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.
, define 
and 
, where 
and 
denote the product and sum of the digits of , respectively. Find all solutions to 
be a triangle such that![\[ \left( \cot \dfrac{A}{2} \right)^2 + \left( 2\cot \dfrac{B}{2} \right)^2 + \left( 3\cot \dfrac{C}{2} \right)^2 = \left( \dfrac{6s}{7r} \right)^2, \]](http://latex.artofproblemsolving.com/6/9/b/69b1964be5f92eb44a36b0b8604bf473fe27e210.png)
and 
denote its semiperimeter and its inradius, respectively. Prove that triangle is similar to a triangle 
whose side lengths are all positive integers with no common divisors and determine these integers.
be a triangle with circumcenter 
. Point 
is the intersection of the parallel line from to 
with the perpendicular line to 
from 
. Let 
be the point where the external bisector of 
intersects with . Let 
be the projection of onto 
. Prove that the lines 
have a common point.
be an acute triangle with 
and let 
and be its orthocenter and circumcenter, respectively. Let 
be the circle 
. The line 
and the circle of radius centered at 
cross at 
and 
, respectively. Prove that , the circle on diameter 
and circle 
are concurrent.
for which the following statement holds: among any five positive real numbers 
(not necessarily distinct), one can always choose distinct subscripts 
such that![\[ \left| \frac{a_i}{a_j} - \frac {a_k}{a_l} \right| \le C. \]](http://latex.artofproblemsolving.com/a/e/e/aeeab8e846634c7922e9312ef139739ad3fbe6eb.png)
be a triangle with circumcircle 
, incenter 
, and -excenter 
. Let the incircle and the -excircle hit 
at 
and 
, respectively, and let 
be the midpoint of arc without . Consider the circle tangent to at and arc 
at . If 
intersects again at 
, prove that 
and 
meet on . be an acute triangle with 
. be the circumcircle circumscribed to the triangle and the midpoint of the smallest arc of this circle. Let and points of the segments and respectively such that 
. Let 
be the second intersection point of the circumcircle circumscribed to 
with . Let 
and be the intersections of lines 
and 
with other than 
, respectively. Let 
and be the intersection points of lines 
and 
with lines and respectively. Show that the 
line passes through the midpoint of
are real numbers, find the smallest possible value of the expression:
and 
be prime numbers. Prove that if 
is a perfect square, then 
is also a perfect square.
and are primes and is a perfect square, prove that 
is a perfect square.
be a positive integer. Aslı and Zehra are playing a game on an 
chessboard. Initially, 
stones are placed on the squares of the board. In each move, Aslı chooses a row or a column; Zehra chooses a row or a column. The number of stones in each square of the chosen row or column must change such that the difference between the number of stones in a square with the most stones and a square with the fewest stones in that same row or column is at most 1. For which values of can Aslı guarantee that after a finite number of moves, all squares on the board will have an equal number of stones, regardless of the initial distribution?
reals, and 
.
?
for 
. It is easy to check that these work.
denote the given assertion. 
gives 
and 
gives 
. From 
we get 
. Thus becomes 
. It follows that![\[
f(x+y)=f(x)+f(y)\ \ \ \ \ \ \ \forall(x,y)\in\mathbb{R}^+\times\mathbb{R}^-.\tag{*}
\]](http://latex.artofproblemsolving.com/9/b/3/9b382f12b114dc17049ff9b665afadca3ebf9ae2.png)
for all 
.
and 
are both positive, 
so we are done. If exactly one of and are positive, we are done by 
. If and are both negative,![\[
f(x+y)=-f(-x-y)=-f(-x)-f(-y)=f(x)+f(y)
\]](http://latex.artofproblemsolving.com/1/b/6/1b6816748309b461ab5a46c76e6e997ac0612a47.png)
by case 
so we are done. 
so 
, as desired. 
for any constant 
, and it's easy to see that this satisfies the equation.
, we get , and plugging in 
, we get 
. From this, we get 
, giving 
. and then , we get 
, which further becomes since . This implies that if 
or 
, we have 
.
, and setting 
gives us that also holds true when both 
. Finally, by negating everything and using , this also holds true when both 
. Hence, 
satisfies Cauchy.
into and use the fact that satisfies Cauchy to get ![\[f((x+1)^2)=f(x^2)+f(x)+f(x+1)=(x+1)f(x+1)=xf(x)+xf(1)+f(x)+f(1),\]](http://latex.artofproblemsolving.com/4/0/2/4027f798cfe379d43cf7f585a710d2d8933d86c6.png)
and by using , we can cancel things to get 
, giving where 
.
. These work since 
. Now we need to show these are the only solutions. Let x = 0, we get 
. Let y = 0, we get 
. By these two we get that f is odd and also that 
. So now using what we got we can write the starting equation as 
. Using this and the fact that f is odd we get that f is additive. Now plugging in x = x + 1 in and using that f is additive we get 
which after clearing things up gives us that 
there are no other solutions and we are ready.
for some constant . It is easy to verify that this satisfies the given condition. Now we will prove that it is the only solution.
.
..
for positive . Since is odd, this is also true for negative .
