 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ia/paper-1/5b/b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Because the sphere passes through the origin and has centre

P

, its radius is

∣ P ∣

. A point

X

lies on it exactly when

∣ X - P ∣^{2} = ∣ P ∣^{2},

or equivalently

2 P \cdot X = ∣ X ∣^{2} .

With

P = α A + βB + γ C

, the three required equations are the Gram matrix system

A \cdot A B \cdot A C \cdot A A \cdot B B \cdot B C \cdot B A \cdot C B \cdot C C \cdot C α β γ = \frac{1}{2} ∣ A ∣^{2} ∣ B ∣^{2} ∣ C ∣^{2} .

For the specified vectors this becomes

α + β = \frac{1}{2}, α + 2 β + γ = 1, β + 5 γ = \frac{5}{2} .

Thus

α = \frac{1}{2}, β = 0, γ = \frac{1}{2},

and the centre is

P = \frac{1}{2} A + \frac{1}{2} C = (\frac{1}{2}, \frac{1}{2}, 1) .

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