be the circumcircle of acute triangle 
are on segments 
and 
respectively such that 
.
and 
intersect minor arcs 
and 
respectively. Prove that lines 
and 
are either parallel or they are the same line.
and 
,
first, then every player sees the last number written and if it is 
then in his turn he writes 
or 
,
be a circle internally tangent to 
and 
,
.
and 
concur on the bisector of 
.
be all positive integers smaller than 
such that 
for all 
is the largest positive integer that divides both 
and 
.
.
and 
are the midpoints of sides 
and 
.
.
,
,
of a hexagon 
inscribed in a circle 
intersect at a point 
,
.
be the radius of the circle tangent to segments 
and 
and internally tangent to 
,
,
,
,
are defined similarly. Prove that![\[
r_{AB}r_{CD} + r_{CD}r_{EF} + r_{EF}r_{AB} = r_{BC}r_{DE} + r_{DE}r_{FA} + r_{FA}r_{BC}.
\]](http://latex.artofproblemsolving.com/0/d/6/0d66d6685bbc97c7553e9ed3ba14ea0dc3328b49.png)
be the circumcentre, 
satisfy the following condition:
is constant.
.
are both perfect squares.
such that ![\[\left(g(m)+n\right)\left(g(n)+m\right)\]](http://latex.artofproblemsolving.com/5/a/c/5ac9fce315e8a524a3a01b345cfe76bc76056d68.png)
is a perfect square for all 
Y by
Let 
be an acute triangle with orthocenter 
, and let 
denote the foot of the altitude from 
to 
. Let the circle 
with diameter 
intersect altitude 
at distinct points 
and 
. Suppose that the circle with center 
passing through point intersects segment 
at 
. Line 
meets at 
. If 
intersects 
at 
, and 
intersects at 
, prove that is tangent to the circumcircle of 
.
be an acute triangle with orthocenter 
, and let 
denote the foot of the altitude from 
to 
. Let the circle 
with diameter 
intersect altitude 
at distinct points 
and 
. Suppose that the circle with center 
passing through point intersects segment 
at 
. Line 
meets at 
. If 
intersects 
at 
, and 
intersects at 
, prove that is tangent to the circumcircle of 
.
.![$[CL]^2 = [CX]^2= [CY]^2 \Rightarrow L$](http://latex.artofproblemsolving.com/3/a/0/3a0d813ceb846745ae927892e0a6bbddb6eec0bd.png)
is the incenter of 
.![$[CL]^2 = [CX]^2 = [CE] \cdot [CB] = [CD] \cdot [CA] \Rightarrow \overline{AL} \perp \overline{CL}\Rightarrow R$](http://latex.artofproblemsolving.com/4/5/6/4567bb6a0790e2821b3a2b4ba1592e30649baf97.png)
and 
belong to the 
mixtilinear incircle of 
.
is the midpoint of arc 
and 
belongs to the 
be the feet of perpendicular from 
meet the line 
at points 
.
centered at 
.
and 
.
.
and similarly 
and we have proved the lemma. 
.
at 
.
,![\[CL'^2 = CD\cdot CA = CX^2 = CL^2.\]](http://latex.artofproblemsolving.com/d/c/9/dc9e1ea07a744020713f6dc1137a2caeb0135af7.png)
Also 
lie on same side of 
lies on 
or 
.
.
is the internal angle bisector of 
.
gives 
and 
m
and 
at 
and 
respectively. We know (well-known) that 
are collinear and 
.
and we are given 
are collinear. This gives 
and 
which further gives that 
is the 
by 
are tangent to 
is tangent to 
.
is parallel to 
,
.
and 
.![\[\measuredangle XRR' = \measuredangle XPB = \measuredangle R'YX,\]](http://latex.artofproblemsolving.com/e/d/0/ed0b04ceb8f66d55c4dd0f96740056f508bbb1a4.png)
so 
is cyclic. Similarly, 
is cyclic as well.
is acute) inversion centered at 
,
,
.
and 
.
is tangent to 
.
and 
are tangents to 
and 
are also tangents to 
form the 
.
,
by 
.
,
and 
are tangent to 
is harmonic. This means that the tangents to 
,
and 
.
,
.
,
lies on 
(as 
and 
)
are collinear (as 
)
,
(as 
and 
)
be the tangent to 
Thus, 
,
.
such that 
so 
is tangent to 
and also 
is tangent to 
and hence by PoP we have 
and so 
.
and see that 
and so 
and hence it is the 
is 
