[/quote]
,
and 
play a game, given an integer 
,
first, then every player sees the last number written and if it is 
then in his turn he writes 
or 
,
be the circumcircle of a scalene triangle 
.
and 
,
.
touch 
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 
respectively. Point 
.
.
,
,
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, 
be the orthocentre, and 
be the centroid.
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 
intersect at 
is 
,
is a parallelogram.
is a perfect square number.
is good if there exists a permutation 
of 
such that 
is perfect square for all 
.
.
of positive integers such that 
and 
are both powers of 
.
Y by aopsuser305, Adventure10, Mango247
A convex quadrilateral 
is circumscribed about a circle 
. Let 
be the diameter of perpendicular to 
. Suppose lines 
and 
intersect at point 
, and lines 
and 
intersect at point 
. Show that the points and lie on the line .
Géza Kós
is circumscribed about a circle 
. Let 
be the diameter of perpendicular to 
. Suppose lines 
and 
intersect at point 
, and lines 
and 
intersect at point 
. Show that the points and lie on the line .Géza Kós
at 
respectively. Clearly 
are concurrent at pole 
of 
intersect 
respectively. Notice that
so projecting through 
are colinear. Hence by Brokard's theorem we can conclude that 
lies on polar of 
be the tangency points on 
respectively. Well-known that 
concur (at 
is polar of 
,
are also collinear because 
is a radical axis of the circles 
.
concur and also 
because of well-known properties of the
.
and on 
.
and 
under dilation at 
passes through the touch-point of 
.
passes through 
;
is on the same side of 
as 
,
is one the same side of 
.
and 
touch 
by Pitot. Then since 
,
.
,
touch to 
.
.
.
are cyclic, so from Radical Axis theroem 
.
lies on polar of 
,
are concurrent. Since 
are collinear, we get triangles 
and 
are perspective. So 
are concurrent, i.e 
are collinear. So 
and 
are perspective 
.
.
and 
and 
,
.
and 
,
and 
at points 
,
,
and 
.
.
is the polar of 
is the polar of 
lies on the polar of 
lies on the polar of 
gives 
,
and 
gives 
.
.
