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by math154, Feb 3, 2012, 5:16 AM
Hmm a bit on the easy side I guess, but still nice.
1. (Russia 2011, 11.4) Ten cars, which do not necessarily start at the same place, are all going one way on a highway which does not loop around. The highway goes through several towns. Every car goes with some constant speed in those towns and with some other constant speed out of those towns. For different cards these speeds can be different. 2011 flags are put in different places next to the highway. Every car went by every flag, and no car passed another right next to any of the flags. Prove that there are at least two flags at which all cars went by in the same order.
Solution
2. (Russia 2002, 11.8) Show that the numerator of
(in reduced form) is infinitely often not a prime power.
Solution
1. (Russia 2011, 11.4) Ten cars, which do not necessarily start at the same place, are all going one way on a highway which does not loop around. The highway goes through several towns. Every car goes with some constant speed in those towns and with some other constant speed out of those towns. For different cards these speeds can be different. 2011 flags are put in different places next to the highway. Every car went by every flag, and no car passed another right next to any of the flags. Prove that there are at least two flags at which all cars went by in the same order.
Solution
Let
denote the time it takes for car
to get to the first flag (choose a starting time arbitrarily). Suppose car
takes time per unit distance
when in and out of town, respectively; let
(not both zero) be the distances from flag
to flag
along in and out of town routes. Then car
takes
to get to flag
. If
for some
, then we must have
for all
(or else we violate the "no passing at flags" rule), whence car
is essentially a "clone" of car
. Since clones are ranked the same at each flag, we can remove all clones to get (WLOG)
pairwise distinct cars (WLOG cars
through
). Then
for all
and
.
Now note that for
,
iff
and
iff
. Graphing
in the Cartesian plane for
(note the symmetry between
and
), we get at most
(where
is the number of lines) distinct regions (well-known, e.g. this). But
and
are in the same region iff the orderings at flags
are equal, so we're done by pigeonhole.
Comment: This is not ridiculous or anything, but doing the 1-d case helps a bit. Indeed, this generalizes to any number of dimensions, and the bound should be basically tight and not hard to find.






















Now note that for


![\[(x_k,y_k)\in S_{i,j}=\{(x,y)\mid L_{i,j}(x,y)=(a_i-a_j)x+(b_i-b_j)y+(t_i-t_j)>0\},\]](http://latex.artofproblemsolving.com/f/9/f/f9f56603a6e3049e2438f5de19aa67b1403e9a91.png)











Comment: This is not ridiculous or anything, but doing the 1-d case helps a bit. Indeed, this generalizes to any number of dimensions, and the bound should be basically tight and not hard to find.
2. (Russia 2002, 11.8) Show that the numerator of

Solution
Outline: Bound denominators using powers of
and primes from
to
. Consider
for
, and note that
, so consider
as well. Subtracting dudes, get a contradiction.
Alternatively, use the stronger result of d'Aurizio mentioned here.
Edit: Actually, this paper looks pretty incorrect...







Alternatively, use the stronger result of d'Aurizio mentioned here.
Edit: Actually, this paper looks pretty incorrect...
This post has been edited 6 times. Last edited by math154, Apr 21, 2012, 4:07 AM