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.
a prime number with 
and 
when 
. then 
else 
.
such that 
. Find the minimum value of:
be a triangle such that 
and 
is obtuse. Point 
is the circumcenter of triangle , and 
is the reflection of 
in 
. Let 
be an arbitrary point on line , such that 
is in between and 
. Line 
cuts the circumcircle of in 
. Circumcircles of triangles 
and 
cut in 
respectively. Prove that 
is tangent to the circumcircle of triangle 
.
from the non-zero reals to the non-zero reals, such that ![\[f(x+y) \left(f(x) + f(y)\right) = f(xy)\]](http://latex.artofproblemsolving.com/2/f/e/2fe8a09e628f60faedaa07ceb154891375763a83.png)
for all non-zero reals 
such that 
.
that satisfy the equation
for any 
. Note that 
.
be an acute-angled scalene triangle with circumcircle 
and circumcenter . Suppose 
. Let 
be the orthocenter and 
be the incenter of triangle . Let 
be the midpoint of the arc of the circumcircle of triangle 
, containing .
be a point on the arc 
of not containing , such that 
. Let 
be the circumcenter of triangle 
. Prove that the lines 
and 
meet on .
non-negative real numbers, then we have:
| 
with 
are positive integers and 
is prime (
), we get: 
such that 
and:
of positive integers such that 
is divisible by 
.
be a set consisting of non-negative integers such that:
,
, both 
and 
..
such that![\[
f\left(\frac{x^2 - f(x)}{f(x) - 1}\right) = x
\]](http://latex.artofproblemsolving.com/8/0/0/8009a4709ffd6708023d67db2b544e605d1fbe68.png)
for all real numbers 
satisfying 
.
and 
such that![\[ {a^2+b\over b^2-a}\quad\mbox{and}\quad{b^2+a\over a^2-b} \]](http://latex.artofproblemsolving.com/e/3/b/e3bc96af9f635e1c10e21dc91894c8afbf113a6d.png)
be the midpoints of the diagonals 
Since 
then 
Parallelogram 
is a rhombus 
bisects 
internally. Let 
Since 
and have corresponding parallel sides, it follows that their angle bisectors are parallel as well. By Menelaus' theorem for 
cut by 
keeping in mind that 
and 
we get
is parallel to the bisector of the angle 
, hence the triangles 
and 
are pseudo-similar and our problem is solved then.
be the center of spiral similarity that sends 
, 
. Therefore, it's also the center of spiral similarity that sends 
and 
. Hence, 
are cyclic quadrilaterals, also 
. Hence, there is a spiral similarity center at that sends 
. But 
.
be a convex quadrilateral such that 
.
be the midpoints of 
,
and 
and 
,
.
and 
,
and
Since, 
,
and 
,
and
Since, 
,
Because, 
and 
,
Since 
,
,
.
and 
.