Here “isometry” is used in the standard <Mazur-Ulam theorem> sense of a surjective distance-preserving map. Surjectivity is needed; a merely distance-preserving embedding need not preserve midpoints, as shown by a <nonsurjective isometry need not preserve midpoints>.

Distance preservation and surjectivity give
$$
u(S_1^{vw})=S_1^{u(v)u(w)}.
$$
Assume inductively that
$$
u(S_{n-1}^{vw})=S_{n-1}^{u(v)u(w)}.
$$
Then the two sets have the same diameter, and the universal distance condition defining the next set transfers through the bijection $u$. Hence
$$
u(S_n^{vw})=S_n^{u(v)u(w)}
$$
for every $n$.

Because $u$ is injective, it also preserves the intersection of this nested family. Part (a) therefore gives
$$
\left\{
u\left(\frac{v+w}{2}\right)
\right\}
=u\left(\bigcap_{n\geq1}S_n^{vw}\right)
=\bigcap_{n\geq1}S_n^{u(v)u(w)}
=\left\{
\frac{u(v)+u(w)}2
\right\}.
$$
Thus
$$
\boxed{
u\left(\frac{v+w}{2}\right)
=\frac{u(v)+u(w)}2.
}
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 2 / 22F / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ii/paper-2/22f/b/solution