Advanced book on Mathematics Olympiad

Real Analysis 511

second day of travel with a homeomorphism (continuous bijection) of the time interval
[ 0 , 1 ], we can ensure that the motion of the first car yields the same parametrization of
the road on both days. Letf(t)be the distance from the second car toAwhen the first
is atton the first day, andg(t)the distance from the second car toAwhen the first is
atton the second day. These two functions are continuous, so their difference is also
continuous. Butf( 0 )−g( 0 )=−dist(A, B), andf( 1 )−g( 1 )=dist(A, B), where
dist(A, B)is the distance between the cities.
The intermediate value property implies that there is a momenttfor whichf(t)−
g(t)=0. At that moment the two cars are in the same position as they were the day
before, so they are at distance at most one mile. Hence the answer to the problem is no.

400.We compute

∑n

j= 0

P( 2 j)=

∑n

j= 0

∑n

k= 0

ak 2 kj=

∑n

k= 0

⎛

⎝

∑n

j= 0

2 kj

⎞

⎠ak

=

∑n

k= 0

2 k(n+^1 )− 1

2 k− 1

=Q( 2 n+^1 )−Q( 1 )= 0.

It follows thatP( 1 )+P( 2 )+ ··· +P( 2 n)=0. IfP( 2 k)=0 for somek<n,we
are done. Otherwise, there exist 1≤i, j≤nsuch thatP( 2 i)P ( 2 j)<0, and by the
intermediate value property,P(x)must have a zero between 2iand 2j.
(proposed for the USA Mathematical Olympiad by R. Gelca)

401.Consider the lines fixed, namely thex- and they-axes, and vary the position of the
surface in the plane. Rotate the surface by an angleφ, then translate it in such a way
that thex-axis divides it into two regions of equal area. The coordinate axes divide it
now into four regions of areasA, B, C, D, counted counterclockwise starting with the
first quadrant. Further translate it such thatA=B. The configuration is now uniquely
determined by the angleφ. It is not hard to see thatA=A(φ),B=B(φ),C=C(φ),
andD=D(φ)are continuous functions ofφ.
IfC( 0 ◦)=D( 0 ◦), then the equality of the areas of the regions above and below the
x-axis impliesA( 0 ◦)=B( 0 ◦)=C( 0 ◦)=D( 0 ◦), and we are done.
IfC( 0 ◦)>D( 0 ◦), then the line that divides the region below thex-axis into two
polygons of equal area lies to the left of they-axis (see Figure 67). This means that after
a 180◦-rotation the line that determines the regionsA( 180 ◦)andB( 180 ◦)will divide
the other region intoC( 180 ◦)andD( 180 ◦)in such a way thatC( 180 ◦)<D( 180 ◦).
Similarly, ifC( 0 ◦)<D( 0 ◦), thenC( 180 ◦)>D( 180 ◦).
It follows that the continuous functionC(φ)−D(φ)assumes both positive and
negative values on the interval[ 0 ◦, 180 ◦], so by the intermediate value property there is an
angleφ 0 for whichC(φ 0 )=D(φ 0 ). Consequently,A(φ 0 )=B(φ 0 )=C(φ 0 )=D(φ 0 ),
and the problem is solved.