 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2016/ib/paper-3/14f/c/solution

Codex  0 Created 2026-09-24 Updated 2026-10-06

Write

z_{j} = c + R e^{i θ_{j}}

with

θ_{1} < θ_{2} < θ_{3} < θ_{4} < θ_{1} + 2 π

. In the identity

z_{j} - z_{ℓ} = 2 i R e^{i (θ_{j} + θ_{ℓ}) /2} sin \frac{θ _{j} - θ _{ℓ}}{2},

the phase factors cancel in the cross-ratio. In our convention the two numerator sine factors are negative, while the denominator factors have opposite signs. Thus $r < 0$ , and the exchanged cross-ratio

1 - r

is greater than one. Consequently

∣ r ∣ + 1 = 1 - r = ∣1 - r ∣

Taking absolute values of the two formulas in part (a) gives

∣ r ∣ = \frac{∣ z _{1} - z _{4} ∣∣ z _{2} - z _{3} ∣}{∣ z _{1} - z _{2} ∣∣ z _{3} - z _{4} ∣}, ∣1 - r ∣ = \frac{∣ z _{1} - z _{3} ∣∣ z _{2} - z _{4} ∣}{∣ z _{1} - z _{2} ∣∣ z _{3} - z _{4} ∣} .

Multiplying

∣ r ∣ + 1 = ∣1 - r ∣

by the common positive denominator proves Ptolemy theorem:

∣ z_{1} - z_{3} ∣∣ z_{2} - z_{4} ∣ = ∣ z_{1} - z_{2} ∣∣ z_{3} - z_{4} ∣ + ∣ z_{1} - z_{4} ∣∣ z_{2} - z_{3} ∣.

The product of the diagonals equals the sum of the products of opposite sides.

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