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 the set of all positive integers. Find all functions 
such that for any positive integers 
and 
, the following two conditions hold:
, and
, 
, and 
are equal.
be a triangle with 
. The 
-excircle touches side 
and line 
at 
and 
, respectively. The 
-excircle touches side 
at 
. Let lines 
and 
meet at 
. Prove that 
.
be three diameters of the circumcircle of an acute triangle . Let be an arbitrary point in the interior of 
, and let 
be the orthogonal projection of on 
, respectively. Let 
be the point such that is the midpoint of 
, let 
be the point such that is the midpoint of 
, and similarly let 
be the point such that is the midpoint of 
. Prove that triangle 
is similar to triangle .
of positive integers such that 
.
players. Every player played every other player once, with no draws. In addition, each player had a numerical rating before the tournament began, with no two players having equal ratings. It turns out there were exactly 
games in which the lower-rated player beat the higher-rated player. Prove that there is some player who won no less than 
and no more than 
games.
, 
, and 
be positive integers. Prove that ![\[ \sum_{i = 1}^n \min\left(\left\lfloor \frac{x}{i} \right\rfloor, m \right) = \sum_{i = 1}^m \min\left(\left\lfloor \frac{x}{i} \right\rfloor, n \right). \]](http://latex.artofproblemsolving.com/b/7/a/b7aebfe11f646d604d2e8cb9d5bd70ec796949e8.png)
of points in the plane is balanced if, for any two different points 
and in , there is a point in such that 
. We say that is centre-free if for any three different points , and in , there is no points in such that 
.
, there exists a balanced set consisting of points. for which there exists a balanced centre-free set consisting of points.
such that
are all powers of 
. be an integer and let 
be the complete graph on vertices. Each edge of is colored either red, green, or blue. Let denote the number of triangles in with all edges of the same color, and let denote the number of triangles in with all edges of different colors. Prove![\[ B\le 2A+\frac{n(n-1)}{3}.\]](http://latex.artofproblemsolving.com/6/5/9/659d0199c3d685d43f1ced3704dfa59b02b586b3.png)
(The complete graph on vertices is the graph on vertices with 
edges, with exactly one edge joining every pair of vertices. A triangle consists of the set of 
edges between 
of these vertices.)
such that the equation ![\[x^{2}-(a^{2}+b^{2}+c^{2}+d^{2}+1)x+ab+bc+cd+da=0 \]](http://latex.artofproblemsolving.com/9/0/4/904d1b23d5c744118a7d68df1e2e94bddac8c1cd.png)
has an integer solution. Prove that the other solution is integer too and both solutions are perfect squares.
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 an acute-angled triangle with circumference 
. Let the angle bisectors of 
and 
intersect again at 
and 
. Let 
be the intersection point of these angle bisectors. Let 
and 
be the respective reflections of and in and . Prove that the center of the circle passing through , , lies on the altitude of triangle from ., there is exactly one card in the deck that has written on it. Now two players draw disjoint sets and of 
cards each from this deck. We would like to define a rule that declares one of them a winner. This rule should satisfy the following conditions:
cards: if the cards are laid down in increasing order face down and we are told which card belongs to which player, but not what numbers are written on them, we can still decide the winner.
and 
, and 
for all 
, then beats .
from the deck, beats and beats then also beats . and such that beats according to one rule, but beats according to the other.
and 
for 
. Prove that the sequence 
is increasing.
If you just only draw on your paper (and maybe wrong) and think that wrong, you should prove your statement before saying something. If you draw on computer, please give me the figure.
to 
same with 
, it's will be IMO 2009 problem 2. so, any idea now, I still thinking and cannot find how to apply that problem for this 
be midpoints of 
，
，
，
be midpoints of 
，
be the circumcircle of 
，
be parallelogram，
，
.
，
，the 9-point center is the circumcircle of 
，so， is midpoint of 
.
. On another hand，we can easy found 
is a isosceles trapezoid，and 
， is on perpendicular bisector of 
，the same way， is also on perpendicular bisector of 
. is midpoint of 
，therefore，we get 
. Q.E.D
be midpoints of ，，， be midpoints of ， be the circumcircle of ， be parallelogram，， .，，the 9-point center is the circumcircle of ，so， is midpoint of . . On another hand，we can easy found is a isosceles trapezoid，and ， is on perpendicular bisector of ，the same way， is also on perpendicular bisector of . is midpoint of ，therefore，we get . Q.E.D
,
,
,,
is the food of the perpendicular line from to .
is the line pass throught and paralley 
, meet 
at and meet at , we have is the midpoint of 
,
meet at , apply "Butterfly Theorem" we have is the midpoint of 
so we have 
so 
![\[ Q.E.D\]](http://latex.artofproblemsolving.com/9/9/4/9944bb829d83b6c8c3731e0b66fafd134f718701.png)
is the midpoint of 
is the mid point of , you should read "Butterfly Theorem" again. is the midpoint of because 
and is midpoint of 
be a triangle and its circumcircle. 
. Let be the midpoint of , be the midpoint of 
, be the midpoint of . the circumcircle pass through 
meets at 
. Prove: 
