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.
, 
, 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. be a positive integer. Prove that there exist three permutations 
, 
, and 
of 
such that ![\[\left|\sqrt{a_k}+\sqrt{b_k}+\sqrt{c_k}-2\sqrt{N}\right|<2023\]](http://latex.artofproblemsolving.com/9/8/0/98097fb721722b9103ef65d72322956877e262cb.png)
for every 
.
. Prove that for any given positive integer , the product
has at most distinct prime divisors., let 
be the sum of the digits of . Determines the smallest positive integer 
such that there are infinite positive integers for which you have 
.
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 .
such that if 
and 
are relatively prime divisors of 
, then 
divides .
be the circumcenter of 
and 
. From the parallelogram we get 
and 
, hence 
, 
and so 
. On the other hand, 
since 
from the parallelogram; also 
since is circumcenter. Therefore 
and thus 
, which completes the proof.
,
.
intersects 
at 
at 
passes from circumcenter of 