Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 2 / 26J / f

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- Solution f

Solution

The preceding parts show that an area maximizer among curves of fixed length is a plane section of the sphere, hence a circle. If its smaller cap has angular radius

0 \leq θ \leq π /2

, then

L = 2 π sin θ, A = 2 π (1 - cos θ) .

For this circle,

A (4 π - A) = 4 π^{2} (1 - cos^{2} θ) = 4 π^{2} sin^{2} θ = L^{2} .

No curve of the same length encloses more area than the maximizing circle. Equivalently every curve enclosing the smaller area

A

satisfies the spherical isoperimetric inequality

L^{2} \geq A (4 π - A),

with equality precisely for circles.

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