<s>[INST] Find the minimum volume of a trirectangular tetrahedron circumscribing a spherical ball Assume the corner of a room with the floor and two walls, all three planes meeting each other at $90^0$.
Say, the point where all three meet is considered the origin O and you have X, Y and Z axes along the intersection of different two planes.
Now you place a spherical ball of radius r touching the floor and both walls. Then you have another plane over the ball touching it and meeting the X, Y and Z axes resp. at point A, B and C where $OA = OB = 4, OC = 8$.
Now, suppose we adjust the lengths of OA, OB and OC in a way that does not affect the size of the inscribed ball and you manage to minimize the volume of the tetrahedron OABC.
Find the minimum volume of the circumscribing trirectangular tetrahedron OABC thus formed.
If this minimum volume is $V_{min} = m+n\sqrt p,$ where m, n and p are positive integers and p is square free, determine $m + n + p$.
Here is how I attempted it -
Say, the center of the ball is point Q. Then,
$O = (0,0,0), Q = (r,r,r), A = (4,0,0), B = (0,4,0), C = (0,0,8)$.
Equation of plane ABC is $\frac{x}{4} + \frac{y}{4} + \frac{z}{8} = 1 ==> 2x+2y+z-8 = 0$.
Distance of Q (r,r,r) from this plane is r as the plane touches the ball.
Therefore, $\frac{|2r+2r+r-8|}{\sqrt{2^2+2^2+1^2}} = r$.
We get $r = 1, 4$. As both point O (0,0,) and Q (r,r,r) are on the same side of the plane ABC, we find substituting these values in the plane equation that r = 1.
Now as we change the lengths of OA, OB and OC, say OA = a, OB = b, OC = c.
$V = \frac {1}{6}abc$ has to be minimized.
Equation of plane ABC = $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 ==> (bc)x+(ac)y+(ab) - abc = 0$
Radius of the inscribed ball is 1, as solved earlier. Now, is there a simple way to say when the volume will be minimum or do I have to go for Lagrange or AM-GM method? Also, finding $m + n + p$ seems confusing.
Please guide me from here or provide rest of the solution.
 [/INST] The intercepts of the plane $\frac xa+\frac yb+\frac zc=1$ must satisfy
\begin{align*}
\frac{\left|\displaystyle\sum_{cyc}\frac 1{a}-1\right|}{\sqrt{\displaystyle\sum_{cyc}\frac 1{a^2}}}&=1\\
\left|\displaystyle\sum_{cyc}\frac 1{a}-1\right|^2&=\displaystyle\sum_{cyc}\frac 1{a^2}\\
\left(\displaystyle\sum_{cyc}\frac 1{a}\right)^2+1-2\displaystyle\sum_{cyc}\frac 1{a}&=\displaystyle\sum_{cyc}\frac 1{a^2}\\
\displaystyle\sum_{cyc}\frac 1{a^2}+2\displaystyle\sum_{cyc}\frac 1{ab}+1-2\displaystyle\sum_{cyc}\frac 1{a}&=\displaystyle\sum_{cyc}\frac 1{a^2}\\
\displaystyle\sum_{cyc}\frac 1{a}-\displaystyle\sum_{cyc}\frac 1{ab}&=\frac12\tag{1}\\
abc\displaystyle\sum_{cyc}\frac 1{a}-\displaystyle\sum_{cyc}a&=\frac{abc}2\\
abc&=\frac{\displaystyle\sum_{cyc}a}{\displaystyle\sum_{cyc}\frac 1{a}-\frac 12}\\
\end{align*}
Minimizing $abc$ is easy since the maximum of the denominator and the minimum of the numerator occur simultaneously at equality $a=b=c$. This can be proved by the $AM\ge HM$ inequality as follows
$$\frac{a+b+c}3\ge \frac3{\frac1a+\frac1b+\frac1c}$$
in which $a+b+c$ is maximum when $\frac1a+\frac1b+\frac1c$ is minimum.
So, putting $a=b=c$ in equation $(1)$, we get
\begin{align*}
\frac3a-\frac3{a^2}&=\frac12\\
a^2-6a+6&=0\\
a&=3+\sqrt3&(\because a>2r=2)\\
\end{align*}
Finally, the minimum volume $\frac{abc}6=\frac{(3+\sqrt3)^3}6=9+5\sqrt3\equiv 17.66\ldots$
</s>