Circumsphere of a tetrahedron

A tetrahedron is a equilateral-triangular pyramid. Let the side length $AB=BC=CD=...=x$ . Then the height of the equilateral triangle $h_\triangle = x \sin 60^\circ = \dfrac {\sqrt 3}2 x$ . The area of the triangle $A_\triangle = \dfrac 12 x^2 \sin 60^\circ = \dfrac {\sqrt 3}4 x^2$ .

The altitude of the tetrahedron $AH = h$ has its foot at the centroid of the base triangle $H$ . Hence $CH = \dfrac 23 \times h_\triangle = \dfrac x{\sqrt 3}$ and $h^2 = x^2 - \left(\dfrac x{\sqrt 3}\right)^2$ , $\implies h = \sqrt{\dfrac 23}x$ . Then the volume of the tetrahedron, $V = \dfrac 13 h A_\triangle = \dfrac 13 \times \sqrt{\dfrac 23}x \times \dfrac {\sqrt 3}4 x^2 = \dfrac {x^3}{6\sqrt 2}$ .

We note that $AO$ is the radius of the sphere which is 1. Let $OH = y$ . Then

$\begin{cases} h = \sqrt{\dfrac 23}x = 1 + y & ...(1) \\ OH^2 + CH^2 = CO^2 \\ y^2 + \dfrac {x^2}3 = 1 \\ \dfrac {x^2}3 = 1-y^2 & ...(2) \end{cases}$

Now squaring equation $(1)$ both sides:

$\begin{aligned} \color{#3D99F6} \frac 23 x^2 & = y^2 + 2y + 1 & \small \color{#3D99F6} \text{Note }(2): \ \frac {x^2}3 = 1 - y^2 \\ 2-2y^2 & = y^2 + 2y + 1 \\ 3y^2 + 2y - 1 & = 0 \\ (3y - 1)(y +1) & = 0 \\ \implies y & = \frac 13 & \small \color{#3D99F6} \text{Note that } y > 0 \end{aligned}$

From $(1)$ , $\sqrt{\dfrac 23}x^2 = 1+\dfrac 13$ , $\implies x = \sqrt {\dfrac 83}$ . Then $V = \dfrac {x^3}{6\sqrt 2} = \dfrac {8\sqrt 8}{3\sqrt 3 \cdot 6 \sqrt 2} = \dfrac 8{9\sqrt 3} = \dfrac {8\sqrt 3}{27}$ . Therefore, $a+b+c = 8+3+27 = \boxed{38}$ .

---

Image credit: Wikipedia

$The~six~sides~S~of~a~regular~tetrahedron~are~face~diagonals~of~the~circumscribing~cube.\\ So~the~sides~of~the~cube~are~~\dfrac S {\sqrt2}.\\ But~both~have~the~same~circumsphere~radius~R.\\ And~R~for~cub~=\frac 1 2*(its~diagonal).\\ \therefore~R=\frac 1 2*\sqrt3* {\dfrac S {\sqrt2}}.\\ But~our~R=1,~~\therefore~1=S*\dfrac{\sqrt3}{2\sqrt2},~~\implies~S=\sqrt{\dfrac 8 3 }=\frac 2 3*\sqrt6.\\ Tetrahedron~volume~,V=\dfrac 1 3*\Big (\dfrac S{\sqrt2} \Big)^3=\dfrac 1 3*\Big(\dfrac 2 3 *\dfrac{\sqrt6}{\sqrt2} \Big )^3.\\ \therefore~V=\dfrac 8 {81}*(\sqrt3)^3=\dfrac{8*\sqrt3}{27}= \dfrac{a*\sqrt b} c.\\ So ~a+b+c=8+3+27=\color{#D61F06}{38}.$