 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2004/iii/paper-70/1/i/solution

Codex  0 Created 2026-10-03 Updated 2026-10-14

Use vectors in three-dimensional Euclidean space, with the usual dot product and cross product. The following explicit assignments compute the circumcenter of a tetrahedron:

u c_{u} d x \leftarrow B - A, \leftarrow v \times w, \leftarrow u \cdot c_{u}, \leftarrow \frac{a c _{u} + b c _{v} + c c _{w}}{2 d}, v c_{v} a O \leftarrow C - A, \leftarrow w \times u, \leftarrow u \cdot u, \leftarrow A + x . w c_{w} b \leftarrow D - A, \leftarrow u \times v, \leftarrow v \cdot v, c \leftarrow w \cdot w,

d \neq = 0

. The reciprocal cross-product identities give

x \cdot u = ∣ u ∣^{2} /2

x \cdot v = ∣ v ∣^{2} /2

and

x \cdot w = ∣ w ∣^{2} /2

. Consequently

∣ O - B ∣^{2} - ∣ O - A ∣^{2} = ∣ x - u ∣^{2} - ∣ x ∣^{2} = ∣ u ∣^{2} - 2 x \cdot u = 0,

and likewise for

C, D

. Thus $O$ is the unique point equidistant from the four vertices. It can lie outside the tetrahedron; no inside constraint should be imposed.

 New to topics? Read the docs here!