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 the set of positive real numbers. Determine all functions 
such that for all positive real numbers 
and 
![\[f(x+f(xy))+y=f(x)f(y)+1\]](http://latex.artofproblemsolving.com/c/6/a/c6a968a17f6fd5fd637c1ee5d855bdff11e5423a.png)
and 
such that![\[
1^n + 2^n + 3^n + \cdots + n^n = x!
\]](http://latex.artofproblemsolving.com/6/4/b/64b5f5919fd3710cc15e96724918eae216c8e4f0.png)
+1 w
be 
. The tangents to at 
meet at point 
. For a point 
on line 
which is not on the segment , let the midpoint of 
be 
. Lines 
meet again at points 
respectively. Let 
be the midpoint of . Prove that the points 
lie on a circle.
red cards, 
white cards, and 
blue cards. A player plays all the cards one at a time. With each play he accumulates a penalty. If he plays a blue card, then he is charged a penalty which is the number of white cards still in his hand. If he plays a white card, then he is charged a penalty which is twice the number of red cards still in his hand. If he plays a red card, then he is charged a penalty which is three times the number of blue cards still in his hand. Find, as a function of 
and 
the minimal total penalty a player can amass and all the ways in which this minimum can be achieved.
and a set 
consisting of 
disting positive integers smaller than 
are given. Prove that there exists a positive integer 
that can be written in the form 
, for 
in at least 
different ways.
, point 
lies on side and point 
lies on side 
. Let 
and 
be points on segments 
and 
, respectively, such that 
is parallel to 
. Let 
be a point on line 
, such that lies strictly between and , and 
. Similarly, let 
be the point on line 
, such that lies strictly between and , and 
.
, and are concyclic.
has an incenter 
Suppose 
. Let be the midpoint of . Suppose that 
. 
meets 
again at point 
. Let points and be such that is the midpoint of 
and is the midpoint of 
. Point lies on the plane such that 
is a parallelogram, and suppose the angle bisectors of 
and 
concur on 
.
and 
meet at 
and 
. Prove that 
.
with another positive integer 
, as long as 
is a perfect square. For example, if the number on the blackboard is 
, Carmela can replace it with 
, because 
, then replace it with 
, because 
. If the number on the blackboard is initially , determine all integers that Carmela can write on the blackboard after finitely many operations.
with circumcenter 
and orthocenter 
, let 
be an arbitrary point on the circumcircle of triangle such that does not lie on line 
and that line 
is not parallel to line . Let be the point on the circumcircle of triangle such that 
is perpendicular to , and let be the point on line such that 
. Let and be the points on the circumcircle of triangle such that 
is a diameter, and 
and 
are parallel. Let be the midpoint of 
.
and 
are perpendicular.
and 
are perpendicular.
be positive integers and 
be an odd prime such that:![\[
a^p \equiv 1 \pmod{p^n}.
\]](http://latex.artofproblemsolving.com/3/4/8/34815b8211ded2f37b0653525dc4b436356e6eee.png)
Prove that:![\[
a \equiv 1 \pmod{p^{n-1}}.
\]](http://latex.artofproblemsolving.com/0/7/a/07a4ed2ab3c94f41028080f7789cfb6e2973955c.png)
be a cyclic quadrilateral. A circle centered at passes through and and meets lines 
and again at points and (distinct from 
). Let denote the orthocenter of triangle 
Prove that if lines 
are concurrent, then triangle and 
are similar.
be the diameter of semicircle ,
be points on the arc ,
be respectively the circumcenter of 
and 
.
.![[asy]
import cse5; import olympiad; unitsize(3.5cm); dotfactor=4; pathpen=black;
real h=sqrt(55/64);
pair A=(-1,0), O=origin, B=(1,0),C=shift(-3/8,h)*O,D=shift(4/5,3/5)*O,P=circumcenter(O,A,C), Q=circumcenter(O,D,B);
D(arc(O,1,0,180),darkgreen);
D(MP("A",A,W)--MP("C",C,N)--MP("P",P,SE)--MP("D",D,E)--MP("Q",Q,E)--C--MP("O",O,S)--D--MP("B",B,E)--cycle,deepblue);
D(O);
[/asy]](http://latex.artofproblemsolving.com/7/a/1/7a1abce71ca933e712f09b65e8a4643908e31e3a.png)
be the diameter of semicircle , be points on the arc , be respectively the circumcenter of and ..
then there are two cases.
by using Law of Sines on 
, respectively.
then we get
![$ \frac{1}{4\cos(y)^2}[1+\frac{1}{4\cos(x)^2}+\frac{\cos(x+2y)}{\cos(x)}]=\frac{1}{4\cos(x)^2}[1+\frac{1}{4\cos(y)^2}+\frac{\cos(2x+y)}{\cos(y)}] $](http://latex.artofproblemsolving.com/2/f/e/2fe1ecd5edc44e94a17c4dc58006961e1a651c8f.png)
and 
.Note that 
.So 
.
and 
so 
.Similarly 
.Combining these details we see that 
.Thus 
as desired.
. Similarly 
. Let 
, then 
. Also 
cuts 
at . Prove that radical axis of circles 
and 
passes through intersection of and 
.
cuts at . Prove that radical axis of circles and passes through intersection of and .
From the proof at post #7 we get 
so 
hence 
is parallel to 
Similarly, we can prove 
so 
and 
are pseudo-similar, hence 
where 
From Reim's theorem we get 
are concyclic, so 
such that 
satisfying 
and 
be the midpoints of 
.
,
.
.
,
,
.
and 
then 
.
since they are both isosceles, we get 
.
.
and 
then multiply to get 
so 
.![\[\angle{PAD} = \angle{DAB} - \angle{PAO} = 90^{\circ} - \beta - (90^{\circ}-\alpha) = \alpha - \beta\]](http://latex.artofproblemsolving.com/2/f/5/2f5cef7d623e232286c439eaff6056f3ca9e33a5.png)
![\[\angle{QBC} = \angle{QBO} - \angle{CBA} = 90^{\circ} - \beta - (90^{\circ}-\alpha) = \alpha - \beta\]](http://latex.artofproblemsolving.com/6/5/7/657e0ca544d3f99ae78ab74542b0da6724a85f41.png)
and 
.
be the unit circle. Let 
and 
.
and 
.
.
