Prove that the circumradius of an isosceles tetrahedron (one where opposite sides are equal) with sides $a,b,c$ is given by $R= \sqrt{\frac{1}8(a^2+b^2+c^2)}$.
I know how to show using vectors that the circumcenter of an isosceles tetrahedron is the centroid (i.e. the average of the vertices considered as vectors). Also, I think one approach to this question is to use coordinates. WLOG, let $A=(0,0,0), C=(b,0,0), B=(b_1,b_2,0), D = (d_1,d_2,d_3)$ be the vertices of the tetrahedron, which satisfy the following equations:
$(b_1-b)^2 + b_2^2 = a^2\tag{1}$ $b_1^2+b_2^2=c^2\tag{2}$ $(d_1-b)^2 + d_2^2 + d_3^2 = c^2\tag{3}$ $d_1^2 + d_2^2 + d_3^2 = a^2\tag{4}$ $(d_1 - b_1)^2 + (d_2-b_2)^2 + d_3^2\tag{5}$
One can then simplify the equation $R = \frac{1}4\sqrt{(d_1+b_1+b)^2 +(d_2+b_2)^2 + d_3^2}$, which can be obtained using the fact that the centroid is the circumcenter, but this does not seem like the easiest approach.
Anyways, we have that $R = \frac{1}4\sqrt{a^2 + b^2+c^2 + 2(d_1b_1+b_1b+d_1b+d_2b_2)}$ and $-a^2 - b^2-c^2 = 2(b_1b+d_1b+d_1b_1+d_2b_2)-2(d_1^2 + b_1^2 + b_2^2 + b^2 + d_2^2+d_3^2) = 2(b_1b+d_1+d_1b_1+d_2b_2) - 2(a^2+b^2+c^2)$ from which the formula follows.