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.
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 .
is an infinite sequence of integers satisfying the following two conditions:
divides 
for 
such that 
for all 
such that 
for all .
, and express these only in 
, that is simple because they are only medians and the result follows.
be the centroid of triangle 
, 
the midpoint of 
and 
the midpoint of 
.
, so 
. This implies that 
if and only if 
is orthocenter of trianble 
. This is equivalent to 
. On the other hand, is orthocenter and is centroid of triangle 
. So, if and only if 
is median and altitude from in triangle . This is equivalent to 
, and to 
.
as the origin of the complex plane, and let 
be the coordinates of points 
respectively. Then we have ![\[d=\frac{a+b}{2} \text{ and } e=\frac{a+c+d}{3}=\frac{3a+b+2c}{6}.\]](http://latex.artofproblemsolving.com/8/9/1/8912f2c3e6900cc680fe60c6d132b803830a99fe.png)
Using the real product of complex numbers, if 
is the circumradius of triangle , then we have ![\[a\cdot a=b\cdot b=c\cdot c = R^2.\]](http://latex.artofproblemsolving.com/c/8/9/c899af10e64c453722ecc78b0c941e66cae26454.png)
Also note that 
, i.e. ![\[(a+b-2c)\cdot (3a+b+2c)=0.\]](http://latex.artofproblemsolving.com/f/2/8/f28b3b49b839c2b1ff1fc1b249382c42929499fe.png)
The last relation is equivalent to ![\[3a\cdot a+a\cdot b+2a\cdot c+3a\cdot b+b\cdot b+2b\cdot c-6a\cdot c-2b\cdot c - 4c\cdot c=0,\]](http://latex.artofproblemsolving.com/1/7/8/178640cb748434659678f80cb81cbb9edeb7a853.png)
which implies ![\[a\cdot b=a\cdot c (*)\]](http://latex.artofproblemsolving.com/e/2/8/e283699086fda0be1832ae8f3ce4aefaf91972db.png)
Now, observe that ![\[AB=AC \iff |b-a|^2=|c-a|^2.\]](http://latex.artofproblemsolving.com/0/4/a/04ab5ae490e38a0f82c4c0fc5749a96282b06541.png)
This yields ![\[(b-a)\cdot(b-a)=(c-a)\cdot(c-a)\]](http://latex.artofproblemsolving.com/5/9/e/59e147df80214cb700ad2b6e4d7509b62ec055de.png)
or ![\[b\cdot b-2a\cdot b+a\cdot a=c\cdot c-2a\cdot c+a\cdot c.\]](http://latex.artofproblemsolving.com/f/9/a/f9a8df43ac573f47f9e7b35fd608d6e60d387600.png)
Hence, we have ![\[a\cdot b=a\cdot c (**)\]](http://latex.artofproblemsolving.com/1/4/5/145b86d56e3780ff477becf3e8da3222989ca292.png)
From 
and 
we have 
. 
case, the other one can be done similarly (I swear I didn't read #2 )
, is its circumcenter, is the midpoint of and is the centroid of the triangle 
. Show that the lines 
and 
are perpendicular if and only if .
, where , let 
be the midpoints of 
. Let be a point on 
, such, 
. Let 
be the orthocenter of . Then, 
Back to the Problem: Let 
be midpoints of 
, then 
and 
are parallelogram. If 
and 
, then, 
are midpoints of , . 
and 
are concurrent, say at , which is centroid of and 
. 
is cyclic 
is orthocenter of . Now apply (Lemma)
and 
are perpendicular, then 
and write Pythagoras.
be the midpoint of and be the midpoint of 
.
therefore 
Let 
.
, as desired. 
for reals 
.
,
.
so 
.
is just 
.
is 
.
.
.
where 
.
.
.
and the slope of 
.
,
or
.
or 
.
which is equivalent to 
Notice that 
is pure imaginary. Or
Notice that 
,
and 
.
only if 
,
Now we prove the iff statement. Note that if
Then
If 
,
, or 
.
.
will imply perpendicularity. 
.
.
So we are done