AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
and 
. Prove that
Equality holds when 
or 
be a cyclic quadrilateral whose opposite sides are not parallel. Suppose points 
lie in the interiors of segments 
respectively, such that 
Let 
intersect 
at 
, and 
intersect 
at 
. Prove that lines 
and 
are either parallel or coincide.
. Prove that:
be reals such that 
Prove that 
Let be reals such that 
Prove that 
be a fixed integer. There are 
beads on a circular necklace. You wish to paint the beads using 
colors, such that among any 
consecutive beads every color appears at least once. Find the largest value of 
for which this task is 
possible.
This easy via the binomial theorem for the quantity is just 
, but how do we arrive at this using the I-E-P?
satisfies 
. If 
, find the maximum possible value of 
.
+1 w
and a constant 
, it is given that 
and 
are also friendly polynomials. Prove that 
.
lying outside the circle 
, two tangents are drawn touching at points 
and 
. A point 
is chosen on the segment 
. Let points 
and 
be the midpoints of segments 
and 
respectively. The circumcircle of triangle 
intersects again at point 
(
). Prove that the line 
passes through the centroid of triangle 
.
be an acute-angled triangle with 
and let 
be the foot of the
-angle bisector on 
. The reflections of lines 
and 
in line meet and at points
and 
respectively. A line through meets and at 
and 
respectively such that and 
while lies strictly between 
and . Prove that the circumcircles of
and 
are tangent to each other.
and 
are fixed points on a given circle and 
is a variable diameter of the same circle, determine the locus of the point of intersection of lines 
and 
. You may assume that 
is not a diameter.
and have coordinates 
and 
respectively and and have coordinates 
and 
. Note that these coordinates satisfy 
and 
since these points are on a unit circle. Now we can find equations for the lines:
in terms of 
and 
:
. These coordinates can be parametrized in Cartesian variables as follows:
and to get 
and 
. Then since 
which reduces to 
This equation defines a circle and is the locus of all intersection points . In order to define this locus more generally, find the slope of this circle function using implicit differentiation:
and , this slope expression reduces to 
and 
respectively, values which are identical to the slopes of lines 
and 
. Thus we conclude that the complete locus of intersection points is the circle tangent to lines and at points and respectively.
and 
.
.
Where 
such that :
and 
.
is isosceles, 
.
,
.