Advanced book on Mathematics Olympiad

1.4 Ordered Sets and Extremal Elements 17

Solution.For the solution, assume that such a dissection exists, and look at the bottom
face. It is cut into squares. Take the smallest of these squares. It is not hard to see that
this square lies in the interior of the face, meaning that it does not touch any side of the
bottom face. Look at the cube that lies right above this square! This cube is surrounded
by bigger cubes, so its upper face must again be dissected into squares by the cubes that
lie on top of it. Take the smallest of the cubes and repeat the argument. This process
never stops, since the cubes that lie on top of one of these little cubes cannot end up
all touching the upper face of the original cube. This contradicts the finiteness of the
decomposition. Hence the conclusion. 

By contrast, a squarecanbe dissected into finitely many squares of distinct size. Why
does the above argument not apply in this case?
And now an example of a more exotic kind.

Example.Given is a finite set of spherical planets, all of the same radius and no two
intersecting. On the surface of each planet consider the set of points not visible from
any other planet. Prove that the total area of these sets is equal to the surface area of one
planet.

Solution.The problem was on the short list of the 22nd International Mathematical
Olympiad, proposed by the Soviet Union. The solution below we found in I. Cuculescu’s
book on the International Mathematical Olympiads (Editura Tehnic ̆a, Bucharest, 1984).
Choose a preferential direction in space, which defines the north pole of each planet.
Next, define an order on the set of planets by saying that planetAis greater than planet
Bif on removing all other planets from space, the north pole ofBis visible fromA.
Figure 5 shows that for two planetsAandB, eitherA<BorB<A, and also that
for three planetsA, B, C,ifA<BandB<CthenA<C. The only case in which
something can go wrong is that in which the preferential direction is perpendicular to the
segment joining the centers of two planets. If this is not the case, then<defines a total
order on the planets. This order has a unique maximal elementM. The north pole ofM
is the only north pole not visible from another planet.

C

N

B

A

A

B

Figure 5