Since
each hole is surrounded by either 5 or 6 neighbors,
we can consider the **puzzle 122** as
a tiling problem of a sphere: in fact, we can
imagine that every polygonal face tiling the
sphere contains a hole in its center, as well
as one facing each of its sides (fig. 1).

If the sphere was tiled only with hexagons, the
numbers of edges (*E*) would be (6 x *H*)/2,
in total (since each hexagons’ side/edge
shares 2 faces).

If the sphere was tiled only with pentagons, the
numbers of edges (*E*) would be (5 x *P*)/2,
in total (since each pentagons’ side/edge
shares 2 faces).

As
the sphere is tiled with a combination of both
polygons, we obtain the following equality:

*a)* *E* = (6*H* +
5*P*)/2

And
since each vertex (*V*) shares 3 edges,
we obtain also this second equality:

*b)* *V* = (6*H* +
5*P*)/3; that is [(6 edges x number of Hexagons)
+ (5 edges x number of Pentagons) divided by 3]

The Euler’s
polyhedron formula states for any convex
polyhedron, the number of vertices (*V*)
and faces (*F*) together is exactly
2 more than the number of edges (*E*): *V* + *F* =
2 + *E*

Substituting the variables of the Euler's formula
with *a)* and *b)* above, and taking
into account that *F* = *H* + *P*,
we get:

*c)* (6*H* + 5*P*)/3
+ (*H* + *P*) = 2 + (6*H* +
5*P*)/2

After reorganizing:

(6*H* + 5*P*)/3 + (*H* + *P*)
- (6*H* + 5*P*)/2 = 2

and simplifying, we finally obtain:

*P* = 12

We must then have 12 pentagons; so of the 384 holes
on the radiolarian, at least **12 have only
5 neighbors**. Surprisingly, the result
is INDIPENDENT of the number of holes!

A
convex polyhedron very similar to the one previously
discussed, made of 12 pentagons and 20 hexagons
only, is called truncated
icosahedron and is well known by soccers, see
fig. opposite. For the soccer ball the Euler's
formulla gives:

60 + 32 - 90 = 2