{aligned} A_\triangle & = \left| \frac {A_x(B_y-C_y)+B_x(C_y-A_y)+C_x(A_y-B_y)}2 \right| \\ & = \left|\frac {3.15674 (3.34378-2.58435) + 2.19157 (2.58435 - 3.49665) + 2.80654(3.49665 - 3.34378)}2 \right| \\ & \approx 0.413494759 \\ \implies A_S & = \frac {4\pi}{3\sqrt 3}A_\triangle \approx \boxed{1.00} \end{aligned}$

First, showing that this is a special case.

The Euclidean distances between the three points are equal and are 0.977205 each. Because it is an equilateral triangle, the Steiner ellipse is, in fact a circle. A circle is a special case of an ellipse, by the way. Therefore, the center of the triangle and of the circle is the average of the coordinates of the points' coordinates, $\{2.71828,3.14159\}$ . Yes, I used $\pi$ and $e$ as the center's coordinates. The Euclidean distance to any of the points is $0.56419$ . The answer is $\pi\ 0.56419^2\approx 1.000\times 10^0$ . See, that was easy! ${}^{TM Staples}$

Ok, say that you did not see that the triangle was equilateral. Now, we will use the method in the wiki page mentioned in the problem statement. For the explanation of what is happening see that wiki page and the Singular Value Decomposition wiki page. $\bm{o}=\left( \begin{array}{ccc} 3.15674 & 2.19157 & 2.80654 \\ 3.49665 & 3.34378 & 2.58435 \\ 1 & 1 & 1 \\ \end{array} \right)$ . $\bm{n}=\left( \begin{array}{ccc} 1 & -\frac{1}{2} & -\frac{1}{2} \\ 0 & \frac{\sqrt{3}}{2} & -\frac{\sqrt{3}}{2} \\ 1 & 1 & 1 \\ \end{array} \right)$ . $\bm{a}=\bm{n}\,\bm{o}^{-1}=\left( \begin{array}{ccc} 1.37746 & 1.11544 & -7.24857 \\ -1.11544 & 1.37746 & -1.29532 \\ 0 & 0 & 1. \\ \end{array} \right)$ . Inverse of $\bm{a}$ gives the transformation function: $\left( \begin{array}{cc|c} 0.438458 & -0.355056 & 2.71828 \\ 0.355056 & 0.438458 & 3.14159 \\ \hline 0. & 0. & 1. \\ \end{array} \right)$ . The last column, first two rows, gives the center of the ellipse.

Doing the singular value decomposition of the first two rows and columns of the transformation function gives: $\bm{u}=\left( \begin{array}{cc} -1. & 0. \\ 0. & 1. \\ \end{array} \right), \bm{\sigma}=\left( \begin{array}{cc} 0.56419 & 0. \\ 0. & 0.56419 \\ \end{array} \right), \bm{v}^*=\left( \begin{array}{cc} -0.777146 & 0.62932 \\ 0.62932 & 0.777146 \\ \end{array} \right)$ . From $\bm{u}$ , one can see that affine transform is considered to be a reflection. From $\bm{\sigma}$ , one can see the radii and that they are equal to each other and are $0.56419$ . From $\bm{v}^*$ , one can see, because it is a rotation matrix, that the rotation angle is $39^{\circ}$ . Therefore, the area of the ellipse is $\pi\times 0.56419\times 0.56419\approx 1.$ . $\therefore$

The special case solution ought to be accessible to an advanced placement $9^{th}$ or $10^{th}$ (American) grade student. The general solution is second year or later university mathematics for a undergraduate student majoring in mathematics.