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.
are positive integer.
touches the sides 
, 
and 
at 
, 
and 
respectively. Let 
denote the other point of intersection of 
and the inscribed circle. Prove that 
extended passes through the midpoint of 
if and only if 
.
and a constant 
, it is given that 
and 
are also friendly polynomials. Prove that 
.
with 
has at least two cycles in it.
that satisfy the following conditions:
is an interval.
be an acute scalene triangle inscribed in a circle 
. The angle bisector of 
intersects 
at 
and at 
. The point 
is the midpoint of 
. Let 
be the altitude in 
, and the circumcircle of 
intersects again at 
. Let 
be the midpoint of , and let 
be the reflection of 
with respect to . Prove that the triangles 
and 
are similar.
such that 
divides 
for any positive integers 
and 
.
for which the sequence of real numbers with 
for all 
is periodic from the beginning.
be an acute triangle and 
is orthocenter. Let be the intersection of 
and 
and be the intersection of 
and . The circumcircle of 
cuts the circumcircle of at 
. Prove that the angle bisectors of 
and 
concur at a point on 
is divisibly by 
.![\[
f(x) = x^{2n} + a_1 x^{2n-1} + \dots + a_{n-1} x^{n+1} + a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + 1
\]](http://latex.artofproblemsolving.com/1/3/d/13d8e46cc3e08e2e6fef60b7a18693b7da0df2d1.png)
that have 
real roots and satisfy 
for all 
.
and 
of triangle intersect in . A perpendicular 
from to 
is drawn. Circumcircles and 
intersect at 
and 
.
.
Where is Fermat point, 
is its isogonal conjugate and 
and are circumcenter and orthocenter of 
the second fermat point? Because if it is, then it's not true. As far as I know, If 
are the first and second Fermat points, and 
are the first and second isodynamic points (just think that 
is the isogonal conjugate of 
), then 
.
s.t. 
is a circle having the segment bounded by the foot of the -bisector and the foot of the exterior -bisector. This circle is called the -Apollonius circle. Similarly we define the and 
-Apollonius circles. It can be shown that these three circles are coaxal, and the two points through all three of them pass are called the isodynamic points of the triangle .
that is fermat point and and are citcumcenter and orthocenter.
, 
, 
are the reflections of a point P in the sides BC, CA, AB of a triangle ABC, then the circumcenter of the triangle 
is the isogonal conjugate Q of the point P with respect to the triangle ABC, and we have 
, 
and 
.
, 
, 
.
, 
and 
. Together with , , (from Lemma 3), this yields YZ || EL, ZX || LD and XY || DE. Hence, the triangles XYZ and DEL are homothetic. Since the triangle DEL is equilateral (Lemma 2), it follows that the triangle XYZ is also equilateral. Hence, instead of saying that the circumcenter of the triangle XYZ is the point F, we can simply claim that the center of the triangle XYZ is the point F.
be the image of the point J in the homothety h. Then, since a homothety maps lines to parallel lines, we have 
. But 
(since the point X is the reflection of the point J in the line BC). Thus, 
. But the point D is the center of the equilateral triangle constructed outwardly on the side BC of triangle ABC, and hence lies on the perpendicular bisector of this side BC. Thus, the line 
, passing through D and being perpendicular to BC, must be the perpendicular bisector of this side BC. In other words, the point lies on the perpendicular bisector of the side BC. Similarly, the same point lies on the perpendicular bisectors of the other two sides of triangle ABC. And this shows that our point coincides with the circumcenter O of triangle ABC. Hence, the image of the point J in the homothety h is the point O.
also is a point in triangle that:![\[ <F'BC=FBA , <F'CB=FCA\]](http://latex.artofproblemsolving.com/5/8/9/589cf07ac938e8cf8b1222f7776ecc7dccd69a41.png)
but I remember there are some much harder problems about the Fermat and isodynamic points.
interesting collinearities, the so-called FNI collinearities. For a list of these collinearities, you can look at my Hyacinthos message #6602, or you can also consult four Forum Geometricorum papers by the late Lawrence S. Evans (paper 1, paper 2, paper 3, paper 4). also is a point in triangle that:
is a pair of real numbers such that
.
