[/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 an acute triangle. Let 
be a point on side 
and 
be a point on side 
such that lines 
and 
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 
and 
(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, Ad112358
The circle 
is circumscribed around the quadrilateral 
. The lines 
and 
intersect at the point 
, and the lines 
and 
intersect at the point 
. The line passing through the center of the circle and perpendicular on 
intersects the lines , and at the points 
, 
and 
, respectively. Prove that the lines 
, 
and 
intersect at one point.
is circumscribed around the quadrilateral 
. The lines 
and 
intersect at the point 
, and the lines 
and 
intersect at the point 
. The line passing through the center of the circle and perpendicular on 
intersects the lines , and at the points 
, 
and 
, respectively. Prove that the lines 
, 
and 
intersect at one point.This post has been edited 1 time. Last edited by parmenides51, Sep 28, 2018, 5:11 PM
Reason: name typo
Reason: name typo
meet at 
and now let 
then from either 2nd Isogonality Lemma or DDIT we can see that 
are isogonal in 
,
and 
therefore we get that 
bisects 
or equivalent to 
being isogonal in 
are colinear as desired thus we are done 
.
