Coordinates of circumcentre of an isosceles triangle in 3D

Question

I have an isosceles triangle in 3D and I need to find the coordinates of the circumcentre of this triangle. I know the coordinates of the three vertices. One method I thought of is to solve equation of sphere using the three vertices. I have to code it, so I would like to know if there is any easier way to do this.

Answer 1

Let $\vec{A} = (A_1, A_2, A_3)$, $\vec{B} = (B_1, B_2, B_3)$, $\vec{C} = (C_1, C_2, C_3)$ be the three vertices of a triangle $ABC$ in any dimension. Let $a, b, c$ be the lengths of corresponding sides, i.e.

$$\begin{cases} a &= |\vec{B} - \vec{C}| = \sqrt{(B_1-C_1)^2+(B_2-C_2)^2+(B_3-C_3)^2}\\ b &= |\vec{C} - \vec{A}| = \sqrt{(C_1-A_1)^2+(C_2-A_2)^2+(C_3-A_3)^2}\\ c &= |\vec{A} - \vec{B}| = \sqrt{(A_1-B_1)^2+(A_2-B_2)^2+(A_3-B_3)^2}\\ \end{cases} $$ Define three constants $\alpha,\beta,\gamma$ by $\displaystyle\; \begin{cases} \alpha &= a^2(b^2+c^2-a^2)\\ \beta &= b^2(c^2+a^2-b^2)\\ \gamma &= c^2(a^2+b^2-c^2) \end{cases} $.

The circumcenter of triangle $ABC$ (in the plane holding the triangle) is located at

$$\frac{\alpha \vec{A} + \beta \vec{B} + \gamma \vec{C}}{\alpha + \beta + \gamma} = \left( \frac{\alpha A_1 + \beta B_1 + \gamma C_1}{\alpha + \beta + \gamma}, \frac{\alpha A_2 + \beta B_2 + \gamma C_2}{\alpha + \beta + \gamma}, \frac{\alpha A_3 + \beta B_3 + \gamma C_3}{\alpha + \beta + \gamma} \right) $$ The $(\alpha,\beta,\gamma)$ above is essentially the barycentric coordinates of the circumcenter with respect to $\triangle ABC$. Since it only depends on the side lengths, above formula works in any dimension.

Answer 2

I'd suggest you avoid the square roots solving a sphere equation will probably imply. Instead, I'd intersect two perpendicular bisector planes of the triangle edges with the plane of the triangle. That way, everything is linear.

Personally I'd perform the computation using projective geometry, but that may be a matter of taste and education, and explaining it here isn't shorter than doing things the conventional way, using the normal representation of each plane and solving a set of three linear equations to intersect them.