AoPS Academy
be a cyclic quadrilateral with 
and let 
and 
be the feet of altitudes from 
to the lines 
and 
,.Prove 
bisects 
.
is strictly increasing : 
such that:
be a real numbers such that 
Prove that
Where 
, let 
be the set of positive integers that do not exceed and are coprime to . Define 
as the smallest positive integer that allows to be partitioned into disjoint subsets, each forming an arithmetic progression.
satisfying 
, 
, and 
.
. We have that 
. If is constant, then 
. If not, there exists 
. Take 
such that 
and vary 
such that 
covers 
and for the case where 
just swap 
with 
and take 
. Hence we get . such that 
. Suppose that there exists 
with 
. From 
and 
we get a contradiction. So is injective.
. And from 
with 
we get that 
so from injectivity we have 
. Notice that 
also satisfies this relation so we get that 
which does satisfy the FE.
denote the given assertion. gives 
gives 
and give 
, therefore we check the two cases 
and 
.
gives 
and give 
, hence 
. and 
give 
so 
.
and 
(which in fact also shows surjectivity).
, then from we know 
. is injective.
s.t. 
. Now look at 
, therefore 
and is injective. and using injectivity finally shows 
or 
.
for some constant 
and 
are the only solutions. 
be the assertion of
or 
for some constant 
is obvious
gives us
so Let 
then
Now gives us then 
there exists some 
such 
Now
gives us
proving its bijectivity
then for some constant 
be a solution to this equation.
gives 
.
and 
gives 
. such that 
: we get 
. such that 
, then 
. So if there are two different such that then from this equation these two should be equal, absurd. Therefore there exists at most one such that : we have already established that , so this must be . For 
, we have from the 
equation. So 
. Combining with 
, we can affirm that is of the form 
where 
is a constant.
for all . Therefore, 
. Clearly for , we necessarily have 
. By the way, with 
, we have 
since . Letting 
we have 
so 
. In summary, . and 
where is a constant. We easily check that there are indeed solutions.
and gives that 
is always annihilated by , then 
yields that
giving either 
or 
for each . If there is such that the latter hold, is injective, giving 
, otherwise we have the constant function everywhere except and a quick check reveals at too, both of which works.
, we have that ![\[P(x,x): xf(x+f(x))=0 \implies f(x+f(x))=0, \ \ \ \forall x \in \mathbb{R*}\]](http://latex.artofproblemsolving.com/f/f/b/ffbb628a8c9ba09c1901d3de6ea60a760ef7c0e3.png)
wherein 
is reals without . Now we have ![\[P(-f(y),y): -f(y)f(0)=(y+f(y))f(f(-f(y)))\]](http://latex.artofproblemsolving.com/4/c/1/4c1751dd826066d1bd0456c3e9e046b06a2b4c5b.png)
where letting 
we would have, 
forcing either otherwise 
Now if we have that acts as a solution, assuming there are more we must have some otherwise we have, ![\[xf(x+f(y))=0 \ \forall (x,y) \in \mathbb{R} \implies xf(x+f(f(0)))=0 \implies xf(x)=0 \implies f(x)=0 \ \forall x \in \mathbb{R*}\]](http://latex.artofproblemsolving.com/5/7/a/57afea7567e7ce0faf945910aa4a303d32217dc7.png)
however for some , we must have that 
which indicates identically a direct contradiction to our assumption. Now we have established our claim so let 
then we compare and 
to obtain that is injective. However since ![\[P(0,1): f(f(0))=0 \ \land \ f(x+f(x))=0 \ \ \forall x \in \mathbb{R*}\]](http://latex.artofproblemsolving.com/c/6/9/c69a6532ae1afebc9bcf3d80f2b5a9d6dd8614e5.png)
so we must have 
for some constant and thus 
however we must also have that, 
and so we have 
.
which are trivial to check that they work.
the given assertion. 
is injective
so 
is not injective
such that 
and 
such that 
we have:
then RHS . It follows that 
which is false, by the way we picked .
therefore 
. 
such that .
are .
and 
we get 
If 
we are done. So assume one of them is . If 
:
and by the first observation 
. We also have 
for all 
. So basically, . 
is a solution, so we'll now look for all nonzero solutions. Let be the assertion of the functional equation given., so . For 
, the equation yields 
. Now taking easily gives us that 
, which is 
for . Suppose there exists 
so that 
. Hence, by comparing 
and 
yields 
is injective. So by applying injectivity at 
and 
will implies 
. Now by letting 
for some 
, we easily get 
, which is indeed a solution. Therefore, we are done. 
holds for all . Then if , we have 
. Fix 
. Then 
, and 
for but 
along with 
produce entire set of reals. Hence is . We check easily that this works. gives 
. Pick nonzero, then . Hence, assume that 
for some 
. Substituting that for , we have![\[f(c+f(y))=c'(y-x_0)\]](http://latex.artofproblemsolving.com/e/b/2/eb2e8da0982bd800793834838cf06a12456a72b9.png)
for constants 
all non-zero. is injective as there is an isolated 
on the LHS, noting 
. is also surjective because shifting and scaling the reals still gives the reals on the RHS. Hence, there is an unique such 
. Letting initially, we have , so when , then 
, injectiveness yields 
. Of course, when , then 
too. One checks this works easily.
and . These clearly work, so we will now prove that they are the only functions. Let denote the assertion..
then use to get . If then use 
to get that , which finishes the claim.
for all then 
for all . Then 
gives 
for all . If then . If then take 
to get 
which implies .
for even a single value of then if 
, then comparing 
and 
immediately gives 
, so is injective. Then, by Claim 
, is a constant. We are done.
such that for all 
,![\[xf(x+f(y))=(y-x)f(f(x)).\]](http://latex.artofproblemsolving.com/e/4/a/e4a3bbb8b91d2aa62d699c24df342fa59be71915.png)
be the assertion and 
.
, contradiction.
. Note that , since:
for all 
for 
for all , or 
which satisfies the equation for any 
.
with 
. Note that 
is a solution, else there is some 
with 
.
, contradiction. So no further solutions.
and checking we see that 
truly satisfies the condition
gives us 
where 
for all 
gives 
,
gives a contradiction.
,
and so 
.
gives 
gives 
for 
for any constant 
such that 
.
,
and 
we obtain 
for some constant 
.
.
,
for all 
identically.
yields that 
.
is not always zero, then we easily obtain 
.
.
attains every real value except 
for all 
.
,
for all a 
and 
we get 
for 
but it is impossible then for all 
)
for every x and 
for every x not 0 and y real,but is easy to see it works also for x=0. But then let a be any real. Put 
to get 
is not 0. If we put x=0 we get that 
to get that f is bijective
and 
are solutions. Let's prove they are the only ones.
such that 
: applying 
give us 
we have 
we can notice that 
: 
then 
absurde, so 
for any 
gives 
.
.
,
.
imply 
then 
so there's at most one such 
.
So either 
,
,
,
where 
yields: 
for 
yields: 
,
which implies that 
such that 
and 
yields that 
.
now gives 
.
implies that 
for 
gives 
(due to the injectivity of 
which works for all constants 
gives 
to get 
,
,
,
.
for all 
,
for some 
since 
or 
for which 
since c is not zero and 
or 
.
.
be arbitrary real numbers such that 
,
So 
)
such that 