[/quote]
,
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.
be a point on side 
be a point on side 
are parallel. Let 
be an interior point of 
.
and 
meet side 
and 
,
and 
.
and 
intersect at a point 
.
,
are collinear.
.
,
as the greatest exponent 
such that 
divides 
.
and infinitely many 
such that 
.
and 
.
is given. Let 
to 
be the intersection point of 
lies on the arc 
.
.
be a trapezium inscribed in a circle 
with diameter 
. The circle with center 
meets 
(where 
at 
.
is perpendicular to 
.
![\[ x + y^2 = p^m \]](http://latex.artofproblemsolving.com/e/c/f/ecf553036f22327dd457dc06611b6bc1b0ef072c.png)
![\[ x^2 + y = p^n \]](http://latex.artofproblemsolving.com/5/b/d/5bd672934a395aabd6865ca4b50d8da75c0fe07e.png)
For 
positive integers and 
being a prime.
and 
.
Equality holds when 
or 
.
at 
intersect again at 
.
Y by Adventure10
Incircle of a triangle 
touches the sides 
and 
at points 
and 
, and the excircle corresponding to the side 
, touching the extensions of these sides in points 
and 
respectively. Suppose that incircle 
touches the side at the point 
. Denote by 
and 
the centers of the cicrumcircles of the triangles 
and 
respectively . Prove that the line 
passes through the midpoint of the segment .
touches the sides 
and 
at points 
and 
, and the excircle corresponding to the side 
, touching the extensions of these sides in points 
and 
respectively. Suppose that incircle 
touches the side at the point 
. Denote by 
and 
the centers of the cicrumcircles of the triangles 
and 
respectively . Prove that the line 
passes through the midpoint of the segment .This post has been edited 1 time. Last edited by parmenides51, Sep 12, 2018, 8:52 PM
Reason: name edit
Reason: name edit
and 
and let 
.
this is equivalent to showing that 
.
and 
.
.
is the center of the excircle we have 
so we are left to prove that ![\[\angle B_2PA_2=360^\circ-\angle A_2PK-\angle KPB_2=360^\circ-(90^\circ+\frac{\beta}{2})-(90^\circ+\frac{\alpha}{2})=180^\circ-\frac{\alpha}{2}-\frac{\beta}{2}=180^\circ-\angle A_2LB-\angle ALB_2=\angle B_2LA_2.\]](http://latex.artofproblemsolving.com/b/1/4/b14b3c43228312b359827b98b096ed10b0832b34.png)
