[/quote]
,
be the set of positive real numbers. Determine all functions 
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)
.
meet at point 
.
on line 
which is not on the segment 
.
meet 
respectively. Let 
be the midpoint of 
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 
,
in at least 
different ways.
lies on side 
lies on side 
.
and 
be points on segments 
and 
,
is parallel to 
.
be a point on line 
,
.
be the point on line 
,
.
,
such that 
satisfying 
has an incenter 
.
.
meets 
again at point 
.
and 
.
is a parallelogram, and suppose the angle bisectors of 
and 
concur on 
.
and 
meet 
.
and 
,
is a perfect square. For example, if the number on the blackboard is 
,
,
,
,
.
and orthocenter 
,
be an arbitrary point on the circumcircle of triangle 
and that line 
is not parallel to line 
is perpendicular to 
.
is a diameter, and 
and 
are parallel. Let 
.
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)
and 
)
Prove that if lines 
are similar.
Y by FreakLM10, MatBoy-123, Adventure10, Rounak_iitr
Let 
be an acute-angled triangle with 
. Let 
be the incentre of triangle , and let 
be the points where the incircle touches the sides 
respectively. Let 
meet the line 
at 
respectively. Further assume that both 
and 
are outside the triangle . Prove that
are concyclic.
is also the incentre of triangle 
.
be an acute-angled triangle with 
. Let 
be the incentre of triangle , and let 
be the points where the incircle touches the sides 
respectively. Let 
meet the line 
at 
respectively. Further assume that both 
and 
are outside the triangle . Prove that
are concyclic.
is also the incentre of triangle 
.
being the orthic triangle of 
and more angle chase in right-angled cyclic quadrilaterals
This was stupidly way too easy for RMO.
and 
.
.
to 
.
and 
,
which means that 
is cyclic. Similarly, we get that 
is also cyclic. Thus, 
This means that 
,
,
,
bisects 
,
.
bisects 
,
.
.
,
is its orthic triangle, and so by a well known property, we get that 
,
.Done 
and 
are concylic.
.
thus 
.
.
is cycli
.
is cycli
is cyclic, as desired.
.
.
is cyclic 
.
is cycli
.
bisects 
bisects 
.
,
is cyclic. Thus, since 
,
as well. Now let 
be the midpoint of 
is right, we have 
,
,
are collinear.
and so, 
is cyclic. Also, as 
are cyclic with 
by fact 5 we get 
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as 
has its orthocenter as 
and 
.
a homothety sending 
,
and we complete our first part.
(
is cyclic). Hence 
bisects 
and by Iran Lemma we have 
.
externally bisects 
i.e. internally bisects 
also bisects 
internally bisects 
in projective terms.
And since 
we get our desired conclusion
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, so we get 
, also we have 
, hence 
are concyclic points, hence 
are concyclic points , hence 
hence we have points 
concyclic 
and we also quadrilateral 
to be cyclic quadrilateral hence we have 
and similarly 
, by using the fact that quadrilateral 
is also cyclic , it is not hard to observe that 
hence we have 
.
are concyclic, So 
and 
.
,
.
,
(say). 
(say).
,
,
.
,
.
and 
are tangents to the incircle of 
.
(vertically opposite). Thus, by angle chasing in 
,
is cyclic. Part (i) proved.
,
is cyclic.
is cyclic 
.
is cyclic 
.
is cyclic and 
.
.
then 
which has orthic triangle 
and by Incenter-orthocenter duality we are done.
and 
is cyclic since 
and similarly 
.
and 
implies 
and analogously 
implies that 
