Steiner Ellipse, Minimal Area Through Three Points

The Steiner ellipse has the minimal area surrounding a triangle. It is characterized by havingits centercoincident with thetriangle's centroid.

The Steiner ellipse can be extended to higher dimensions with one more point than the dimension. It is useful in a number of fields, such as statistics for determining which data points are outliers.

For a given chord or triangle base, the maximal area triangle occurs when the third vertex of the triangle is at the intersection of the chord's perpendicular bisector. Since that needs to be true for all three sides of the triangle, the triangle must be equilateral.

Using a calculus-based approach,

$\text{Maximize}\left[\left\{\text{Area}\left[\text{Triangle}\left[\left( \begin{array}{cc} 1 & 0 \\ \cos (2 \pi \theta ) & \sin (2 \pi \theta ) \\ \cos (2 \pi \phi ) & \sin (2 \pi \phi ) \\ \end{array} \right)\right]\right],0\leq \theta \leq 1\land 0\leq \phi \leq 1\right\},\{\theta ,\phi \}\right] \Rightarrow$ $\left\{\frac{3 \sqrt{3}}{4},\left\{\theta \to \frac{2}{3},\phi \to \frac{1}{3}\right\}\right\}.$

The area within the circle is $\pi$. The area within the triangle is $\frac{3 \sqrt{3}}{4}$. The ratio of the areas is $\frac{4 \pi }{3 \sqrt{3}}$or about $2.41839915231229$.

Affine transformationspreserve length ratios and linearities. This means they also preserve area ratios as those are just lengths squared, volume ratios in the 3-dimensional case as those are just lengths cubed, etc.

Affine transformations can be described by linear equations and therefore by matrix algebra. These matrix algebra operations can be eased byaugmenting the matrices. Using $n$为new, $o$为old, and $a$为the augmented affine transformation matrix, $o$and $n$像

$\left( \begin{array}{ccc} x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3 \\ 1 & 1 & 1 \\ \end{array} \right).$

The augmented affine transformation matrix $a$is

$\left( \begin{array}{ccc} a_{1,1} & a_{1,2}& b_1 \\ a_{2,1} & a_{2,2} & b_2 \\ 0 & 0 & 1 \\ \end{array} \right).$

The augmented affine transformation matrix equation is $n=a\cdot o$, where the dot $\cdot$ismatrix multiplication.

Using this problemIs it a Circle?as an example, solve the following problem:

Find the forward augmented affine transformation matrix $\bm{a}$of $\bm{o}$to $\bm{n}$, from the problem's triangle to the equilateral triangle within the unit circle.

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The $\bm{a}$matrix is as above.

The $\bm{o}$matrix is $\left( \begin{array}{ccc} -2 & 0 & 2 \\ 2 & -1 & 0 \\ 1 & 1 & 1 \\ \end{array} \right).$

The $\bm{n}$matrix is $\left( \begin{array}{ccc} 1 & -\frac{1}{2} & -\frac{1}{2} \\ 0 & \frac{\sqrt{3}}{2} & -\frac{\sqrt{3}}{2} \\ 1 & 1 & 1 \\ \end{array} \right).$

The matrix equation is $\bm{a}\cdot\bm{o}=\bm{n}.$

Invertthe $\bm{o}$matrix. If the matrix is singular, i.e. is not invertible, then the triangle is collinear and has a zero area.

Rightmatrix multiplyboth sides of the matrix equation above by the inverse of $\bm{o}$above:

$\bm{a}\cdot\bm{o}\cdot\bm{o}^{-1}=\bm{n}\cdot\bm{o}^{-1} \implies \bm{a} = \bm{n}\cdot\bm{o}^{-1},$

which gives $\bm{a}$as follows:

$\left( \begin{array}{ccc} -2 & -\frac{2}{\sqrt{3}} & 0 \\ \frac{5}{3} & -\frac{1}{\sqrt{3}} & \frac{1}{3} \\ 0 & 0 & 1 \\ \end{array} \right).$

Byinvertingthat transformation matrix, one may go in the opposite direction.

The inverse of $\bm{a}$is $\left( \begin{array}{ccc} -\frac{3}{16} & \frac{3}{8} & -\frac{1}{8} \\ -\frac{1}{16} \left(5 \sqrt{3}\right) & -\frac{1}{8} \left(3 \sqrt{3}\right) & \frac{\sqrt{3}}{8} \\ 0 & 0 & 1 \\ \end{array} \right).$

One method follows immediately. The other follows after the first method.

Unfortunately, which diameters of the unit circle map to the major and minor axes of the ellipse are still unknown.

