 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ia/paper-1/5c/a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

Expanding the squared Euclidean distances gives

∣ x - a ∣^{2} = ∣ x - b ∣^{2} ⟺ 2 (b - a) \cdot x = ∣ b ∣^{2} - ∣ a ∣^{2} .

Thus one may take

n_{A B} = \frac{b - a}{∣ b - a ∣}, p_{A B} = \frac{∣ b ∣ ^{2} - ∣ a ∣ ^{2}}{2∣ b - a ∣} .

The normal is parallel to

A B

, so

L_{A B}

x

lies on both

L_{A B}

and

L_{BC}

, then

∣ x - a ∣ = ∣ x - b ∣ = ∣ x - c ∣

, hence it also lies on

L_{C A}

. Geometrically, the three perpendicular bisectors of the noncollinear triangle concur at its circumcenter.

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