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 sequence of positive numbers satisfying, for any positive integers 
such that 
, 
Show that there exist positive numbers 
so that 
for any positive integer 
.
be a triangle with incentre 
and circumcentre 
. Let 
be the touchpoints of the incircle with 
, 
, 
respectively. Prove that 
is the Euler line of 
.
be a scalene triangle with orthocenter 
and medial triangle 
. Let 
be a point on 
such that 
. If 
is the midpoint of segment 
, show that 
is isoceles.
such that the sequence 
defined by ![\[ a_0 = M + \frac{1}{2} \qquad \textrm{and} \qquad a_{k+1} = a_k\lfloor a_k \rfloor \quad \textrm{for} \, k = 0, 1, 2, \cdots \]](http://latex.artofproblemsolving.com/2/1/b/21b2a1f93b11a94b84e7b55f4b4f679aa20e36c6.png)
contains at least one integer term.
non-negative real numbers,but 
0.Prove that![\[1\leq \frac{a+b}{a+4b+c}+\frac{b+c}{b+4c+a}+\frac{c+a}{c+4a+b}\leq \frac{4}{3}\]](http://latex.artofproblemsolving.com/8/b/7/8b7d7b8cbcab8f2eed31e17aac717fa3c70c92f1.png)
be an acute triangle with orthocenter . Let 
be the point such that the quadrilateral 
is a parallelogram. Let be the point on the line 
such that bisects 
. Suppose that the line intersects the circumcircle of the triangle 
at 
and 
. Prove that 
.
that satisfy the following property: if 
, 
, 
, 
are all the positive divisors of with 
, then 
divides 
for every 
.
be an isosceles triangle with 
. The circle 
, passing through 
and , intersects segment at 
. The circle 
is tangent to at and passes through 
. Let and 
be the midpoints of segments and , respectively. The line 
intersects and at points 
and 
, respectively, where and are the intersections closer to . Prove that 
., 
, 
is the point on such that 
. Let be the circle through tangent to 
at , and the circle through tangent to at . Let 
be the second intersection of and . Prove that lies on .
be an acute-angled triangle with 
. Let 
be the circumcircle of . Let 
be the midpoint of the arc 
of containing 
. The perpendicular from to meets 
at and meets again at 
. The line through parallel to meets line 
at . Denote the circumcircle of triangle 
by 
. Let meet again at 
. Prove that the line tangent to at meets line on the internal angle bisector of 
.
be natural numbers. As the New Year arrives, it is a cherished tradition among Iranian families to call their relatives and exchange Nowruz greetings. Just moments after the New Year begins, a family with members calls a related family with 
members to celebrate the occasion. Since the two families are close, every member of one family wishes to speak with every member of the other family to exchange greetings. Thus, during this single call, all 
distinct one-on-one conversations must take place exactly once. be the set of black cells at the end of the game. Ali and Shayan respectively aim to minimize and maximise the perimeter of the shape by playing optimally. (The perimeter of shape is defined as the total length of the boundary segments between a black and a white cell.), assuming both players play optimally?
be an acute triangle with 
. Let 
be its circumcircle, its orthocenter, and the foot of the altitude from . Let be the midpoint of . Let be the point on 
such that 
and let be the point on such that 
. Assume that the points , , , and are all different and lie on in this order.
and 
are tangent to each other.
be a convex quadrangle, 
the intersection of lines 
and 
, 
the intersection of lines 
and 
and 
the intersection of diagonals 
and 
. Show that if 
then 
is the bisector of 
and 
is the bisector of 
.
then is the bisector of and is the bisector of .
such that 
and 
. By Menelaus theorem on 
, line 
and Ceva theorem on , cevians 
, we discover that 
. So, 
divide harmonically the segment 
.
, we use Ceva theorem on , cevians and we use 
to discover that 
, so, again, 
divide harmonically the segment (here, 
is the point of infinity corresponding to the lines parallel to ). dividing harmonically the segment . So, the lines 
divide harmonically the lines 
. Suppose that 
. Choose a line 
parallel to through . intersects the lines 
in the points 
. And the points 
should divide harmonically the segment 
. So, it’s clear that 
, which gives us 
, and that the lines 
bissect the lines 
.
I'm so sorry. thanks tdl 
