[/quote]
,
be real numbers satisfying 
Prove that 
and determine all cases of equality.
be the set of rational numbers. A function 
is called aquaesulian if the following property holds: for every 
,![\[ f(x+f(y)) = f(x) + y \quad \text{or} \quad f(f(x)+y) = x + f(y). \]](http://latex.artofproblemsolving.com/4/0/e/40e7d033daf1bd62c2bfd5d022eeaea3f860c5b3.png)
Show that there exists an integer 
such that for any aquaesulian function 
there are at most 
for some rational number 
,
lines
such that for all 
,
be a fixed integer. There are 
beads on a circular necklace. You wish to paint the beads using 
colors, such that among any 
consecutive beads every color appears at least once. Find the largest value of 
for which this task is 
possible.
,
,
,
denote the feet of the altitudes from 
,
,
,
denote the circumcenter of 
.
and 
denote the projections of 
,
.
such that 
and 
.
and 
lies on 
such that 
,
bisects 
.
,
coin from the left and moves all coins in that chain to the left end of the row. For example, if 
and 
,
would be 
with 
such that for every initial ordering, at some moment during the process, the leftmost 
and 
Prove that
Let 
Prove that
Let 
Prove that
prove that
such that for all naturals 
the function satisfies:![\[a + b \mid a^{f(a)} + b^{f(b)} \]](http://latex.artofproblemsolving.com/e/b/b/ebb44041cedd29de516192d983a334d46e7fe7d6.png)
Bob wants to find all possible functions Alice could have. Help Bob and find all functions that Alice could have.
-
Y by Adventure10, Mango247
Three nonnegative real numbers 
, 
, 
are written on a blackboard. These numbers have the property that there exist integers 
, 
, 
, not all zero, satisfying 
. We are permitted to perform the following operation: find two numbers 
, 
on the blackboard with 
, then erase and write 
in its place. Prove that after a finite number of such operations, we can end up with at least one 
on the blackboard.
, 
, 
are written on a blackboard. These numbers have the property that there exist integers 
, 
, 
, not all zero, satisfying 
. We are permitted to perform the following operation: find two numbers 
, 
on the blackboard with 
, then erase and write 
in its place. Prove that after a finite number of such operations, we can end up with at least one 
on the blackboard.This post has been edited 2 times. Last edited by worthawholebean, May 1, 2008, 8:55 PM
,
so that 
continues to hold. (For example, if we replace 
then we replace 
.
is strictly decreasing (this took me three pages of casework, but presumably there's a nicer way). Since it is a strictly decreasing sequence of positive integers, after a finite number of operations we have 
,
is some permutation of 
and 
and 
,
.
.
and 
to make 
smaller.
,
.
,
and 
,
.
.
and 
.
and always operate on the largest and smallest 
.
and 
.
,
.
such that 
are positive integers and that 
.
,
,
,
;
.
.
than there actually are when using these arguments. Not saying there isn't a correct one, though, but I imagine it would be largely similar to the 
![\[ a_1 x_n + a_2 y_n + a_3 z_n = 0 \]](http://latex.artofproblemsolving.com/3/4/e/34ecc8974cb37f4a49c5db1727fc9c9e44e75fbb.png)
.
,
themselves, not to the 
.
as long as 
.
and 
without loss of generality and consider two cases.
or 
;
and get ![\[ 0 = a_1r_1 + a_2r_2 + a_3r_3 > a_1r_3 + a_3r_3 \implies a_1 + a_3 < 0 \]](http://latex.artofproblemsolving.com/d/7/c/d7c4d10bc4e8bfc3b63eb61215757a1828bedfb7.png)
so 
,
.
and 
are less than zero. Assume for contradiction that 
and 
both hold (if either fails then we use 
and 
and 
are both positive in this case, so we get 
and 
;
.![\[ 0 = a_1r_1 + a_2r_2 + a_3r_3 > a_1r_2 + a_2r_2 + a_3r_2 = r_2(a_1+a_2+a_3) \implies a_1+a_2+a_3 < 0 \]](http://latex.artofproblemsolving.com/0/5/d/05d8a2193416474be46b235d6a9bf1ae4e803f72.png)
which is a contradiction.
at all, nonetheless with that quotient. So you can't completely get rid of one of your numbers and use the Euclidean algorithm on them that quickly.
,
,
.
with 
.
then we will go 
.
.
can't be possible, since it would imply that the original triplet was 
and 
is zero, then we are already done since it would imply the nonzero (WLOG 
.
and 
.
is given. After a step, if we do 
then we can do simple change 
to 
is strictly decreasing.
.
to either 
or 
or 
.
.
.
or 
;
or 
;
or 
.
or a result, which gives the result.
be the numbers on the blackboard after 
with 
minimal such that 
.
-
moves, the numbers on the blackboard satisfy 
where 
are nonnegative integers and 
equals the signature. We will also assume that 
are distinct, otherwise are done immediately.
first, and consider the equation ![\[a_k(x_k-z_k) + b_k y_k - (c_k-a_k)z_k = 0.\]](http://latex.artofproblemsolving.com/f/7/6/f765d38bb991416a639ea75ebd7100baa2bb4fbc.png)
Letting 
,
.
,![\[(a_k-c_k)x_k + b_k y_k - c_k(z_k-x_k) = 0\]](http://latex.artofproblemsolving.com/e/d/c/edc4860971b9f5bf84316dcc22fa5cce5da7a871.png)
and note that by defining 
,
.
,
and 
,
as all the terms are positive; if simultaneously we have 
and 
,
,
,
,
equals zero.
.
,
and check that 
,
has a smaller signature than 
.
to zero, in which case the remaining 
or 
and 
represent the values of 
.
at all times, so replacing 
would transform 
.
can be strictly decreased unless one of 
equals 
.
.
or 
;
gives 
,![\[a_1 < a_1+a_3 < \frac{a_3}{2} < -a_1 \implies |a_1+a_3| < |a_1|.\]](http://latex.artofproblemsolving.com/d/a/c/dacfd0e07cfa1edc13ce11521428b34699acdd4a.png)
,
,
and 
,
