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.
and 
are standing in front of the room in which there is 
. They know that there is a chessboard in the room and that on every square there is a coin. Every coin is black on one side and white on the other side and is flipped randomly. enters the room and then points at exactly one square on the chessboard. After that, must flip exactly one coin of his choice on the chessboard to the other side and leave. Finally, enters the room ( and haven't met again after entered the room) and he has to guess which square did point at. and have that will make this happen every time?
such that for any real value of 
we have:
be a triangle with incenter 
and circumcircle 
with circumcenter 
. The incircle touches 
at 
respectively. 
is another intersection point of external bisector of 
with , and 
is 
incircle touch point to . Let 
be points lie on . 
intersect 
at 
. Assume that 
. Suppose that 
and 
are concyclic, and 
are concurrent.
are concurrent. 
, tangent line to at and 
are concurrent.
be a real number. Gugu has a napkin with ten distinct real numbers written on it, and he writes the following three lines of real numbers on the blackboard:
, where 
and 
are two (not necessarily distinct) numbers on his napkin.
, where and are
, where 
are four (not necessarily distinct) numbers from the first line. such that, regardless of the numbers on Gugu's napkin, every number in the second line is also a number in the third line.
be a positive integer. Show that there exist a polynomial 
with integer coefficient that satisfy the following is at most 
for each 
be the number of positive divisors of . Let 
be the number of positive divisors of which have remainders 
when divided by 
. Find all positive integral values of the fraction 
.
be a scalene triangle with the orthocenter 
. Let 
be the reflection of over 
and 
be the reflection of over 
. Let the tangents to the circumcircle of at points and meet at a point 
. Suppose that the lines 
and 
meet at a point . Prove that 
is perpendicular to 
.
be three mutually tangent circles such that 
are externally tangent at 
, 
are internally tangent at , and 
are internally tangent at . Let 
be the centers of , respectively. Given that is the foot of the perpendicular from to , prove that 
. be a scalene triangle with circumcircle , and suppose the incircle of touches at 
. The angle bisector of 
meets and at 
and 
. The circumcircle of 
intersects the -excircle at 
, 
, and at 
. Prove that line passes through either or ., let and be the circumcenter and orthocenter, respectively. Let 
and 
be the midpoints of and , respectively, and let and be the feet of the perpendiculars from and to the opposite sides, respectively. Show that if is the intersection of 
and 
, then 
is perpendicular to 
.
. Prove that:
. Prove that:![$(a+b+c)^4(ab+bc+ca) - 9\sum{\frac{a}{c}} \ge 54[(a+b)(b+c)(c+a)+abc-1]$](http://latex.artofproblemsolving.com/d/3/c/d3cf439c79cb84ca23083e51b6190ab5118f34e8.png)
be a scalene triangle with point and 
on the plane such that 
is an equilateral . Let intersect 
and 
at and 
respectively and intersect 
and 
at and 
respectively .
is obtuse with 
to avoid configurational confusion. Now, note that by the Law of Sines on 
,![\[\frac{AW}{AB} = \frac{\sin \angle ABW}{\sin \angle AWB} = \frac{\sin (120-B)}{ \sin (60-C)}\]](http://latex.artofproblemsolving.com/2/0/3/203dccd87d72f76d2e5ff42fb09a421fe0b393a3.png)
Similarly,![\[\frac{AZ}{AC} = \frac{\sin (60+C)}{\sin (B-60)}\]](http://latex.artofproblemsolving.com/0/5/b/05bb3d5943ae753afa5df4a97be3bf93f3acdd6c.png)
![\[\frac{AY}{AB} = \frac{\sin (B-60)}{\sin (C+60)}\]](http://latex.artofproblemsolving.com/f/2/5/f252c986da7fbabd35ae53a6a58a9ff90f4044a8.png)
and![\[\frac{AX}{AC} = \frac{\sin (60-C)}{\sin (120-B)}\]](http://latex.artofproblemsolving.com/f/8/5/f8518212751d1933a335126f99ba94848d87921d.png)
Now it is easy to see that,![\[\frac{AW}{AZ} = \frac{\sin (120-B)}{ \sin (60-C)} \cdot \frac{\sin (B-60)}{\sin (60+C)} \cdot \frac{AB}{AC}\]](http://latex.artofproblemsolving.com/0/8/8/0885b3a57024fc6384b8cbe633a3d741e60c8e21.png)
and![\[\frac{AY}{AX} = \frac{\sin (B-60)}{\sin (C+60)} \cdot \frac{\sin (120-B)}{\sin (60-C)} \cdot \frac{AB}{AC}\]](http://latex.artofproblemsolving.com/3/3/2/332cdf6eda4a15c3d88391eb296aeb4319cc0e3b.png)
from which we may conclude that![\[\frac{AW}{AZ} = \frac{AY}{AX}\]](http://latex.artofproblemsolving.com/1/d/3/1d3cc00b856005181a33e51b8e54745e27f9834c.png)
implying that 
as desired.
. As 
and 
are equilateral, 
.
such that 
coprime, 
and 
.
,such that 
P
