The vertices of a tetrahedron lie on a sphere I am struggling a bit with the following (elementary) question:
How to prove that every regular tetrahedron admits a circumsphere, i.e. there exist a sphere on which all four vertices lie.

I would like to find a slick  elegant proof, which is elementary and  "computation-free" as possible.
One possible way is perhaps to prove that all $4$ medians intersect and that the intersection point has equal distance to all vertices. (some triangles should be congruent, I guess).
But I am not sure how to do that.
In particular I prefer a proof that do not mention the notion of inner product, and do not use explicit coordinates of a particular tetrahedron. (of course, if one considers a specific set of vertices in $\mathbb{R}^3$, this can be verified directly. But I want a more conceptual geometric proof).
The best would be an elementary proof you can present to high-school students:)

A more algebraic approach would be to consider  $x_1,x_2,x_3,x_4 \in \mathbb{R}^3$ to be the vertices of a regular tetrahedron, i.e. $|x_i-x_j|$ is constant for $i \neq j$. After translating we may assume that $\sum x_i=0$, and we need to prove that $|x_i|^2$ does not depend on $i$. But I am not sure how to do it.
 A: A coordinate solution.
Let $(a_1,a_2,a_2), (b_1,b_2,b_3), (c_1,c_2, c_3)$, and $(d_1,d_2, d_3)$ be four (non-coplanar) points.  Consider the determinant equation
$$
\left|\begin{array}1
x^2+y^2+z^2 & x & y & z &1
\\
a_1^2 + a_2^2 + a_3^2 & a_1 & a_2 & a_3 & 1
\\
b_1^2 + b_2^2 + b_3^2 & b_1 & b_2 & b_3 & 1
\\
c_1^2 + c_2^2 + c_3^2 & c_1 & c_2 & c_3 & 1
\\
d_1^2 + d_2^2 + d_3^2 & d_1 & d_2 & d_3 & 1
\end{array}\right|
= 0 .
\tag1$$
Note that the four points $(a_1,a_2,a_2), (b_1,b_2,b_3), (c_1,c_2, c_3)$, and $(d_1,d_2, d_3)$ satisfy the equation.  Expanding by the first row, we see that $(1)$ is an equation of the form
$$
A(x^2+y^2+z^2) + Bx + Cy + Dz + E = 0.
$$
Since the four points to not lie in a plane, $A \ne 0$.  So the graph of $(1)$ is: a sphere, or a single point, or empty.  Because the four points satisfy the equation, the only possibility is a sphere.
A: A very simple non-mathematical proof:
How many coordinates are necessary to determine a sphere?
Four points are needed to define a sphere, a tetrahedron has four vertexes.  Thus, by definition a sphere must exist assuming a non-degenerate case.
A: Using plane geometry, you get that the three line segment bisector planes of the triangle $\mathcal T = \{x_1,x_2,x_3\}$ intersect in a line $\mathcal L$ orthogonal to the plane containing $x_1,x_2,x_3$ and passing through the circumcenter of $\mathcal T$.
Now consider the three line segment bisector planes of $[x_1,x_4]$, $[x_2,x_4]$ and $[x_3,x_4]$ and the three respectives intersection points $P_1, P_2, P_3$ with $\mathcal L$. As $ P_1 \in \mathcal L$, $P_1$ is equidistant of $x_1, x_2, x_3$. It is also equidistant of $x_4$. Repeating this observation to $P_2,P_3$, you get that $P_1=P_2=P_3$ and the desired conclusion.
A: We'll show that a tetrahedron $ABCD$ can be inscribed in a sphere. Let $O_1$ and $O_2$ be the circumcenters of $\triangle ABC$ and $\triangle ABD$, respectively. Draw line $\ell_1$ through $O_1$ perpendicular to face $ABC$ and line $\ell_2$ through $O_2$ perpendicular to face $ABD$. Since $O_1A=O_1B=O_1C$, the points on line $\ell_1$ are equidistant from vertices $A,B,C$. Similarly, the points on $\ell_2$ are equidistant from vertices $A,B,D$. So the points on both $\ell_1$ and $\ell_2$ are equidistant from $A$ and $B$, hence lines $\ell_1$ and $\ell_2$ lie in a plane (the perpendicular bisector plane of segment $AB$). Thus, $\ell_1$ and $\ell_2$ are either parallel or they intersect. Suppose they're parallel. Since $\ell_1$ is perpendicular to face $ABC$, $\ell_2$ is also perpendicular to face $ABC$. So we get that faces $ABD$ and $ABC$ are perpendicular to line $\ell_2$, which implies that the two faces are parallel, a contradiction. Therefore, $\ell_1$ and $\ell_2$ intersect. Let $O$ be their intersection point. We have $OA=OB=OC$ since $O$ is on $\ell_1$ and $OA=OB=OD$ since $O$ is on $\ell_2$. Then $OA=OB=OC=OD$ and the tetrahedron can be inscribed in a sphere with center $O$ and radius $OA$.
A: Consider the center of mass of the tetrahedron. Because the tetrahedron is regular, you can't distinguish any vertex from any other - they all look exactly the same and can be carried to each other by an isometry that leaves the tetrahedron unchanged. So in particular, "distance from the center of mass" is not a property you could use to distinguish vertices from one another - they must all have the same distance, and thus lie on a sphere centered at that point.
A: Inscribe the regular tetrahedron in a cube, then circumscribe a sphere on the cube.

A: Here's a sketch of a proof that is visually obvious, if you have the right pictures in your head, but to make it precise it seems like it requires some fancier ideas (continuity and some limiting processes). I'll just share it in case it's useful somehow.
Given 3 non-collinear points in 3-dimensional space, then there is a circle that contains those three points (there is a plane that contains the three points, and inside that plane they are the vertices of a triangle, so we can take its circumcircle).
Given a circle in 3-dimensional space, consider the line that passes through its center and that is orthogonal to the plane that contains the circle. For every point on this line (except the circle's center) there is a sphere that contains the circle and this point. (We can construct it by considering the sphere as a surface of revolution about the line, then reasoning about the 2D cross-section.) Here's an illustration, showing a sphere containing the circle and its two points of intersection with the orthogonal line, and observe that, holding the circle constant, the sphere is purely a function of the intersection point P.

Now, given four points in 3D space that form a tetrahedron (it does not have to be a regular tetrahedron, just non-degenerate: every triple of points is non-collinear and the four points do not lie in a plane), take three of them and form the circle and its orthogonal line as above. The fourth point lies to one side of the plane that runs through the circle -- cutting the line at the center point of the circle, take the ray that is on the same side of the plane as the fourth point. For a point on this ray close enough to the circle's center point, the fourth point is outside the corresponding sphere, and for a point on this ray far enough away from the circle's center point, the fourth point lies inside the corresponding sphere. Therefore, there is some intermediate point such that the fourth point lies on the corresponding sphere. This is the circumsphere of the tetrahedron.
To be able to argue in this way about "close enough to one end of the ray or another," what you need is to see that the limit of the spheres in both cases is exactly the plane that contains the circle, and in either case what's "inside" the "sphere" is an entire half space bounded by the plane.
