AoPS Academy
+1 w
[/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.
, where 
is a prime number. Show that if 
is an even number, then the polynomial ![\[-1+\prod_{k=0}^{n-1} P\left(X^{p^k}\right)\]](http://latex.artofproblemsolving.com/5/d/f/5df80f885e63852e4f53a6145dafd18d63dc3526.png)
is divisible by 
.
reals, and 
.
?
be a triangle with 
. Let 
be the orthocenter of . Point
is the midpoint of 
and point 
lies on the side 
such that 
. Prove that
.
be a regular polygon, and let 
be its set of vertices. Each point in is colored red, white, or blue. A subset of is patriotic if it contains an equal number of points of each color, and a side of is dazzling if its endpoints are of different colors. is patriotic and the number of dazzling edges of is even. Prove that there exists a line, not passing through any point in , dividing into two nonempty patriotic subsets. be an integer greater than 
. For a positive integer 
, let 
. Suppose that there exists a 
-element set 
such that is an -element subset of 
; shares at most one common element; is contained in exactly two elements of . in terms of .
, define 
and 
, where 
and 
denote the product and sum of the digits of , respectively. Find all solutions to 
be a point inside a triangle . Let 
meet at 
, let 
meet 
at 
, and let 
meet 
at 
. Let 
be the point such that is the midpoint of 
, let 
be the point such that is the midpoint of 
, and let 
be the point such that is the midpoint of 
. Prove that points 
cannot all lie strictly inside the circumcircle of triangle .
lies inside triangle so that 
. The point 
is such that 
is a parallelogram. Prove that 
.
be an acute-angled triangle with 
. Let be the orthocenter of triangle , and let 
be the midpoint of the side . Let be a point on the side and a point on the side such that 
and the points , , are on the same line. Prove that the line 
is perpendicular to the common chord of the circumscribed circles of triangle and triangle 
.
balls to 
boxes as she wishes. After that, Aslı and Zehra make alternated moves which consists of taking a ball in any wanted box starting with Aslı. One who takes the last ball from any box takes that box to herself. What is the maximum number of boxes can Aslı guarantee to take herself regardless of Zehra's moves?
be a triangle with 
, and let be the foot of the altitude from 
. A point is chosen inside the triangle 
so that 
bisects 
. Let be the intersection point of the lines 
and . Let 
be the semicircle with diameter that meets the segment 
at an interior point. A line through is tangent to at . Prove that the lines 
and meet on .
is a rational number.
and other permutations.
![\[ f\left( f(x) y + \frac xy \right) = xyf(x^2+y^2). \]](http://latex.artofproblemsolving.com/b/6/5/b656324b6499f7457b7b9eccbf5eaa01b6ad6e59.png)
for every 
such that 
which is the polynomial 
, sufficently small we can get that value to be negative, and as the coefficient of 
is positive there exists a positive value of which satisfies the equation. Pluging that in gives 
.
there must exist a value of such that 
, this is true because for fixed positive , 
, and 
so the result follows from IVT. Now these two equal implies 
or 
. So plugging 
to 
gives 
which works.
Pick such that 
Note that by Descartes rule of signs, we know that there exists a positive root, . Take 
to get that 
Thus, 
If we plug this back into our old equation, we see that 
However, 
and so 
. Let 
be defined by 
. Note that 
is a continuous function of 
. Then 
while:
so there is some 
such that 
. Now rearranges into 
so:
Then 
, so from we get 
, which is indeed a solution.
and 
. Prove that
and 
. Prove that
denote the given assertion.
in 
.
to be nonpositive.
contradiction.
,
But then 
satisfies 
,
Since this function works, it comprises the full solution set.
which clearly works.
.
This can be checked to exist by IVT. Taking such a 
which gives the desired result
which works.
.
,![\[ Q(y) = y^3 - f(t)y^2 + t^2y - t. \]](http://latex.artofproblemsolving.com/a/e/b/aeb44e83e41538ce323a15898543ec7d60ceaf8e.png)
Since 
implies 
,
with![\[ f(t)y + \frac ty = t^2 + y^2. \]](http://latex.artofproblemsolving.com/c/1/a/c1ad460abb84b04a78eca7b5c38441975f0d1e46.png)
By 
,
,
.![\[ \left(\frac 1t\right)^3 - \frac{f(t)}{t^2} = 0 \implies f(t) = \frac 1t \]](http://latex.artofproblemsolving.com/b/5/1/b511a7bcfed7e18991bb8be2c8f1e80bf6ef3b43.png)
for all