Pyramid in Sphere

Suppose the vertices of the given tetrahedron are $(0, 0, 0) , (a, 0, 0) , (0, b, 0) , (0, 0, c)$ . And suppose the specified point is $(x, y, z)$ , then

$x^2 + y^2 + z^2 = 9 \hspace{12pt} (1)$

$(x - a)^2 + y^2 + z^2 = 5 \hspace{12pt} (2)$

$x^2 + (y - b)^2 + z^2 = 6 \hspace{12pt} (3)$

$x^2 + y^2 + (z - c)^2 = 7 \hspace{12pt} (4)$

Subtracting $(2), (3), (4)$ from $(1)$ yields,

$2 a x - a^2 = 4 \hspace{12pt} (5)$

$2 b y - b^2 = 3 \hspace{12pt} (6)$

$2 c z - c^2 = 2 \hspace{12pt} (7)$

from which,

$(x, y, z) = ( a/2 + 2/a , b/2 + 3/(2b) , c/2 + 1/c ) \hspace{12pt} (8)$

Plugging this into the equation $(1)$

$(a/2 + 2/a)^2 + (b/2 + 3/(2b) )^2 + (c/2 + 1/c)^2 = 9 \hspace{12pt} (9)$

Now from the AM-GM inequality, we know that

$(a/2 + 2/a) \ge 2 , \hspace{12pt} (b/2 + 3/(2b) ) \ge \sqrt{3} , \hspace{12pt} (c/2 + 1/c) \ge \sqrt{2} \hspace{12pt} (10)$

Hence,

$(a/2 + 2/a)^2 + (b/2 + 3/(2b) )^2 + (c/2 + 1/c)^2 \ge 4 + 3 + 2 = 9 \hspace{12pt} (11)$

Hence, for equality to occur, we must have

$(a/2 + 2/a) = 2 , \hspace{12pt} (b/2 + 3/(2b) ) = \sqrt{3} , \hspace{12pt} (c/2 + 1/c) = \sqrt{2} \hspace{12pt} (12)$

Again from AM-GM , equality for each of these three occurs when the two terms are equal, hence, we want,

$a/2 = 2/a \Rightarrow a = 2$

$b/2 = 3/(2b) \Rightarrow b = \sqrt{3}$

$c/2 = 1/c \Rightarrow c = \sqrt{2}$

Now we have the only possible dimensions of the tetrahedron. The center of the circumsphere (also called outsphere) of this special tetrahedron is given by

$O = \dfrac{1}{2} (a, b, c)$

And, hence, the circumradius is $R = \dfrac{1}{2} \sqrt{a^2 + b^2 + c^2 } = \dfrac{1}{2} \sqrt{ 4 + 3 + 2 } = \dfrac{3}{2} = \boxed{1.5}$