and 
meet at two distinct points 
and 
.
through 
(other than 
.
be the midpoint of the minor arc 
(i.e., the point on 
subtends the arc 
be the midpoint of the minor arc 
.![\[
MN \parallel \text{(radical axis of } \omega_1, \omega_2).
\]](http://latex.artofproblemsolving.com/6/6/4/6647f0f7182bdf5096ec5f2956a272803082f4f7.png)
is on 
and 
is on 
.
.
is the circumcenter of 
,
is the circumcenter of 
intersects with 
at 
be a triangle with orthocentre 
.
be the foot of altitudes of 
onto the opposite sides, respectively. Consider 
,
about line 
.
cut 
,
be the orthocenter of 
.
are concyclic.
be a continuous function such that 
show that 
is even.
Y by nguyendangkhoa17112003, anantmudgal09, abdelkrim, Hexagrammum16, Plasma_Vortex, Seicchi28, The_Maitreyo1, Carpemath, sriraamster, SHREYAS333, TheMathLife, samuel, Vietjung, lifamily, AwesomeDude86, Toinfinity, OlympusHero, A-Thought-Of-God, L567, RudraRockstar, samrocksnature, Palindrome_, centslordm, megarnie, Math_Is_Fun_101, Lamboreghini, veirab, EpicBird08, Adventure10, deplasmanyollari, lian_the_noob12, ItsBesi, buddyram, MathIQ., Rounak_iitr, Akacool, cubres, NicoN9
Let 
be the set of integers. Determine all functions 
such that, for all integers 
and 
, 
Proposed by Liam Baker, South Africa
be the set of integers. Determine all functions 
such that, for all integers 
and 
, 
Proposed by Liam Baker, South AfricaThis post has been edited 2 times. Last edited by djmathman, Jul 18, 2019, 4:41 AM
and 
for some 
constant.
,
for all 
.
for this one, we quickly obtain that 
,
.
by symmetry and so the condition becomes 
It is equivalent to 
This means 
is linear on even numbers. Notice that 
,
,
,
for all 
for all 
(divide the case when 
is negative and when 
or 
for some constant 
gives 
, or for every 
in the image of 
.
A simple variable-switching and some translation gives the well-known Cauchy equation. Hence 
after we reach the last step in @above's solution. Cauchy ends it.
=>
(
(
and 
in (1) and use (2), we have 
,
an
,
.
.
.
to get 
,
.
this is just 
over integers implying 
and 
.
or 
and 
or 
for an arbitrary integer 
.
for all positive integers 
be the assertion, then we take 
and subtract to get 
,
gives 
.
for 
,
. Thus 
gives 
,
,
.
denote assertion into the original equality.
we get that 
,
.
here, we get 
.
or 
.
(using the evaluations 
)
for all 
,
for any constant 
and 
,
for all 
.
.
yields 
,
yields 
,
for all 
are the only linear solutions. Compare 
to get that 
is constant and obviosly 
.
and 
.
to get 
.
to show that it is indeed linear which finishes.
while 
Therefore the assertion becomes 
so the function 
thus 
for integer constants 
Plugging this back into the assertion yields either 
for some constant 
These are the only solutions, and it is not hard to see that these work.
for any constant 
,
,![\[2 f(a) + 2 f(b) = f(f(a + b)) + f(0). \]](http://latex.artofproblemsolving.com/1/0/c/10cf4ab93a27cecdc6e0936399afb5c3ab8e1e40.png)
Now asserting 
gives us 
,
yields![\[ 2 f(x + 1) + 2f(-x) = 2 f(x + 1) - 2 f(x) + f(f(0)) + f(0) = f(f(1)) + f(0) \]](http://latex.artofproblemsolving.com/e/9/a/e9a0478988c22d6064309a9fd26072f3f9b47658.png)
![\[\implies f(x + 1) - f(x) = \dfrac{f(f(1)) - f(f(0))}{2}. \]](http://latex.artofproblemsolving.com/c/6/4/c64114b46552fc4ac82623f07954f3aef1b43812.png)
Since the RHS is constant, 
as our claimed solution set. We are done.
for any integer 
.
term. This motivates us to set 
.
.
we find that 
so 
.
.
we notice is always a constant, so this implies 
.
.
,
.
,
.
whose solution is only 
.
,
.
and 
gives 
.
gives 
gives 
Thus, 
implies the only solutions are 
where 
,
is any integer. It is straightforward to check that these functions are indeed solutions, so now we prove that they are the only ones. Let 
,
,
.
denote this assertion. By 
,
,
.
is Cauchy; hence 
,![\[2k(a+b)+3\ell = k^2(a+b) + \ell(k+1).\]](http://latex.artofproblemsolving.com/c/2/7/c274150eb3948a4216e46476cf224840e624a3e2.png)
This implies 
and 
,
,
or 
and thus we have that 
Plugging in 
we see that 
and thus our equation simplifies to 
The solution to this is all linear equations, 
.
and 
Let us shift things to the origin by defining a new function 
as,
This equation is quite easy to deal with and I can think of two ideas one could use
and 
Notice the above holds for positive integers but one can deal with negative integers in a similar manner. So clearly 
,
is constant.
.
.
.
.
.
,
.
.
.
or 
.
![\[f(0)+2f(b)=f(f(b)).\]](http://latex.artofproblemsolving.com/7/9/b/79b5461182ca29d3e67ac8e313588f56f899a91d.png)
Thus, ![\[f(f(a+b))=f(0)+2f(a+b)=f(2a)+2f(b).\]](http://latex.artofproblemsolving.com/0/d/1/0d13b0c067e7717387cb8c111ee1eae032765b64.png)
Therefore, ![\[f(2a)-f(0)=2f(a+b)-2f(b).\]](http://latex.artofproblemsolving.com/7/c/6/7c66f407304d2680fc75467f2873a37d826f40b2.png)
Fix 
to obtain that 
is constant, so the function is linear or constant by induction (since the function is over integers). Plug these back into the original equation to obtain the solutions described above.
