 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2002/ia/paper-3/8b/a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

The three known vertices have pairwise squared distances

8

, so the regular tetrahedron must have every squared edge length equal to

8

. Comparing the squared distance from

d

b

with that from

d

a

gives

∣ d - b ∣^{2} - ∣ d - a ∣^{2} = 4 + (x + 1)^{2} - (x - 1)^{2} = 4 + 4 x .

Equality forces

x = - 1

. Similarly,

∣ d - c ∣^{2} - ∣ d - a ∣^{2} = 4 + (y + 1)^{2} - (y - 1)^{2} = 4 + 4 y,

y = - 1

. The resulting point has squared distance

8

from each of the other three vertices. Therefore

x = y = - 1, d = (1, - 1, - 1) .

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