Determine the inscribed ellipsoid within a cube with given ratios of axes

Question

Given a cube centered at the origin, with side length $2a$, determine the length of the semi-axes of the ellipsoid inscribed in the cube, touching all its $6$ faces, such that the semi-axes lengths are of the form $ b , b , \dfrac{b}{2} $. Determine $b$ in terms of $a$.

My attempt:

The ellipsoid can be oriented, with its shortest semi-axis along one of the four major diagonals of the cube. Let pick one such diagonal to be the diagonal extending from $(-a, -a, -a)$ to $(a,a,a)$. Then the rotation matrix giving the orientation of the ellipsoid, is found by attaching a reference frame to the ellipsoid, with its $z'$ axis along the shortest semi-axis (although other choices are possible), and then the rotation matrix can be chosen as

$ R = [ r_1, r_2, r_3 ] $

where

$ r_3 = [ \sin \theta \cos \phi , \sin \theta \sin \phi, \cos \theta ] $

with $\theta = \cos^{-1} \left( \dfrac{1}{\sqrt{3}} \right) = \tan^{-1}( \sqrt{2} ) $

And $ \phi = \dfrac{\pi}{4} $

And we can then choose $r_1, r_2$ to be

$ r_1 = [ \cos \theta \cos \phi , \cos \theta \sin \phi, - \sin \theta ]^T $

$ r_2 = [- \sin \phi, \cos \phi , 0 ]^T $

The equation of the ellipsoid will be

$ r^T R D R^T r = 1 $

where $ r = [x,y,z]^T$, and $R$ is the rotation matrix given above, and

$ D = \begin{bmatrix} \dfrac{1}{b^2} && 0 && 0 \\ 0 && \dfrac{1}{b^2} && 0 \\ 0 && 0 && \dfrac{4}{b^2} \end{bmatrix} $

Let $A$ be the tangency point, between the ellipsoid and the cube, at the plane $x = -a$, then we can write using the expression for the gradient of the ellipsoid,

$ R D R^T A = - k \mathbf{e_1} $

where $\mathbf{e_1}$ is the unit vector along the positive $x$ axis, i.e. $ \mathbf{e_1} = [1, 0, 0]^T $

This implies that

$ A = - k R D^{-1} R^T \mathbf{e_1} $ where $ k \gt 0 $

Since $A$ is on the ellipsoid, then substituting this expression into the equation of the ellipsoid, gives us

$ k = \dfrac{1}{\sqrt{ \mathbf{e_1}^T R D^{-1} R^T \mathbf{e_1} } } $

Furthermore, if we now pre-multiply the above equation by $\mathbf{e_1}^T $, we get

$ \mathbf{e_1}^T A = - a = -\sqrt{ \mathbf{e_1}^T R D^{-1} R^T \mathbf{e_1} } $

So that

$ \mathbf{e_1}^T R D^{-1} R^T \mathbf{e_1} = a^2 $

To evaluate the left hand side, first note that

$ R^T \mathbf{e_1} = \begin{bmatrix} \cos \theta \cos \phi \\ - \sin \phi \\ \sin \theta \cos \phi \end{bmatrix} = \begin{bmatrix} \dfrac{1}{\sqrt{6}} \\ - \dfrac{1}{\sqrt{2}} \\ \dfrac{1}{\sqrt{3}} \end{bmatrix} $

Therefore,

$ \mathbf{e_1}^T R D^{-1} R^T \mathbf{e_1} = \dfrac{b^2}{6} + \dfrac{b^2}{2} + \dfrac{b^2}{12} = \dfrac{3}{4} b^2 $

Hence, finally, we have

$ \dfrac{3}{4} b^2 = a^2 $

Giving,

$ b = \dfrac{2}{\sqrt{3}} a $

And my question now, is this analysis sound? and is the final result correct?

Your verification of my attempt, comments, and alternative answers are appreciated.

Answer 1

A completely different approach, less straightforward, but worth to be presented because it uses techniques that can be interesting in other circumstances.

Fig. 1 : The ellipsoid and its tangent cone.

Let us consider the ellipsoid :

$$x^2+y^2+4z^2=b^2$$

and the point $X_1(0,0,a \sqrt{3})$ (a vertex of the cube).

