AoPS Academy
table. It is permissible to reverse the signs of all the numbers in any row or column. Prove that after a number of these operations, we can make the sum of the numbers along each line (row or column) nonnegative.
be an acute triangle and 
it's incenter. Let 
, 
and 
be the projections of onto 
, 
and 
respectively. 
and 
is the projection of onto 
. Let 
be the middle of the arc not containing 
and 
the second intersection of 
and the circumcircle of 
. If 
is the midpoint of 
, 
, 
and 
, prove that 
is cyclic.
such that the following equation holds for every real 
:![\[ f(f(x) + y) = \lfloor x + f(f(y)) \rfloor. \]](http://latex.artofproblemsolving.com/3/f/b/3fbbb68e9f9c104fe0f701c8ad07e7fbbf583f1e.png)
Note: 
denotes the greatest integer not greater than 
.
such that![\[ f(f(x) + y) = \lfloor x + f(f(y)) \rfloor \]](http://latex.artofproblemsolving.com/3/b/5/3b58250769729fc386ba9e8ea5ce104a257a944f.png)
for any 
.
for all , for some fixed 
. To see this, we see that for all ,![\[ f(f(x) + y) = \lfloor \lfloor x \rfloor + y \rfloor + 2k = \lfloor x \rfloor + \lfloor y \rfloor + 2k = \lfloor x + \lfloor y \rfloor + 2k \rfloor = \lfloor x + f(f(y)) \rfloor \]](http://latex.artofproblemsolving.com/1/3/7/137d3ce388f9388cf5aef2f46fdb9dd9d04c7291.png)
We'll now prove that there are no other solutions. Let 
be the assertion of and 
to the given functional equation.
.
. 
gives us![\[ f(\alpha) = f(f(0) + \alpha - f(0)) = \lfloor f(f(\alpha - f(0))) \rfloor \in \mathbb{Z} \]](http://latex.artofproblemsolving.com/0/3/8/03877910b91cf9bdc7cb9d4dd37058465af7ae12.png)
.
from the first claim, 
and 
gives us![\[ \lfloor a \rfloor + f(f(0)) = f(f(a)) = f(f(b)) = \lfloor b \rfloor + f(f(0)) \implies \lfloor a \rfloor = \lfloor b \rfloor \]](http://latex.artofproblemsolving.com/0/d/c/0dcd346a3aa8b0cb133537c53a3f75e875ad0a26.png)
![\[ f(f(0) + x) \stackrel{P(0,x)}{=} \lfloor f(f(x)) \rfloor \stackrel{\text{Claim 01}}{=} f(f(x)) \stackrel{\text{Claim 02}}{\implies} f(x) \stackrel{\text{Claim 01}}{=} \lfloor f(x) \rfloor = \lfloor x + f(0) \rfloor \stackrel{\text{Claim 01}}{=} \lfloor x \rfloor + f(0) \]](http://latex.artofproblemsolving.com/8/e/7/8e76736789a1f52969ddadb4a0e11478cd67f2a0.png)
as desired.
be the assertion, clearly 
gives us 
.
gives us
By substituting it to , we will get
From here, it's already clear what the answer is.
be the assertation ofFrom 
, we have 
where 
i.e.![\[f(f(x)+y) = \lfloor x + f(f(y))\rfloor = \lfloor x +\lfloor y+c\rfloor \rfloor = \lfloor x \rfloor + \lfloor y+c\rfloor.\]](http://latex.artofproblemsolving.com/4/a/4/4a43bbce0409af97ad69041ae4dd7a1e6bc0cf58.png)
From we must have 
for all 
. If 
, take 
and we have contradiction. Because 
, we have 
. From 
we have 
. From 
we have![\[f(x) = \lfloor x-f(0)+c\rfloor = \lfloor x-k+2k\rfloor = \lfloor x \rfloor +k,\;k\in \mathbb{Z}\]](http://latex.artofproblemsolving.com/c/2/8/c285ff95d3a837b8e499af51dd7de0f30e7ed7ec.png)
and we can check if this solution to the equation.
as the assertion of and to the given functional equation on the problem. From 
, we obtain 
, so .
we obtain![\[ f(f(f(x))) = \lfloor f(x) + f(f(0)) \rfloor = f(x) + f(f(0)) \text{ }(\star) \]](http://latex.artofproblemsolving.com/1/a/5/1a53598b9dffc318160143bc1f0830e14dc1327d.png)
And therefore, from comparing 
and 
we obtain:
for some constant 
. Since 
and are both integers, then must be an integer as well, so the solutions are of the form 
, where 
. It should be easy to check this into the original equation.
be the assertion
for each real y the function is integer
this implies f surjective over integer
proof
and 
then we are done
(1)
(From equation (one) we get )
such that the following equation holds for every real :Note: denotes the greatest integer not greater than ., and the assertion 
.
for all 
(since 
)
so if 
is defined by 
for all 
we have 
for all . Then rewrites as (using 
as well):
for all 
. Now 
is injective, and swapping 
we get 
, so 
. Then we can write 
for some constant 
(where 
) and all , all such functions work.
![\[ f(f(x))=\lfloor x+f(f(0)) \rfloor.\]](http://latex.artofproblemsolving.com/f/2/c/f2c84353899600a033c0dae582b35fcd1052bf29.png)
This implies that 
is surjective over 
.
with 
.
,![\[ f(y)= \lfloor c + f(f(y-c)) \rfloor, \]](http://latex.artofproblemsolving.com/a/b/4/ab4dd421feaff76bce4496957a73b2dc16f8641e.png)
which implies that 
for all ![\[ f(f(x)+y)=\lfloor x + \lfloor y +f(f(0)) \rfloor \rfloor = \lfloor x \rfloor + \lfloor y \rfloor + f(f(0)). \]](http://latex.artofproblemsolving.com/0/1/9/019a5f029de5cb8852f954cbf02f4ee6869812bc.png)
in the last expression, and switching the variables, we have![\[ \lfloor x\rfloor+ \lfloor f(y) \rfloor+f(f(0))=\lfloor y \rfloor + \lfloor f(x) \rfloor+ f(f(0)) \iff f(x)-\lfloor x \rfloor = f(y) - \lfloor y \rfloor. \]](http://latex.artofproblemsolving.com/2/4/a/24a15b248a41f6f2d90f394b3a4f5611a1f2f065.png)
The last equation implies ![\[ f(x)=\lfloor x \rfloor +K \]](http://latex.artofproblemsolving.com/b/e/a/beabb402b29f90b1563204a2654f7dc3a57fe80c.png)
for all 
,