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.
such that for all reals 
, we have![\[f(x)^2+f(y)^2+f((x+y)^2)=2x^2+2y^2+f(xf(y)+y(f(x))\]](http://latex.artofproblemsolving.com/6/b/9/6b93f24d3061853a70965c4b57d3565d6646702f.png)
Shortly, if we define 
, the statement becomes:![\[g(x+y)+g(x-y)=2g(x)+2g(y)\]](http://latex.artofproblemsolving.com/3/b/4/3b4840c15e0cb0c3f61cf14b166fcd1d73a5369e.png)
Where 
(which makes 
bounded on some interval). I have proved that 
. From here I have to prove that 
, and I want to use the fact that is bounded on some interval, but the form of 
is not the form of regular Cauchy Equation. Does anyone know how to transform the form become Cauchy? So we can use the information of is bounded?
lies on side 
of triangle 
From points 
and 
, perpendiculars 
and 
are dropped to the line 
. Let 
be the midpoint of segment . Prove that 
be an acute triangle. Let 
be its incenter, 
and 
such that 
, 
, 
are collinear. Let 
such that 
. 
is defined similarly. Let 
such that 
. Prove that 
.
such that it satisfies 
where 
is a prime number.
: If then 
where 
. This implies that 
where 
. The equation becomes 
. If 
then has no solution because divides LHS but it does not divide RHS.
: Then 
. This implies that 
which implies 
. Hence 
. Now the equation 
can be written as 
. Since is prime and 
, this implies that 
. Hence the equation has no solution in this case.
: The equation becomes ![\[\frac{a^p}{p}=(p-1)!(p+1)(p+2)\cdots (2p-x)+1\]](http://latex.artofproblemsolving.com/d/6/7/d67a0510c1c944e6903be6d095e5450e709950b6.png)
where 
. From LHS we conclude that 
. Let 
be the highest power of that divides 
. Now the equation becomes ![\[p^{tp-1}k^p=(p-1)!(p+1)(p+2)\cdots (2p-x)+1\]](http://latex.artofproblemsolving.com/7/b/6/7b634dc78af920e362dbef7d0de7367ca68ce094.png)
where and not divisible by . From the above equation we conclude that 
for all 
. If it is not so, then the value of the expression 
, because 
will become a common factor. Therefore either 
, or 
and for all . then 
. Hence, this case is not possible. Hence 
must be 
.
then 
. This case is also not possible.
. In this case the equation becomes ![\[p^{p-1}-1=(p-1)!(p+1)\cdots (2p-x)\]](http://latex.artofproblemsolving.com/5/0/6/506174df72726dd23d39d3dbde1645dcad2e7513.png)
where . and 
be different points on a circle 
such that 
is not a diameter. Let 
be the tangent line to at . Point 
is such that is the midpoint of the line segment 
. Point 
is chosen on the shorter arc of so that the circumcircle 
of triangle 
intersects at two distinct points. Let be the common point of and that is closer to . Line 
meets again at 
. Prove that the line 
is tangent to .
be positive real numbers such that 
. If 
is the maximal value ![\[3\sqrt[3]{a} + 6\sqrt[3]{b} - \dfrac{c^2}{250}\]](http://latex.artofproblemsolving.com/7/0/8/708b68beacaa697ae96ec7da695d5d2c8ccb783a.png)
then find the value of 
points, where no 3 points are collinear. Each move consists of selecting 2 of the points and drawing a new line segment connecting them. The first player to draw a line segment that creates an odd cycle loses. (An odd cycle must have all its vertices among the 
points from the start, so the vertices of the cycle cannot be the intersections of the lines drawn.) Find all such that the player to move first wins.
be three pairwise distinct real numbers with 
be the foot of perpendicular from to the Euler line (the line passing through the circumcentre and the orthocentre) of an acute scalene triangle 
. A circle 
with centre passes through and , and it intersects sides 
and 
at 
and 
respectively. Let 
be the foot of altitude from to 
, and let be the midpoint of . Prove that the circumcentre of triangle 
is equidistant from and .
. Prove that:
be a scalene triangle with incircle 
tangent to segment at .Line 
intersects segment at 
.
lie on segment such that 
.Denote as the midpoint of arc 
of 
.Suppose 
intersects 
again at . is the reflection of over line 
. is the midpoint of segment 
.
intersects again at .Prove that: 
is tangent to .
Show that 
satisfying
Prove
![\[a^2+b^2+16c^2 = 9k^2 + 1.\]](http://latex.artofproblemsolving.com/5/4/5/545f8edbdebe4e759c5a70bd3d489e5d53c9147c.png)
yields 
.
,
must be divisible by 
,
are primes; thus, two of them must equal exactly 
![\[18+16c^2=9k^2+1\implies 17=9k^2-16c^2=(3k-4c)(3k+4c).\]](http://latex.artofproblemsolving.com/7/1/b/71bef806084778dee85166f201f023c8f79e611a.png)
We must then have 
and 
- solving yields 
and 
.
is a solution.
as well. Then ![\[3^2+b^2+16\cdot 3^2=9k^2+1\implies 152=9k^2-b^2=(3k-b)(3k+b).\]](http://latex.artofproblemsolving.com/5/6/0/5600ae0d937f7d9c246cc8748b9dd36d9371b59f.png)
The prime factorization of 
is 
.
and 
must have the same parity yields only two cases to check.
and 
,
and 
.
is prime, the triples 
work.
and 
,
and 
.
is prime, the triples 
both work.
that work subject to 
