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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.
a number of 
distinct points, so that the distance (measured on the chord) between any two points (from the considered points) is a rational number?
, and consider a chord [a,b] on the circle. Then if there is a point with rational distance to a and to b, by ptolemy's theorem it has rational distance to every other point in the structure. (since the quadrilateral is cyclic)
, and represent a second point as ![$[\cos(t),\sin(t)]$](http://latex.artofproblemsolving.com/d/7/5/d75f3abd1173f85d678788a68a1722b0fc3eb518.png)
. Then we find that 
has to be rational, so there are an infinity of points at rational distance of that one point, more specifically all points with rational y-ordinate... but then I miss the rigorous argument to complete the proof... anyone able to finish?
s.t. 
are both rational. Call such angles good. By angle-sum formulas, if 
are good, then 
are also good.
on the circle 
, let 
be the angle subtended by 
. Now inductively construct points on s.t. all angles formed by them are good; for 1,2 take any good angle. If there are points chosen, pick a good angle and a marked point 
s.t. the point on with 
is distinct from the points. Since there are infinitely many good angles but finitely many marked points, such exists. For a previously marked point 
we have 
for suitable choices for the two 
. Since 
are both good, it follows that 
is good, which finishes induction by adding .
work: since 
for on the circle, it follows that is rational, and so we're done.
s.t. 
s.t. 
s.t. 
and 
was arbitrary , so for any no. of points we can have that distance between any two points is rational