[/quote]
,
.
and 
be the feet of the altitudes from 
and from 
to 
respectively.
and 
as the reflections of 
be the circumcircle of 
.
with 
with 
is the circumcenter of 
can be inscribed within a circle.
be a triangle whose incircle is tangent to 
,
,
at 
,
and 
,
and 
lie on 
so that 
and 
.
and 
intersect on the circumcircle of 
,
lie on a line in this order and satisfy 
.
,
is acute, and let 
be its orthocentre. Points 
and 
,
and 
and 
and 
.
intersect the circumcircle of 
be the midpoint of 
and let the perpendicular bisector of 
intersect 
at 
.
,
.
be an increasing sequence of nonnegative integers such that every nonnegative integer can be expressed uniquely in the form 
,
and 
are not necessarily distinct. Determine 
.
be a cyclic quadrilateral with 
.
not containing 
and 
.
,
.
and find out the set of numbers 
(unordered, of course). Find the minimal number of questions, which are necessary to find out all numbers 
.
and 
.
?
be the number of positive integers that divide the integer 
.
,
?
is a square of an integer.
be all positive integers smaller than 
such that 
for all 
is the largest positive integer that divides both 
and 
.
.
,
,
are odd primes and![\[2^ap^b=(p+2)^c-1.\]](http://latex.artofproblemsolving.com/5/8/5/58518a8ebed14836b72af743ff100eae2efba5e2.png)
let 
be the circumcircle of 
and let 
be its radius. Point 
such taht 
and point 
is the foot of the perpendicular from 
mets 
such that 
and 
lie on a line in that order. Finally, let 
and 
.
Y by Adventure10, Mango247
Let 
and 
be points on the side 
of the triangle 
(with between 
and ) such that 
,and let 
be the angle bisector of 
. Line meets lines 
and 
at 
and 
respectively. Given that 
and 
, prove that triangles 
and 
have the same area.
Thanks in advance.
and 
be points on the side 
of the triangle 
(with between 
and ) such that 
,and let 
be the angle bisector of 
. Line meets lines 
and 
at 
and 
respectively. Given that 
and 
, prove that triangles 
and 
have the same area.Thanks in advance.
for the sake of convenience.
and that of 
.
;
,
and we conclude that 
.
.
;
.
,
,
.
and 
,