我们可以利用一对共轭semidi吗ameters. One end of the semidiameters is the shared center of the triangle and of the Steiner ellipse. The centroid of the triangle is easily computed as it is the average or mean of the triangle's vertices' coordinates. A pair of conjugate semidiameters are orthogonal radii of the unit circle which when reverse affine transformed to the original environment become a set of conjugate semidiameters. One member of a set of conjugate semidiameters is readily at hand as one of the triangle's vertices was mapped to $x$-axis unit vector position, $(1,0)$. The other member of that set of conjugate semidiameters is not much harder as the base of the equilateral triangle is parallel to the $y$-axis and its length is $\sqrt{3}$in the unit circle environment. Therefore, another member of a conjugate semidiameter pair is one half of the Euclidean distance between the base vertices divided by $\sqrt{3}$in the original environment projected from the triangle's centroid parallel to the triangle's base, in either direction from the centroid.

一个方法getting the major and minor axes and then the equation of the Steiner ellipse is a method calledRytz的建设.

Here is the original triangle:

Here is the same triangle decorated with the circle projected into the original environment, the solution ellipse which is the orange ellipse which surrounds the projected circle (they are the same), the two conjugate semidiameters in red and the parallelogram which is tangent to the ellipse where the conjugate semidiameters touch the ellipse, the corners of which are projections of the $(1,1),\,(1,-1),\,(-1,-1),\,(-1,1)$points in the unit circle environment.

The computations are done more easily with the ellipse center and triangle centroid at the origin.

Sorting the conjugate semidiameters by their distance from the origin gives $\left(\frac{2}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)$and $\left(-2,\frac{5}{3}\right)$.

Rotate the shorter conjugate semidiameter by $90^o$in the direction that puts the end closer to the end of the longer conjugate semidiameter to get $\left(-\frac{1}{\sqrt{3}},\frac{2}{\sqrt{3}}\right)$and call that point $P'.$

调用之间找到平衡 $P'$and the end of the longer conjugate semidiameter $D:$ $\bigg(\frac{1}{2} \left(-2-\frac{1}{\sqrt{3}}\right),\frac{1}{2} \left(\frac{5}{3}+\frac{2}{\sqrt{3}}\right)\bigg).$

Intersect an infinite line passing through $P'$and the end of the longer conjugate semidiameter $D$and a circle centered at $D$and radius such that it passes through the origin (the green circle). $a=\Big(\frac{2}{39} \left(4 \sqrt{3}+3\right),\frac{1}{13} \left(4 \sqrt{3}+3\right)\Big)$and $b=\Big(\frac{1}{13} (-7) \left(\sqrt{3}+4\right),\frac{14}{39} \left(\sqrt{3}+4\right)\Big).$

Intersect an infinite half-line starting at the origin and in the direction of point $a$immediately above with a (red) circle centered at the origin and of a radius of the Euclidean distance between the end of the longer conjugate semidiameter and point $b$immediately above. That is the end of the ellipse's semi-minor axis: $\left(\frac{4}{\sqrt{39}},2 \sqrt{\frac{3}{13}}\right).$

Intersect an infinite half-line starting at the origin and in the direction of point $b$immediately above with a (blue) circle centered at the origin and of a radius of the Euclidean distance between the end of the longer conjugate semidiameter and point $a$immediately above. That is the end of the ellipse's semi-minor axis: $\left(-\frac{8}{\sqrt{13}},\frac{16}{3 \sqrt{13}}\right).$

Therefore, the ellipse will be tangent to the red and blue circles mentioned in the previous two paragraphs.

The general equation for an ellipse centered at the origin is $\text{ga}\,\text{xs}^2+\text{gb}\,\text{xs}\,\text{ys}+\text{gc}\,\text{ys}^2=1$, using two character variable names.

Using the coordinates of the ends of the semi-major and semi-minor axes and one of the triangles' vertices, three equations in three unknowns can be written: $\frac{64 \text{ga}}{13}-\frac{128 \text{gb}}{39}+\frac{256 \text{gc}}{117}=1\land \frac{16 \text{ga}}{39}+\frac{8 \text{gb}}{13}+\frac{12 \text{gc}}{13}=1\land \frac{4 \text{ga}}{3}+\frac{2 \text{gb}}{3}+\frac{\text{gc}}{3}=1$. Solving those equations gives $\text{ga}\to \frac{21}{64},\text{gb}\to \frac{9}{16},\text{gc}\to \frac{9}{16}$. This results in the equation: $21\, x^2+36\, x\, y+36 y^2=64$. Moving the center from the origin to its proper location gives the final ellipse equation: $7\,x^2+12\, x\,y+12 y^2=4 (x+2 y+5)$.

