AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
be a cyclic quadrilateral with circumcenter 
, such that 
is not a diameter of its circumcircle. The lines 
and 
intersect at point 
, so that 
lies between 
and , and 
lies between 
and . Suppose triangle 
is acute and let 
be its orthocenter. The points 
and 
on the lines and , respectively, are such that 
and 
. The line through , perpendicular to , intersects at 
, and the line through , perpendicular to , intersects at 
. Prove that the points , , are collinear.
such that 
. Show that 
. When will equality hold?
which satisfy the condition![\[ f(m + n) \geq f(m) + f(f(n)) - 1
\]](http://latex.artofproblemsolving.com/4/b/d/4bddce124cf0da82cb8bb6c6ae2325239354f27c.png)
Find all possible values of 
.
be a positive integer and 
be a positive integer coprime to 
. Let 
, and for 
, define
Find, in terms of and 
, the greatest positive integer 
for which there exists an index 
such that 
is divisible by 
.
table are initially white. Alice and Bob play a game. First Alice paints of the fields in red. Then Bob chooses 
rows and columns from the table and paints all fields in them in black. Alice wins if there is at least one red field left. Find the least value of such that Alice can win the game no matter how Bob plays.
be an interior point of the acute triangle 
with 
so that 
The point on the segment 
satisfies 
the point on the segment 
satisfies 
and the point 
on the line satisfies 
Let 
and 
be the circumcenters of the triangles 
and 
respectively. Prove that the lines 
and 
are concurrent.
, define 
as the product of the digits of (in decimal). Find all positive integers such that 
.
be a function satisfying the following conditions:
, if 
and 
, then 
;
is surjective.
. be a triangle with 
and 
. Let and be the midpoints of 
and 
respectively. The point is inside of 
such that 
is equilateral. Let 
and 
. Prove that 
.
. This has to be covered by three types of tiles - red, blue and black. The red tiles are of size 
, the blue tiles are of size and the black tiles are of size 
. Let 
denote the number of ways this can be done. For example, clearly 
because we can have either a red or a blue tile. Also 
since we could have tiled the plot as: two red tiles, two blue tiles, a red tile on the left and a blue tile on the right, a blue tile on the left and a red tile on the right, or a single black tile.
for all 
.
for all 
.![\[ \binom{m}{r} = \begin{cases}
\dfrac{m!}{r!(m-r)!}, &\text{ if $0 \le r \le m$,} \\
0, &\text{ otherwise}
\end{cases}\]](http://latex.artofproblemsolving.com/b/8/d/b8d653661e421db1fce7d93837e7bce31337bb30.png)
for integers 
.
in the given equation gives 
Call this statement 
. This immediately gives is injective.
and 
gives 
Here is some constant. Also, comparing 
and 
and using injectivity, we have 
Here 
is another constant. Now this gives 
holds for all 
, which forces 
. So in fact 
Now 
gives 
Call this new statement 
.
gives 
Comparing 
and 
and mutilplying y 
gives 
Using 
in each of these terms, we have 
, and replacing 
, we get 
Now we use the statements 
to simplify that into 
Setting 
, there, we get 
, and since 
because of injectivity, we get 
, which in turn implies 
can be written as 
, which becomes Cauchy's equation after setting 
. Since the codomain of is bounded from below, we must have 
, and only 
fits.
It is clear that we can replace 
by 
and get:
Suppose that 
for some 
Then 
gives us 
contradiction. So 
But this means that
Combining the two boxed results gives us 
such that
for all positive real numbers and 
.
for all 
. Put 
thus 
for all 
. Observe that as 
we see 
has no upper bound. Thus, for any 
it is possible to pick with 
; hence 
is a subset of , consequently is surjective over 
. If 
then plugging 
and 
subsequently, we conclude 
or is injective. Thus, is a bijection on positive reals.
yielding 
for all 
. Swapping 
fixes the LHS, hence 
for some constant and all 
. Hence 
for all . Playing the Devil's trick again, we put 
in the original equation; so 
and swap 
; injectivity of then yields that 
is a constant function. Thus, 
and so 
for all 
. Now plug 
in 
to conclude that 
for all 
. Now we immediately get 
for all and so 
(putting 
); hence is additive too. Now is strictly increasing so if 
then 
and vice-versa. Thus, 
for all 
and is the identity function. It also works. 
to note that is injective and 
. Now take with 
gives 
. Therefore is bijective.
Replace 
by 
in the above relation and consider symmetric equation with 
swapped we deduce
Fix and vary to get using bijective that 
so 
, i.e. 
.
Replace by 
to get
Now use injectivity to get
. Taking a similar consideration as before with 
we see that 
. Therefore 
, so 
becomes Cauchy FE. Extend by 
for all 
and note that extended satisfies Cauchy on the reals, and 
for all 
, so is linear, from which we conclude that 
, as desired.
such thatfor all positive real numbers and .
is the only solution to the given equation. Note that it indeed works.
denote the given assertion.
