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.
, inscribed in the circle 
, satisfies 
and 
, and 
is the intersection point of the diagonals 
and 
. The circle with center 
and radius 
intersects in two points 
and 
. Prove that the line 
is tangent to the circles with diameters 
and 
.
be a triangle inscribed in a circle, where the tangents to the circle at points 
and 
intersect at the point 
. Let 
be a point on the arc (not containing ) such that 
and 
. Let the lines 
and 
intersect at point 
. Let 
be the reflection of with respect to the line . The lines 
and 
intersect at point 
, and intersects the circumcircle of again at point 
. lies on the circumcircle of 
.
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 
, if and only if 
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 of each 
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.
. From the fourth term of the sequence onwards, each term of the sequence is the average of the previous three. Given that the first six terms of the sequence are natural numbers and that the sixth number written was 
, what is the fifth term of the sequence?
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 :
for all prime numbers 
.
for all positive integers 
such that 
, draw exterior equilateral triangles on sides 
and to obtain 
and 
, respectively. Let 
be the intersection of the altitude through and the median through . Let 
be the intersection of the altitude through and line 
. Let 
be the intersection of the median through and the line 
. Prove that , , and lie on a common line. be an even positive integer. A sequence of real numbers is called complete if for every integer with 
either the sum of the first terms of the sum or the sum of the last terms is integral. Determine numbers.
be an acute triangle such that 
Let be the midpoint 
the orthocenter of 
the midpoint of 
and 
the circumcenter of 
Prove that 
is a parallelogram.
with orthocenter 
and be the midpoint of , if 
is the reflection of over , then 
is diameter of 
.”
, we get that 
. is the midpoint of 
and 
is the midpoint of 
.
.
be 
.
is .
so 
. Hence, 
, which proves the Lemma. 
. By Lemma 1, 
and 
. Hence, 
.
.
and 
be on 
respectively such that 
and 
. Since 
is cyclic, 
so 
(ratios of diameters of circumcircles=ratios of sides). 
. Hence, 
and 
so is a parallelogram. 
and 
bisect each other but just note that is the orthocenter of and that is the diameter of the nine-point circle of . It's well known that the circumcenter of the nine-point circle is the midpoint of the line that joins the circumcenter and the orthocenter, in this case , and we're done.
and 
are both perpendicular to and thus parallel. Further we have 
, so we're done.
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)
to 
and 
because 
,
and 
.
and 
.
because 
,
,
and all of them have modulus equal to 
so