Finding the minimum volume of a Tetrahedron.

Question

Suppose you have the surface $\xi$ defined in $\mathbb{R}^3$ by the equation: $$ \xi :\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1 $$ For $ x \geq 0$ , $ y \geq 0$ and $ z \geq 0$. Now take any point $P \in \xi$ and consider the tangent plane ($\pi_t)$ to $\xi$ at $P$. Calculate the minimum volume of the region determined by the $xy$, $yz$, $xz$ planes and $\pi_t$.

Answer 1

Assume for the moment $a=b=c=1$. The surface $\xi$ then is the unit sphere $S^2$. A typical point ${\bf n}$ of $S^2$ in the first octant then has coordinates $(n_1,n_2,n_3)$ with $n_i>0$ and $\sum_i n_i^2=1$. Furthermore the tangent plane to $S^2$ at ${\bf n}$ is given by the equation ${\bf n}\cdot{\bf x}=1$, i.e., $n_1x_1+n_2x_2+n_3x_3=1$. Intersecting this plane with the three coordinate axes gives the points $\bigl({1\over n_1},0,0\bigr)$, $\bigl(0,{1\over n_2},0\bigr)$, $\bigl(0,0,{1\over n_3}\bigr)$. It follows that the simplex $S$ in question has volume $${\rm vol}(S)={1\over 6 n_1n_2n_3}\ .$$ This volume is minimal when its reciprocal is maximal. By the AMG inequality $$\root3\of {n_1^2 n_2^2n_3^2}\leq{1\over3}(n_1^2+n_2^2+n_3^2)={1\over3}\ ,$$ with equality iff $n_i=3^{-1/2}$ for all $i\in[3]$. It follows that $${\rm vol}(S)\geq{1\over6} 3^{3/2}={\sqrt{3}\over2}\ .$$ In the case of arbitrary $a$, $b$, $c>0$ we therefore have $${\rm vol}(S)\geq{\sqrt{3}\over2}abc\ ,$$ by standard geometric principles.

Answer 2

Calling

$$ \begin{cases} p_0 = (x_0,y_0,z_0)\\ p_1 = (x_1,y_1,z_1)\\ \Lambda = \mbox{diag}[\frac{1}{a^2},\frac{1}{b^2},\frac{1}{c^2}]\\ \vec n_0 = 2\Lambda\cdot p_0\\ f(p_1^*) = \frac 16 x_1^* y_1^* z_1^*\\ x_1^* = \frac{a^2}{x_0}\\ y_1^* = \frac{b^2}{y_0}\\ z_1^* = \frac{c^2}{z_0}\\ C(p_0) = p_0\cdot\Lambda\cdot p_0-1 \end{cases} $$

Here $(x_1^*, y_1^*,z_1^*)$ are the intersections of the plane $(p_1-p_0)\cdot \vec n_0$ with the axis $(x_1,y_1,z_1)$ so the problem can be stated as

$$ \min f(p_1^*)\ \ \mbox{s.t.}\ \ (p_1-p_0)\cdot \vec n_0 = 0, \ C(p_0) = 0 $$

The Lagrangian is

$$ L(p_0,p_1,\lambda,\mu) = f(p_1^*)+\lambda(p_1-p_0)\cdot \vec n_0+\mu (C(p_0)-1) $$

and the stationary points are determined by solving

$$ \nabla L = 0 $$

or

$$ \left\{ \begin{array}{rcl} \frac{2 \mu x_0+\lambda (x_1-2 x_0)}{a^2}-\frac{a^2 b^2 c^2}{6 x_0^2 y_0 z_0} &=&0\\ \frac{2 \mu y_0+\lambda (y_1-2 y_0)}{b^2}-\frac{a^2 b^2 c^2}{6 x_0 y_0^2 z_0} &=&0\\ \frac{2 \mu z_0+\lambda (z_1-2 z_0)}{c^2}-\frac{a^2 b^2 c^2}{6 x_0 y_0 z_0^2} &=&0\\ \frac{\lambda x_0}{a}&=&0 \\ \frac{\lambda y_0}{b}&=&0 \\ \frac{\lambda z_0}{c}&=&0 \\ \frac{x_0 (x_1-x_0)}{a^2}+\frac{y_0 (y_1-y_0)}{b^2}+\frac{z_0 (z_1-z_0)}{c^2}&=&0 \\ \frac{x_0^2}{a^2}+\frac{y_0^2}{b^2}+\frac{z_0^2}{c^2}&=&1 \\ \end{array} \right. $$