, and is odd, we can directly use Cauchy to finish 
and 
. It is easy to verify these both work. Now, we will show that these are the only functions satisfying the given conditions. Plugging in 
,![\[f(x^2) = xf(x)\]](http://latex.artofproblemsolving.com/d/7/8/d784a31622ac92abd632d594076ec4c054df76be.png)
Plugging in 
,![\[f(-y^2) = -yf(y) = -f(y^2)\]](http://latex.artofproblemsolving.com/8/f/5/8f502d7303fdc1659acd551919d8cfe6e2cddc62.png)
So, 
for all 
, thus it is odd. Now, we prove the following claim: is additive.![\[f(x^2 - y^2) = f(x^2) - f(y^2) \implies f(a - b) = f(a) - f(b)\]](http://latex.artofproblemsolving.com/6/3/b/63b96640347c304a638d22b2829295552979b029.png)
for all nonnegative reals 
. Now, set 
, for nonnegative 
. This gives![\[f(y) = f(x+y) - f(x) \implies f(x) + f(y) = f(x + y)\]](http://latex.artofproblemsolving.com/b/c/6/bc6e20512b330e88a99c1b414fb6b1b117c2f51e.png)
for all 
. Since is odd, we have![\[-f(-x-y) = f(x + y) = f(x) + f(y) = -f(-x) - f(-y) \implies f(-x-y) = f(-x) + f(-y)\]](http://latex.artofproblemsolving.com/b/6/9/b69fa62cd1336afe1422f32c8a9fea2ef40b6dee.png)
so is also additive over all the nonpositive reals. Now, rewrite![\[f(x) - f(-y) = f(x + y) \implies f(x) = f(x + y) + f(-y)\]](http://latex.artofproblemsolving.com/f/3/1/f3112d0d7d22c07376e163921b03e3074db3a0f2.png)
We can set 
for any 
, therefore the above equation rewrites as![\[f(a -y) = f(a) + f(-y)\]](http://latex.artofproblemsolving.com/7/d/f/7df76e980385a4b9984df6a66eb89d2ba634068d.png)
So, 
is also true for all pairs 
with 
. is additive over all the real numbers. Thus, we have![\[f((x + 1)^2) = f(x^2) + f(2x) + f(1) = xf(x) + 2f(x) + f(1)\]](http://latex.artofproblemsolving.com/f/3/3/f33b06811523575e439f27b19d791f52441d6ac3.png)
![\[f((x + 1)^2) = (x + 1)f(x + 1) = (x + 1)f(x) + xf(1) + f(1)\]](http://latex.artofproblemsolving.com/7/d/4/7d4eb1e6de5ca74c8303a4d1663dd507105790e3.png)
Setting the two equations equal, we get![\[(x + 2)f(x) = (x + 1)f(x) + xf(1) \implies f(x) = xf(1) \implies f(x) = cx\]](http://latex.artofproblemsolving.com/9/0/7/907af73bb2cfd7c2138784a7262ae84d6bc1f711.png)
for some 
, so we are done.
be the assertion that the functional equation holds for 
. Plugging 
gives and plugging 
gives is odd. Plugging gives for all real . So for nonnegative real 
gives 
, and we can extend this to all negatives because is odd. So for all real . Plugging this into the original equation gives 
, so is the only solution.
denote the assertion. Then 
Therefore 
for all nonnegative 
Therefore, for positive we have 
As 
is odd, it follows that is additive. Hence 
This solution clearly works.
for any constant which clearly works. Now we prove they are the only ones.
and 
for nonzero implies is odd. Now gives 
so in fact . Imposing 
, replacing 
yields for all 
.
for 
, we have![\begin{align*}
2 f(x) + f(1) &= f(2x + 1) \\
&= (x + 1) f(x + 1) - x f(x) \\
&= (x + 1) [f(x) + f(1)] - x f(x) \\
&= f(x) + x f(1) + f(1) \\
\end{align*}](http://latex.artofproblemsolving.com/6/7/b/67b8eee6689b4cf05c8d019adca40964625e3f8b.png)
so 
for . But is odd, so for some constant as needed.
.
as seperate cases, we see that 
Thus, we also see that 
or 
For 
we see that the domain is all positive, and thus, 
Now we substitute 
to get that 
Thus, we are done.
be the set of real numbers. Determine all functions 
such that ![\[ f(x^2 - y^2) = x f(x) - y f(y) \]](http://latex.artofproblemsolving.com/6/8/f/68f78e0df627f6ac04052433c01627bf16f0af8e.png)
for all pairs of real numbers 
.
(1)
(2).
be positive reals. Then
,
.
(3).
(4)
(5).
can be positive or negative.
,
to get 
for 
.
,
to get 
.
,
to get 
.
is constant, let 
.
for all real 
.
for all 
for every real x.
.
such that 
and 
we find 
,
.
(see Paladin8's solution) we have 
,
.
,
,
.
and using 
you get the result...
a
,
are linear functionals, all such functions f would satisfy the functional equation. I do have a question, though: do there exist linear functionals from R to R that are not linear functions?
.
,
and similarly, 
.
.
and 
,
.
is a continuous function. Hence, we can take the derivative of it. Letting 
,
is an infinitely small number, we have that 
.
is constant, so 
is constant. It follows that 
is linear for positive 
.
,
.
,
.
.
is negative, then 
.
.
"
and 
are collinear since any two points are collinear. We prove through induction that 
is on this line. By the inductive hypothesis, 
is on the line. Now, 
,
,
given a rational and 
and 
to our conjecture from Cauchy's functional equation that 
.
.
.
for one of 
.
gives us 
,
.
when at least one of 
and thus 
after performing the same manipulation on the 
term. Therefore, as both terms can be negative, this finishes off the second case.
which is also equal to 
.
so the solutions are 
over all real constants 