be distinct natural numbers each of which has exactly 
positive integer factors. For instance, 
has exactly 
.
divides any of the 
'
be an integer. Prove that 
can be written as a polynomial with integer coefficients in the variables 
,
and 
.
,
are real numbers such that 
is an integral multiple of 
.
,
for all positive integers 
.
be the orthocenter of an acute-angled triangle 
.
centered at the midpoint of 
and passing through 
and 
.
,
,
and 
.
with real coefficient. It is known that 
and 
.
can't have 
distinct real roots.
to meet 
meets the midpoint of 
.
,
.
with side length of 
.
is selected on the diagonal 
.
as 
.
.
as 
respectively, compute the area of 
given the area of 
.
.
is equilateral. Given that 
and the midpoint of 
be reals with 
a
.
be non-negative reals such that 
.
people seated around a circular table, and 
cookies
be one of these people. Find the least 
Y by Rounak_iitr
Let 
be an acute-angled triangle with 
, with incircle 
centered at 
which touches 
and 
at points 
and 
, respectively. The point 
on is such that 
and and 
lie on the same halfplane with respect to the angle bisector of 
. Let 
and 
be the intersections of with 
and 
different from , respectively. Let 
be a point on the line such that 
. Let 
be the intersection of 
and different from . Prove that 
.
be an acute-angled triangle with 
, with incircle 
centered at 
which touches 
and 
at points 
and 
, respectively. The point 
on is such that 
and and 
lie on the same halfplane with respect to the angle bisector of 
. Let 
and 
be the intersections of with 
and 
different from , respectively. Let 
be a point on the line such that 
. Let 
be the intersection of 
and different from . Prove that 
.
.
is tangent to the incircle, then 
.
.
is the midpoint of 
because we know that 
are tangents to the incircle and 
are collinear. We're done!
is the line passing through 
perpendicular to 
be coplanar distinct points. Let 
be any point in the same plane.
by finding convenient expressions for both segments.
.
-- then 
is isosceles and 
by symmetry. Now Stewart's theorem (which for the isosceles 
with radius 
)
where 
,
,
.
from Power of a Point and 
from the incircle.) On the other hand, Power of a Point also gives 
,
(Also, 
.
.
and so 
so we no longer need 
we apply Thales' theorem on 
,
(wow, so 
seems to be the midpoint of 
and 
,
(as usual 
)
,
,
.
and the Law of Cosines in 
we focus on 
.
For 
focus on triangle 
.
and use the Law of Cosines, but in my opinion it is better to use that 
Therefore 
,
,
.
and we want 
being cyclic. We have that 
(where 
is midpoint of 
and it lies on 
by the harmonic quadrilateral 
)
.
and 
)
is isosceles trapezium, therefore by symmetry we have 
.
,
.
Since 
.
Moreover, 
.
to be the intersection of 
and 
and 
are isosceles trapezium, then 
and 
.
.
,
This means that 
,
is isosceles trapezium, implying that 
we get that 
.
are both perpendicular to 
,
is a trapezium. Finally we have![\[\sphericalangle{HDE} =\sphericalangle{MD'J} = \sphericalangle{D'JF}=\sphericalangle{MFK}\implies EH=MK \implies HK || EM.\]](http://latex.artofproblemsolving.com/9/5/5/9555bee2ecf2cd68dc900fb4ab1290dae769d2a3.png)
