,
be the real root of equation 
.
;
.
be an integer, 
a finite set of integers (not necessarily positive) and 
subsets of 
,
is 
.
elements.
Prove that
Where 
Prove that
Let 
Prove that
Let 
Prove that
be an acute triangle. Define 
as its incenter. Let 
and 
be the incircle's tangent points to 
and 
,
be the midpoint of 
.
be the intersection point of a perpendicular line passing through 
.
intersects the circumcircle of 
at 
.
intersects line 
at 
.
is cyclic.
for which there exists a sequence of distinct positive integers 
,
,
,
satisfying ![\[s_1^{s_2}=s_2^{s_3}=\cdots=s_{n-1}^{s_n}.\]](http://latex.artofproblemsolving.com/a/f/5/af594ff07aa2f89cb70ae669263558049556cf0e.png)
.
parts is a representation of n as a sum 
,
belong to 
.
parts with 
.![$\sqrt[3]{6n}$](http://latex.artofproblemsolving.com/2/0/1/201489b1f95ad200036698a606f712da7bb465b9.png)
different parts.
other knights where friendship is mutual. Prove that King Arthur can place some of his knights around a round table in such a way that every knight is friends with the 
people adjacent to him and between them there are at least 
friendships of knights that are not adjacent to each other.
is an integer. Some pairs of cities are connected by direct (two-way) flights. For two cities 
we define:
A 
between 
,
,
and 
for every 
;
A 
between 
A 
between 
,
.
.
are fixed positive integers.
be the set of all functions 
with the following properties:
the following equality holds: 
such that
,
.
Y by MathLuis, mpcnotnpc, JollyEggsBanana, Rounak_iitr
Let 
be a triangle with incenter 
, and let 
be a point on side 
. Points 
and 
are chosen on lines 
and 
respectively such that 
is a parallelogram. Points 
and 
are chosen on side such that 
and 
are the angle bisectors of angles 
and 
respectively. Let 
be the circle tangent to segment 
, the extension of 
past , and the extension of 
past . Prove that is tangent to the circumcircle of triangle .
be a triangle with incenter 
, and let 
be a point on side 
. Points 
and 
are chosen on lines 
and 
respectively such that 
is a parallelogram. Points 
and 
are chosen on side such that 
and 
are the angle bisectors of angles 
and 
respectively. Let 
be the circle tangent to segment 
, the extension of 
past , and the extension of 
past . Prove that is tangent to the circumcircle of triangle .
and 
,
be midpoint of arc 
on 
,
be 
respectively now let reflection of 
over 
and reflection of 
over 
.
and 
are diameters on 
respectively, now let 
be the 
excenters of 
then notice we have 
which implies 
cyclic and similarily 
is cyclic however since 
we also get that 
is cyclic and similarily 
is cyclic, however it doesn't end here...
so 
on 
however 
are symetric in 
are reflections of 
over 
are colinear, and similarily 
are colinear.
which means 
is cyclic and similarily we have 
cyclic, but also note that 
which means that 
are colinear.
which does in fact show that 
is isosceles and therefore 
,
,
where 
(clearly then 
is A-excenter of 
)
which means that 
is cyclic but by summing arcs we end up realising 
is diameter and in fact now this means 
is 
be the miquelpoint of 
then 
colinear and then Reim's twice gives 
and from double Reim's once again we have that 
and this is excellent news because now we can note that 
which shows that 
are tangent at 
and 
.
and 
be the projections of 
and 
be the projections of 
respectively, and let 
be the tangency point of 
is tangent to 
is a parallelogram, by projecting onto 
.
Simplifying yields 
,
But 
With Thales' Theorem, 
so 
thus getting 
and the conclusion follows.
are in centres, 
meets 
in 
and tha
.