The equation of the right circular cone with apex $X_1$ tangent to the ellipsoid, is (using a classical formulation : see here for example )

$$(X^TAX_1)^2-(X^TAX)(X_1^TAX_1)=0\tag{1}$$

where (we use homogeneous coordinates with fourth coordinates equal to $1$) :

$$A=\pmatrix{1&&&\\&1&&&\\&&4&\\&&&-b^2}$$

(1) can be expanded in this way :

$$(4a\sqrt{3}z-b^2)^2-(x^2+y^2+4z^2-b^2)(12a^2-b^2)=0\tag{2}$$

Besides, a small calculation shows that a right circular cone having 3 mutually orthogonal generatrices (see the answer by Angina Seng to this question) has an angle (sometimes called "half angle") $\alpha$ such that $\cos \alpha = \frac23$.

The equation of this cone can be obtained by identifying the two ways to express a dot product :

$$(0 \ \ 0 \ \ -1)\pmatrix{x\\y\\z-a \sqrt{3}}=1 . \sqrt{x^2+y^2+(z-a \sqrt{3})^2} \cos \alpha \tag{3}$$

finaly transformed by squaring into :

$$x^2+y^2-\frac12 (z-a \sqrt{3})^2=0\tag{4}$$

As (2) and (4) must be equivalent expressions, with the help of a C.A.S., it is not difficult to see that a necessary and sufficient condition is that :

$$3b^2=4a^2$$

i.e., indeed : $b=\frac{2a}{\sqrt{3}}$.

Answer 2

Here's a more geometric approach. Let $ABCDEFGH$ be a cube of edge $2$, with $A=(0,0,0)$, its opposite vertex $G=2\sqrt3(1,1,1)$ and center $O=\sqrt3(1,1,1)$.

If $AG$ is the rotational axis of the ellipsoid, then the ellipsoid is tangent to the three diagonals of the faces from $A$ and to the three diagonals of the faces from $G$. Let $P=(0,t,t)$ be the point of tangency on $AH$, with parameter $t$ to be determined.

If we take any plane containing $AG$, such as $ABGH$, the intersection of this plane with the ellipsoid is an ellipse whose semiaxes $OV$ and $OU$ have the same length of the semiaxes of the ellipsoid. We must then find $t$ such that $OU=2OV$.

To compute $OV$ we must find first of all the reflection of $P$ about line AG: $$P'={t\over3}(4,1,1),$$ and the midpoint of $PP'$: $$ M={2t\over3}(1,1,1). $$ Semiaxis $OV$ is given then by the usual formula: $$ OV=\sqrt{OM\cdot OA}=\sqrt{3-2t}. $$ To find the other semiaxis we can find the reflection of $P'$ about $O$: $$ Q'=\left(2-{4t\over3},2-{t\over3},2-{t\over3}\right), $$ the midpoint of $PQ'$: $$ N=\left(1-{2t\over3},1+{t\over3},1+{t\over3}\right) $$ and the intersection of $ON$ with $AH$, which is the intersection of the tangents at $P$ and $Q'$: $$ K=\left(0,{3\over2},{3\over2}\right). $$ As before: $$ OU=\sqrt{OK\cdot ON}=\sqrt{t}. $$ The request $OU=2OV$ gives then: $$ t={4\over3}, \quad\text{whence:}\quad OU={2\over\sqrt3},\quad OV={1\over\sqrt3}. $$

Answer 3

Any conic curve has a so-called orthoptic curve (set of points from which one "sees" the curve under a right angle, more formally from which can be issued a system of right angle axes tangent to the given curve).

One dimension above : Any quadric curve has an associated orthoptic surface (i.e., set of points from which is issued a system of 3 orthogonal axes). In the case of an ellipsoid, it is a sphere whose radius is known : see Theorem 4.1 here.

In the case of our ellipsoid with equation :

$$\frac{x^2}{b^2}+\frac{y^2}{b^2}+\frac{z^2}{(b/2)^2}=1,$$

this theorem gives the following radius for this sphere :

$$\sqrt{b^2+b^2+(b/2)^2}=\frac{3b}{2}$$

to be identified with $a \sqrt{3}$ (half the longest diagonal of a cube with side $2a$), confirming your result.

Remark : I just saw the remark of @intelligenci pauca refering to a previous question of yours with a common result with Theorem 4.1 above.