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 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 .
is an infinite sequence of integers satisfying the following two conditions:
divides 
for 
such that 
for all 
such that 
for all .
with complex coefficients for which all complex roots of the polynomials and 
have absolute value 1.
Show that there exists a constant 
such that for all positive integer 
![\[\sum_{k\le n^{\varepsilon}}(n\text{ mod } k)>cn^{2\varepsilon}.\]](http://latex.artofproblemsolving.com/f/e/9/fe933e886e28c2ecf1fa880d0e650ac7a2998aaa.png)
Proposed by Cheng Jiang
such that 
be an isosceles trapezoid with 
. Let be the midpoint of . Denote by 
and the circumcircles of the triangles 
and 
, respectively. Let be the crossing point of the tangent to at with the tangent to at . Prove that 
is tangent to .
, adjacency iff the Euclidean distance between the points is 
) and this will follow as a consequence.
occur as unit distance graphs, then so does 
( here $\G\times H$ is described as 
or 
).
by the complex numbers 
and for 
by the numbers 
. we claim that for some choice of 
, 
will do the job(a suitable rotation).
iff either 
or 
. clearly if the condition holds then the adjacency is satisfied. suppose 
and that the corresponding complex numbers are at a distance from one another. Then this gives a quadratic in 
and hence 
can take only finitely many values here.ruling this out for each such set of 
hence rules out finitely many values of and therefore a suitable choice exists.
for any given . since we can form the graph 
, we can form 
(the unit cube) and that solves the problem.
has the form 
where 
is unknown set of distinct unit vectors in 
. We can build 
inductively, starting from the empty set and adding vectors to it, one by one. We just need to make sure that two thing are respected:
vectors in 
are distinct;
in the sum changes from 
to ). points at will be at unit distance from exactly points corresponding to sums at which one and only one of coefficients differs from coefficients of this point. However, respecting these conditions is not hard because 
and for each new vector being added to there is at most some finite set of forbidden endpoints given by sums/differences of already determined vectors but the rest of the (infinite) unit circle is permissible.
