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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-2/26i/b/ii/solution

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Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 2 / 26I / b / ii / Solution by

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

The original perimeter is

P (Ω) = \int_{x_{0}}^{x_{1}} (1 + u_{+}^{'} (x)^{2} + 1 + u_{-}^{'} (x)^{2}) d x .

The two boundary graphs of the symmetrized domain are

\pm h /2

, so

P (S_{L} (Ω)) = \int_{x_{0}}^{x_{1}} 4 + (u_{+}^{'} - u_{-}^{'})^{2} d x .

Apply the triangle inequality in

R^{2}

to the vectors

(1, u_{+}^{'}), (1, - u_{-}^{'}) .

It gives pointwise

1 + u_{+}^{'}^{2} + 1 + u_{-}^{'}^{2} \geq 4 + (u_{+}^{'} - u_{-}^{'})^{2} .

After integration,

P (S_{L} (Ω)) \leq P (Ω) .

Equality in the Euclidean triangle inequality holds exactly when the two displayed vectors are nonnegative scalar multiples. Their first coordinates are both one, so this means

u_{+}^{'} = - u_{-}^{'} .

Equality of perimeters therefore holds exactly when

u_{+} (x) + u_{-} (x) = c

is constant. In that case the midpoint of every vertical chord lies on

y = c /2

, and

Ω

has the horizontal line

y = c /2

, parallel to

L

, as an axis of symmetry. The converse is immediate.

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