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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)
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 
Prove that
Let and 
Prove that
Let and 
Prove that
Let and 
Prove that
with 
. The points 
lie on 
respectively, such that 
. Extend 
to meet the circumcircle of at 
respectively. Let the circumcircles of 
meet at points 
. Extend 
to at , given 
, find ![$[ABC]$](http://latex.artofproblemsolving.com/d/3/3/d33cc80fa8f093e155c5be46d2e5d9da3d7e1ef5.png)
.
with 
, denote the incenter of the triangle as 
. Extend 
to meet the circumcircle of 
at 
, find the length of 
.
and 
. Prove that
Equality holds when 
Equality holds when 
Equality holds when 
and 
touch each other externally, and they have a common tangent at points 
and 
respectively, where lies between 
and . Prove that 
.
and touch each other externally, and they have a common tangent at points and respectively, where lies between and . Prove that .
![[asy]
real y=0.9876543209876389;
size(8cm);
pointpen=pathpen=black;
D(CR(D((0,5)),5));D(CR(D((8,3.2)),3.2)); D(CR(D((40/9,y)),y)); D(D("A",(0,0))--D("B",(40/9,0))--D("C",(8,0))--(8,3.2)--(0,5)--cycle); D((0,5)--(40/9,y)--(8,3.2)); D((0,y)--(8,y),dashed);
[/asy]](http://latex.artofproblemsolving.com/9/0/d/90dccdd7ec238d7809932b10c02543ec26811e37.png)
. Then by Pythagoras we have 
From 
and 
, we can get 
, so the inequality becomes 
We can rearrange to 
By SOS and 
, it suffices to show that 
But squaring and rearranging, this drops to 
which is true by Muirhead. We get equality when 
.
and similalrly 
, 
, 
. 
, so 
, and similar results for 
and 
.
(just rearrange).
, we get that we need to prove 
, and it has been said before somewhere in the post that 
, so we need to prove that 
for positive 
.
(WLOG, since inequality is homogenous), we need to prove that 
for all . But 
.
= 
. Then by AM-GM, 
. So for all we need to prove that 
.
. if 
, then f(x) is increasing, since 
which is obviously greater than 0 (it's greater than 15/16), so the gradient is positive. So the minimum of the function is at the lower bound of x, i.e. 
. So the minimum is at 9/4 and we are done.
by rearrangement, and 
by AM-GM.