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

PutVertical slicing and Fubini's theorem giveThe vertical slice of $S_L(\Omega )$ is $(-h(x)/2,h(x)/2)$ and has the same length $h(x)$. Therefore

The original perimeter isThe two boundary graphs of the symmetrized domain are $\pm h/2$, soApply the triangle inequality in ${\mathbb{R}}^2$ to the vectorsIt gives pointwiseAfter integration,

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 meansEquality 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 $\Omega$ has the horizontal line $y=c/2$, parallel to $L$, as an axis of symmetry. The converse is immediate.

Fix any direction and choose a line $L$ in that direction lying outside the bounded domain. Steiner symmetrization in that direction preserves convexity and, by part (i), preserves area. Since $\Omega$ minimizes perimeter among admissible convex domains of that area,Part (ii) gives the reverse inequality, so equality holds. Its equality characterization says that $\Omega$ has an axis of symmetry parallel to $L$. Since the direction was arbitrary, $\Omega$ admits axes of symmetry in every direction.