 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ia/paper-1/5c/c/solution

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

a, b, c, d

. Suppose

(b - a) \cdot (d - c) = 0, (c - a) \cdot (d - b) = 0.

Subtracting the second scalar product from the first gives

- (d - a) \cdot (c - b) = 0,

Put

S_{1} = ∣ b - a ∣^{2} + ∣ d - c ∣^{2}, S_{2} = ∣ c - a ∣^{2} + ∣ d - b ∣^{2}, S_{3} = ∣ d - a ∣^{2} + ∣ c - b ∣^{2} .

Direct expansion gives, cyclically,

S_{1} - S_{2} = - 2 (d - a) \cdot (c - b),

with the other two differences equal to minus twice the scalar products of the other opposite-edge pairs. All three scalar products vanish, hence

S_{1} = S_{2} = S_{3} .

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