[/quote]
,
that satisfy the equation 
?
and 
Find the value of 
such that 
be a triangle with a circle 
with center 
tangent to sides 
at 
respectively. Suppose the circle with diameter 
intersects the circumcircle of 
is the reflection of 
over 
lie on 
are parallel to 
.
lie on the circumcircle of 
![\[f(xf(y) + x) = xy + f(x)\]](http://latex.artofproblemsolving.com/9/9/f/99f580ebc50846e6bc2c004667559922749a4dfa.png)
be a sequence of positive real numbers such that for every 
,![\[
a_n = \max_{1 \leq i \leq 2025} a_{n-i} - \min_{1 \leq i \leq 2025} a_{n-i}
\]](http://latex.artofproblemsolving.com/e/3/0/e30e9e533eb9a41cf847484e676ac4522edda665.png)
Prove that there exists a natural number 
such that for all 
,![\[
a_n < \frac{1}{1404}
\]](http://latex.artofproblemsolving.com/0/1/f/01f1a81eed2230332d51a15501ef2a6ad3a7ec82.png)
be an acute triangle with 
and let 
be its orthocenter. Let 
be a point on the perpendicular bisector of 
such that 
and 
are on different sides of 
,
a point on the perpendicular bisector of 
such that 
and 
a point on the perpendicular bisector of 
such that 
and 
lie on the opposite side of 
.
and 
square is filled with numbers 
.
squares is summed,and arranged in an increasing order. Is it possible to obtain the following sequences as a result of this operation?
lie on the perpendicular bisector of 
lie on a circle. Suppose 
are perpendicular to sides 
at points 
The tangent lines from points 
intersects at point 
Prove that: 
are parallel.
,
be four points in the plane, with 
and 
.![\[\frac{AB \cdot CD}{AC \cdot BD}, \]](http://latex.artofproblemsolving.com/f/3/c/f3cf9729b32b210faeea6e2d23da3582be2ae061.png)
and 
are orthogonal. (Intersecting circles are said to be orthogonal if at either common point their tangents are perpendicuar. Thus, proving that the circumcircles of the triangles 
such that for any positive real numbers 
and 
,
Proposed by Athanasios Kontogeorgis, Grecce, and Dorlir Ahmeti, Kosovo
of positive real numbers satisfies ![\[a_{k+1}\geq\frac{ka_k}{a_k^2+(k-1)}\]](http://latex.artofproblemsolving.com/0/3/9/03909a5f60d944063182ec97cf73f283178e8d4d.png)
for every positive integer 
.
for every 
.
,
and 
be the respective intersections with BC of the internal angle bisector, external angle bisector, and the median from A. The circumcircle of 
intersects 
a second time at point 
different from A. Define 
and 
analogously. Prove that the circumcenter of 
lies on the Euler line of ABC.
Y by Wave-Particle, Ankoganit, dgrozev, Adventure10, Mango247, TheHimMan, ohiorizzler1434
Petya and Vasya play the following game. Petya conceives a polynomial 
having integer coefficients. On each move, Vasya pays him a ruble, and calls an integer 
of his choice, which has not yet been called by him. Petya has to reply with the number of distinct integer solutions of the equation 
. The game continues until Petya is forced to repeat an answer. What minimal amount of rubles must Vasya pay in order to win?
(Anant Mudgal)
(Translated from here.)
having integer coefficients. On each move, Vasya pays him a ruble, and calls an integer 
of his choice, which has not yet been called by him. Petya has to reply with the number of distinct integer solutions of the equation 
. The game continues until Petya is forced to repeat an answer. What minimal amount of rubles must Vasya pay in order to win?(Anant Mudgal)
(Translated from here.)
This post has been edited 2 times. Last edited by anantmudgal09, Feb 5, 2018, 1:11 PM
Reason: Titular reference to "Game of Thrones"
Reason: Titular reference to "Game of Thrones"
rubles. We'll use the following notation: for an integer 
is the number of distinct integer roots of 
or more distinct integer root for some ![$P(x)\in\mathbb Z[x]$](http://latex.artofproblemsolving.com/e/6/6/e66709403885b04838c3e6b96cf89eacdbecdc44.png)
and 
,
and 
have no integer root.
.
for integers 
'
'![$Q\in\mathbb Z[x]$](http://latex.artofproblemsolving.com/5/5/8/5582e4242918643944e8d9790f4e73e49b9d2052.png)
.
,
for each 
;
in the first three moves. If 
,
by the lemma and Vasya wins. If one of 
and 
is 
,
;
in the next move and win, because then 
by our lemma. So suppose 
are all 
.
are 
in some order. Then at least one of 
.
can't be 
;
,
,
.
in the next two moves and hopes to win, then Petya could smash his hopes by conceiving the polynomial 
Indeed, 
has two roots 
and 
and 
has no roots at all, because for 
,
.
in this order. To see his answers are consistent, let 
be the numbers asked by Vasya in this order. Then, for 
where 
is a positive integer, so that 
and 
and 
so we are done.
and 
.
,
and 
be such that 
.
,
,
,
.
,
that 
has no solutions and we are done in at most 
or 
(
or 
has no integer roots),by the same logic in the first case.
.
,
has 
solutions respectively.
,
where 
.
.
simultaneously. Since 
and 
,
if the answer is nonzero. 
,
fail because then 
,
.
and 
.
.
which for similar reasons as before must either be 
,
which forces 
unless 
,
has at least 
has no integer solutions.
for integers 
and integer polynomial 
,
,
are distinct. 
first. If Petya responds with any integer at least 
and 
and be sure that both answers will be 
.
integers. Without loss of generality assume 
for some integer 
.
has precisely one, and as the second term is at least 
when it is positive, 
has no solutions.
.
attains 
,
,
never. Thus, Vasya does not win.
is sufficent, that is, no integer polynomial attains 
.
.
.
,
instead, we only have two spots that are valid spots for 
,
.
,
and 
,
so 
are consecutive and in particular 
,
with 
so 
,
both return 
,
.
return the same value or 
,
if 
otherwise.