A copy of the inverse of the transformation matrix $\bm{a}$, which is called $\bm{b},$is

$\left( \begin{array}{cc|c} -2 & -\frac{2}{\sqrt{3}} & 0 \\ \frac{5}{3} & -\frac{1}{\sqrt{3}} & \frac{1}{3} \\ \hline 0 & 0 & 1 \\ \end{array} \right).$

The first two columns and first two rows areaffine matrix(the return rotation matrix).

The third column first two rows of $\bm{b}$is the translation from the origin of the center of the ellipse.

The parametric form of the ellipse equation is $\left\{-\frac{2 \sin (\theta )}{\sqrt{3}}-2 \cos (\theta ),-\frac{\sin (\theta )}{\sqrt{3}}+\frac{5 \cos (\theta )}{3}+\frac{1}{3}\right\}$.

ThesIngular-value decomposition, also onWikipediaandWolfram MathWorld, of the first two columns and the first two rows, a square 2 $\times$2 matrix, is

$\begin{aligned} \bm{u}&=\left( \begin{array}{cc} -\frac{3}{\sqrt{13}} & -\frac{2}{\sqrt{13}} \\ \frac{2}{\sqrt{13}} & -\frac{3}{\sqrt{13}} \\ \end{array} \right)\\ \bm{\sigma}&=\left( \begin{array}{cc} \frac{8}{3} & 0 \\ 0 & \frac{2}{\sqrt{3}} \\ \end{array} \right)\\ \bm{v}&=\left( \begin{array}{cc} \frac{7}{2 \sqrt{13}} & -\frac{\sqrt{\frac{3}{13}}}{2} \\ \frac{\sqrt{\frac{3}{13}}}{2} & \frac{7}{2 \sqrt{13}} \\ \end{array} \right). \end{aligned}$

These three matrices are

• $\bm{u}$, the rotation of the ellipse into its final orientation
• $\bm{\sigma}$, the scaling that establishes the lengths of the semi-major and semi-minor axes, respectively
• $\bm{v}$, the rotation of what will become the semi-major and semi-minor axes onto the $x$- and $y$-axes, respectively.

Map $x$and $y$onto the unit circle coordinate system using the $\bm{a}$affine transform:

$\left\{-\frac{3 x}{16}+\frac{3 y}{8}-\frac{1}{8},-\frac{5 \sqrt{3} x}{16}-\frac{1}{8} 3 \sqrt{3} y+\frac{\sqrt{3}}{8}\right\}.$

Substitute those transformed $x$and $y$definitions into the equation of the unit circle $x^2+y^2=1$:

$7 x^2+12 x y+12 y^2=20+4 x+8 y.$

Create an affine transformation from $\bm{v}$used to transform the unit vectors and their negatives into the unit circle coordinate system to locate the ends of the ellipse's axes:

$\left(\bm{t}= \begin{array}{cc|c} \frac{7}{2 \sqrt{13}} & -\frac{\sqrt{\frac{3}{13}}}{2} & 0 \\ \frac{\sqrt{\frac{3}{13}}}{2} & \frac{7}{2 \sqrt{13}} & 0 \\ \hline 0 & 0 & 1 \\ \end{array} \right).$

Compose $\bm{b}$and $\bm{t}$. Note, the application order is $\bm{t}$and then $\bm{b}$:

$\left( \begin{array}{cc|c} -\frac{8}{\sqrt{13}} & -\frac{4}{\sqrt{39}} & 0 \\ \frac{16}{3 \sqrt{13}} & -2 \sqrt{\frac{3}{13}} & \frac{1}{3} \\ \hline 0 & 0 & 1 \\ \end{array} \right)$

Transforming {1, 0}, {-1, 0}, {0, 1}, {0, -1} using the transformation immediately above gives

$\left\{-\frac{8}{\sqrt{13}},\frac{1}{3}+\frac{16}{3 \sqrt{13}}\right\},\left\{\frac{8}{\sqrt{13}},\frac{1}{3}-\frac{16}{3 \sqrt{13}}\right\},\left\{-\frac{4}{\sqrt{39}},\frac{1}{3}-2 \sqrt{\frac{3}{13}}\right\},\left\{\frac{4}{\sqrt{39}},2 \sqrt{\frac{3}{13}}+\frac{1}{3}\right\}.$

The black dots are the original points, the ellipse is plotted from the equation provided above in upper and lower sections, and the red lines are the semi-major and semi-minor axes of the ellipse:

This method readily can be extended to higher dimensions.

The following two animated GIFs illustrate the same ellipsoid. The difference is what is kept constant.

In this clip, the visual image size is kept constant and the axes are allowed to change their length to do that:

In this clip, the axes are kept constant size and the visual image size changes to do that:

Note well that the ratio of the ellipsoid volume divided by the tetrahedron volume remains constant: $\frac{3 \sqrt{3} \pi }{2} \approx 8.16209713905398.$