Therefore, let 
. Then we have 
and some 
.
is injective . Now 
imply that . So 
and hence is bijective .
.
. Then, 
.
. (By induction on this, we get 
and observe that:
Now using 
and 
.
. So , iterating on both sides we get 
. is an involution. Now comparing and 
we see that
Plugging this back in we get 
and 
.
.
, we get unbounded values of , and for a fixed value of , we get that is surjective over 
. So, letting 
tend towards infinity gives surjectivity.
, 
gives 
, so is bijective.
. As varies for fixed , we get that the image of 
is 
. Thus, if 
, 
, so is increasing. As it is both increasing and bijective, our function must be continuous.
. This gives that 
, so it is fixed as varies. Note that we must have 
to make sure the argument is positive, and as approaches this lower bound, the LHS gets arbitrarily close to 
. The RHS cannot be less than , since it will otherwise be exceeded as approaches 
. Likewise, it can't be more than . So, we must have 
. As 
, LHS approaches , so we can get by similar logic that it must always be . Thus, 
.
, which gives 
. As , LHS approaches 
while RHS is always larger, but approaches 
by above. Using a similar argument, if 
, then we eventualy have RHS>LHS, and if 
, we have the opposite. Hence, 
. is both an involution and increasing. If it is nonconstant, though, we can find 
such that 
. Thus, is the only solution.
such that
for all positive real numbers and . is injective function.
the given assertion.
: 
so 
, so is injective.
. is injective we have that 
.
so 
.
so 
.
it's simple to show that 
so 
. we have that 
so 
.
, thus 
. and 
.
etc.
denote the given assertion, by comparing ![\[P\left(\frac{1}{x}, \frac{1}{f(1)}\right) \implies \frac{1+xf\left(\frac{1}{f(1)}\right)}{f\left(\frac{1}{f(1)}\right)}=f(f(1)+f(x))\]](http://latex.artofproblemsolving.com/8/1/1/811b4efc48217026e70946539a8d3ea55150d4b8.png)
and ![\[P\left(1,\frac{1}{f(x)}\right)\implies \frac{1+f\left(\frac{1}{f(x)}\right)}{f\left(\frac{1}{f(x)}\right)}=f(f(x)+f(1))\]](http://latex.artofproblemsolving.com/1/9/8/198edb0bf4d14a9b3080c10c46ecc4ef6a437e0a.png)
we will get 
where 
From here, it's also clear that is injective. Also, by comparing ![\[P(f(x),1) \implies \frac{f(x)+f(1)}{f(x)f(1)}=f\left(1+f\left(\frac{1}{f(x)}\right)\right)\]](http://latex.artofproblemsolving.com/b/c/6/bc6babd50b111a2e007f9b1a01d9e36524efd067.png)
and ![\[P(f(1),x) \implies \frac{f(1)+f(x)}{f(1)f(x)}=f\left(\frac{1}{x}+f\left(\frac{1}{f(1)}\right)\right)\]](http://latex.artofproblemsolving.com/e/7/d/e7dd89e6fa05585f0b6a8978c7df2c4c89855805.png)
we will get 
where 
Therefore, we have 
or equivalently, 
Since this holds for all positive real , it must be the case where 
which implies 
Now notice that ![\[P\left(\frac{1}{x}, \frac{1}{f(x)}\right) \implies f(2f(x))=2x\]](http://latex.artofproblemsolving.com/7/2/f/72f36b565fb2f8b0259131176877377cbd21bd04.png)
and so 
Then, ![\[P\left(\frac{1}{x}, \frac{1}{f(3x)}\right)\implies f(f(3x)+f(x))=4x=f(2f(2x)) \implies f(3x)+f(x)=2f(2x)\]](http://latex.artofproblemsolving.com/2/4/b/24b751b1a41faa66d24c744f27e7432a880a5a01.png)
and ![\[P\left(\frac{1}{2x}, \frac{1}{f(4x)}\right)\implies f(4f(x)+f(2x))=f(f(4x)+f(2x))=6x=f(2f(3x)) \implies 4f(x)+f(2x)=2f(3x)\]](http://latex.artofproblemsolving.com/f/0/6/f0653891e21e5989dcf4f515163269a8c0da583d.png)
give us ![\[4f(2x)-2f(x)=2f(3x)=4f(x)+f(2x) \implies f(2x)=2f(x) \ \ \forall x\in \mathbb{R}^{+}.\]](http://latex.artofproblemsolving.com/1/7/2/1724321d59be8c5e7e5cd64552f1d69de27b4675.png)
Hence, 
and 
Lastly, ![\[P\left(f\left(\frac{1}{x}\right), \frac{1}{y}\right)\implies f(x+y)=\frac{f\left(\frac{1}{x}\right)+f\left(\frac{1}{y}\right)}{f\left(\frac{1}{x}\right)f\left(\frac{1}{y}\right)}=f(x)+f(y)\]](http://latex.artofproblemsolving.com/7/2/f/72fae9afa4e92a0b01b59a72782717e96881952c.png)
implies is additive over the positive reals, thus is linear and letting 
where 
are constants, we see that since 
, 
is the only solution. 
