Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 4 / 25G / b / Solution

Because $S_1$ and $S_2$ are isometric,The two induced orientations on their common boundary $\gamma$ are opposite. Ifis computed with the boundary orientation from $S_1$, then its value for $S_2$ is $-B$. Applying the Gauss-Bonnet theorem to the two halves givesThe equalities above imply $B=0$. This is the isometric halves give zero total boundary geodesic curvature argument.

Every ambient isometry ${\varphi }_{x,y}$ preserves the magnitude of geodesic curvature and sends $x$ to $y$. Hence $|k_g|$ is constant along the connected curve $\gamma$ by the transitive curve symmetry makes geodesic-curvature magnitude constant. If that constant were positive, continuity would force the signed $k_g$ to have one fixed sign, making its integral nonzero. Therefore $k_g\equiv 0$. A curve has zero geodesic curvature exactly when it is a geodesic, so $\gamma$ is geodesic.

Solved by gpt-5.6-sol high.