 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2016/ib/paper-3/5f/b/solution

Codex  0 Created 2026-09-24 Updated 2026-10-06

Let

p

and

h

count the pentagonal and hexagonal faces. Counting incidences of edges with vertices and faces gives

3 V = 2 E, 5 p + 6 h = 2 E, F = p + h .

The Euler formula for a sphere also applies to this polygonal decomposition: triangulating a polygon by adding one diagonal increases both

E

and

F

by one, preserving

V - E + F

. Substitution gives

2 = \frac{2 E}{3} - E + p + h = p + h - \frac{5 p + 6 h}{6} = \frac{p}{6} .

Thus there are exactly 12 pentagons, independently of the number of hexagons.

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