Now put instead of in equation. Let 
.
then 
implies that is injective.
Let 
be .
which holds for any 
which implies that is linear so 
.
which since are constant but 
are not implies that 
so 
so 
which clearly works.
, which also implies injective as a result
, we get 
for some constant over all values of 
, so 
implies 
, we have 
gives us 
, Since is surjective, varying the value of , we have that is a strictly increasing function
where 
Then, note 
contradiction. The same holds for when 
for all values of 
such thatfor all positive real numbers and . be the assertion of the given FE, which is 
Assume that , then 
and 
give or is injective.
and 
, we easily get 
Similarly, comparing 
and 
, we get 
for all . Let , we get 
, so 
and gives 
gives 
Plugging by 
into the above FE and changing the role of , by the injectivity, we get 
If 
, 
and the injectivity give 
, absurb. Thus 
, so 
gives 
immediately implies is additive, so after checking, we get for all is a solution.
. is bijective.
. For surjectivity, observe we can hit any value above 
, since we can make as small as we want we are done.
, we have 
as is a positive real. Then we have 
, so letting approach zero gives 
. Now substitute 
, so we have 
for 
. Now observe each pair 
results in exactly one value of , which is the same as the value of given by 
, but also this value is uniquely determined by so is constant, thus 
, but since the left hand side gets as small as we want we must have and 
.
, then we have 
, so 
giving 
. is the identity. Assume 
but 
. Now we have 
, giving 
. Clearly, cannot be 
, so 
for 
. Likewise, assume 
, which gives 
, so iff 
. This fact carries the rest of the solution, we can now proceed with a standard rational extension. over rationals. Let 
be the assertion 
. Now 
gives 
, then we can use the assertion 
, being 
to get 
. Now we can always induct, do 
to get 
and 
to get 
. Now to get all rationals we induct on the denominator, we can use 
to get all rationals with denominator , then use 
and 
to get all rationals with denominator 
, and so one and so forth we win.
for all reals. First take 
. Assume 
. Then there exists some rational 
with 
, contradiction. Symmetrical argument proves for . Now consider 
, this gives 
, giving 
by injectivity.
![\[\frac{1}{f(y)}+x=f(\frac{1}{y}+f(x))\]](http://latex.artofproblemsolving.com/d/6/1/d61b799f90102cf14a4dda741b04b6fd11dfb855.png)
Answer is which holds. Let be the assertion. is bijective., compare 
and 
to get contradiction. Fix and increase to see that takes all sufficiently large positive values. We can pick 
sufficiently small thus, is surjective..
in order to observe that![\[\frac{1}{f(\frac{1}{f(y)})}+x=f(f(x)+f(y))=\frac{1}{f(\frac{1}{f(x)})}+y\implies x-\frac{1}{f(\frac{1}{f(x)})}=y-\frac{1}{f(\frac{1}{f(y)})}\]](http://latex.artofproblemsolving.com/e/0/e/e0ec6817e4de007d6120947e82374c8453bdbf2a.png)
Since this implies 
is constant and it's smaller than any positive , it must be a nonpositive constant. Let 
or 
where 
. By symmetry and injectivity we have![\[f(\frac{1}{x}+\frac{1}{y+c})=f(\frac{1}{x}+f(\frac{1}{f(y)}))=\frac{1}{f(y)}+\frac{1}{f(x)}=f(\frac{1}{y}+f(\frac{1}{f(x)}))=f(\frac{1}{y}+\frac{1}{x+c})\]](http://latex.artofproblemsolving.com/0/2/2/022f9c23f19b6522c4a79de2defeed855d4c8212.png)
Thus, 
is constant which requires . So 
as we have claimed. is involution and 
. gives 
. We get 
and injectivity implies 
is independent of . Let 
. Since 
we observe 
. Thus, 
.![\[f(f(x))+f(\frac{1}{y})+d=f(f(x)+\frac{1}{y})=x+\frac{1}{f(y)}\]](http://latex.artofproblemsolving.com/6/6/2/662168d5b320e079781cb5a5f7d3c51f3fe62441.png)
and 
. Since 
, we must have 
which implies 
. Thus, is an involution and . is additive.
yields 
. is additive and takes values on positive reals, is Cauchy function hence which implies 
. So is the only solution as desired.
,
.
,
forall 
forall 
,
,
,
,
(from here, we easily get that: 
)
,
,
,
,
,
,
,
,
and let 
.
.![$P(x,\frac{1}{f(y)})\implies f(f(x)+f(y))=x+y\implies [a+b=c+d\iff f(a)+f(b)=f(c)+f(d)]\implies \exists g:\mathbb{R}\rightarrow \mathbb{R},g(a)=f(x+a)-f(x)$](http://latex.artofproblemsolving.com/7/5/8/7580abc760b6c61ee8fad23138f5e465924202db.png)
.
.
and so 
,
.
which clearly satisfied the condition.