giving

$$ \begin{array}{cccc} x_0&y_0&z_0& f(p_1^*)\\ \frac{a}{\sqrt{3}}&\frac{b}{\sqrt{3}}&\frac{c}{\sqrt{3}}&\frac{1}{2} \sqrt{3} a b c \end{array} $$

Answer 3

Consider a point $P(x_0,y_0,z_0)\in \Bbb R_{>0}^3$ on the given ellipsoid $(E)$ with equation $$ \frac {x^2}{a^2} + \frac {y^2}{b^2} + \frac {z^2}{c^2} =1\ . $$ Then the plane tangent in $P$ to $(E)$ is given by the linear part of the Taylor polynomial around $(x_0,y_o,z_0)$ of first order of the polynomial of second degree in the above equation, equated to zero. To isolate this polynomial of degree one, write $$ \begin{aligned} &\frac {x^2}{a^2} + \frac {y^2}{b^2} + \frac {z^2}{c^2} -1 \\ &\qquad= \frac 1{a^2} ((x-x_0)+x_0)^2+ \frac 1{b^2} ((y-y_0)+y_0)^2+ \frac 1{c^2} ((z-z_0)+z_0)^2 -1 \\ &\qquad= \underbrace{ \left( \frac 1{a^2} x_0^2+ \frac 1{b^2} y_0^2+ \frac 1{c^2} z_0^2 -1 \right)}_{=0} \\ &\qquad\qquad+ \frac 2{a^2} x_0(x-x_0)+ \frac 2{b^2} y_0(y-y_0)+ \frac 2{c^2} z_0(z-z_0) \\ &\qquad\qquad\qquad\qquad+ \text{higher order terms} \\ &\qquad\qquad\qquad\qquad \text{ containing factors $(x-x_0)^2$, and $(y-y_0)^2$, and $(z-z_0)^2$ .} \end{aligned} $$ So the equation of the tangent plane in $P$ to $(E)$ is (given by "dedoubling"): $$ \frac 1{a^2} x_0(x-x_0)+ \frac 1{b^2} y_0(y-y_0)+ \frac 1{c^2} z_0(z-z_0)=0\ . $$ Simpler maybe: $$ \frac 1{a^2} x_0x+ \frac 1{b^2} y_0y+ \frac 1{c^2} z_0z =1\ . $$ This plane hits the $Ox$ axis in the point with $y=z=0$ and $x=a^2/x_0$. Similar formulas for the intersections with the other axes.

So we have to minimize the volume: $$ V =\frac 16\frac{a^2\;b^2\;c^2}{x_0\;y_0\;z_0}\ . $$ For this we have to maximize the product $x_0y_0z_0$. The inequality between the arithmetic and geometric mean solves the problem. $$ 1 = \frac {x_0^2}{a^2} + \frac {y_0^2}{b^2} + \frac {z_0^2}{c^2} \ge 3\left(\frac {x_0^2y_0^2z_0^2}{a^2b^2c^2}\right)^{1/3} = 3\left(\frac {x_0y_0z_0}{abc}\right)^{2/3}\ , $$ so $$ x_0y_0z_0\le 3^{-3/2}abc\ . $$ The quality holds for the point $P^*\left(\frac a{\sqrt 3},\frac b{\sqrt 3},\frac c{\sqrt 3}\right)$. We get the minimal volume: $$ V^* = \frac 16\frac{a^2b^2c^2}{3^{-3/2}abc}= \frac {\sqrt 3}2\; abc\ . $$