 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ib/paper-3/14e/c/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

K < 0

everywhere, compactness makes the integral negative, so

χ (S) < 0

. For every triangulation,

3 F = 2 E

again yields the general identity

\overline{d} = 6 - \frac{6 χ ( S )}{V} .

Since

V \geq 1

and

χ (S) < 0

6 < \overline{d} \leq 6 - 6 χ (S) .

This supplies bounds depending only on the surface.

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