[/quote]
,
,
,
and 
,
is the intersection point of the diagonals 
and 
.
and radius 
intersects 
and 
.
is tangent to the circles with diameters 
and 
.
satisfying the equation![\[
(f(x+y))^2= f(x^2) + f(2xf(y) + y^2), \quad \forall x, y \in \mathbb{R}.
\]](http://latex.artofproblemsolving.com/2/5/e/25eed911f303ed7cc4e03765a2f943ad4641c451.png)
be a triangle inscribed in a circle, where the tangents to the circle at points 
and 
intersect at the point 
.
be a point on the arc 
and 
.
and 
intersect at point 
be the reflection of 
and 
intersect at point 
,
.
.
be non-negative real numbers such that 
Prove that
such that 
and 
for any real numbers 
and 
.
is written on the blackboard. Every time, Carmela can erase the number 
on the black board and replace it with a new number 
,
is a perfect square. Prove or disprove that all positive integers 
can be written exactly once on the blackboard.
and 
as ![\[\sqrt{2}=1.a_1a_2a_3\dots\quad\quad\sqrt{3}=1.b_1b_2b_3\dots\]](http://latex.artofproblemsolving.com/4/d/5/4d5049d5c3b26757a25c5dd34462c5462228b83d.png)
where each 
or 
is a digit between 0 and 9. Prove that there exist at least 1000 values of 
between 
and 
such that 
.
of positive integers with 
is called good if we can colour each integer red, blue, green or purple, in such a way that
consecutive integers at least one is coloured red;
of each 
consecutive integers at least one is coloured blue;
of each 
consecutive integers at least one is coloured green;
of each 
consecutive integers at least one is coloured purple.
.
,
points in the plane, no three of them belonging to a same line. Each of these points is colored using one of four given colors. Prove that there is a color and at least 
scalene triangles with vertices of that color.
such that :
.
such that 
and 
and 
,
be the intersection of the altitude through 
be the intersection of the altitude through 
.
be the intersection of the median through 
.
either the sum of the first 
.
Y by buratinogigle, anantmudgal09, Adventure10, Mango247
Let 
be angle bisector of acute triangle 
.Point 
choosen on such that 
.point 
lies on the extention of from 
such that 
.Point 
is diametrically opposite the point on the circumcircle of 
.Prove that 
passes through midpoint of arc .(S. Berlov)
be angle bisector of acute triangle 
.Point 
choosen on such that 
.point 
lies on the extention of from 
such that 
.Point 
is diametrically opposite the point on the circumcircle of 
.Prove that 
passes through midpoint of arc .(S. Berlov)
my 100th post This post has been edited 1 time. Last edited by MRF2017, Mar 14, 2016, 4:02 PM
to intersect 
gives us the following equivalence:
of 
.
denote the midpoints of arc 
such that 
as 
such that 
and 
are collinear.
,
in 
such that it corresponds to 
are isogonal wrt 
.
because 
are isogonal. Thus 
.
,
,
.
.
,
similarly. Since 
are angle bisectors in similar triangles 
respectively, therefore![\[\frac {FX}{XH}=\frac {YE}{YH}\implies XY||EF\]](http://latex.artofproblemsolving.com/3/6/b/36b9b833b99a2cbfcaa24a01256541a2ef719bee.png)
It follows that 
are homothetic about 
are collinear and we are done.
(external) meet 
meet 
at point 
.
.
maps 
we have that 
.
and we get the claim. Now, clearly 
implies that 
passes through the anti-pose of 
.
are collinear.
is the intersection of 
and 
.
