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.
is chosen. Prove that there exists a prime number that divides infinitely many terms of the sequence 
, where![\[
b_k = a^{k^k} \cdot 2^{2^k - k} + 1.
\]](http://latex.artofproblemsolving.com/1/7/5/1751a60482d729a36c71b77ac9c978e724f40da0.png)
+1 w
, if any element of the set 
is fantabulous, then all the elements are fantabulous. Does it follow that the number 
is fantabulous?
that satisfies:![\[
\sum_{i=1}^{n} a_{\left\lfloor \frac{n}{i} \right\rfloor} = n^k
\]](http://latex.artofproblemsolving.com/b/b/e/bbe7fe81e366fe390be5d17193d1581cc40dbece.png)
be a positive integer. Prove that for all integers 
, we have:![\[
\frac{c^{a_n} - c^{a_{n-1}}}{n} \in \mathbb{Z}
\]](http://latex.artofproblemsolving.com/0/4/4/04464c1a350e3f9a929ab51534018498680068fc.png)
such that![\[
f(f(x)+y) = x+f(f(y))
\]](http://latex.artofproblemsolving.com/9/d/7/9d780faa5901646f7035e88400a9e3d8023f04ff.png)
for all 
.
be a function from the Euclidean plane to the real numbers such that 
for any acute triangle 
with circumcenter 
, centroid 
and orthocenter 
. Prove that 
is constant.
-
plane, given a parabola 
passing through 3 points 
and 
.
vary in the range of 
, draw the domain swept out by .
is defined by 
and 
Show that there exist infinitely many prime number that divide at least one number in this sequences
be an acute scalene triangle , 
be the orthogonal projection of 
on 
, 
and 
are the midpoints of 
and 
respectively. Let 
and 
are points on the minor arcs 
and 
of the circumcircle of triangle respectively such that 
. Show that the circumcircles of triangles 
and 
are tangent if and only if lies on 
.
be 
given points in the plane, and let 
be a real number. Alice and Bob play the following game. Firstly, Alice constructs a connected graph with vertices at the points , i.e., she connects some of the points with edges so that from any point you can reach any other point by moving along the edges.Then, Alice assigns to each vertex 
a non-negative real number 
, for 
, such that 
. Bob then selects a sequence of distinct vertices 
such that 
and 
are connected by an edge for every 
. (Note that the length 
is not fixed and the first selected vertex always has to be 
.) Bob wins if![\[
\frac{1}{k+1} \sum_{j=0}^{k} r_{i_j} \geq r;
\]](http://latex.artofproblemsolving.com/1/7/9/1795db8e4a509dd465f6ff462093ae75b04de2b4.png)
otherwise, Alice wins. Depending on 
, determine the largest possible value of 
for which Bob has a winning strategy.
such that![\[
f(x^2 \cdot f(x) + f(y)) = f(f(x^3)) + y
\]](http://latex.artofproblemsolving.com/8/6/5/865b10bbc5ba7721078b1c60eaeede2abfeeb1a2.png)
.
be an acute triangle and let be an interior point for it such that 
. Prove that
When does equality hold?
, with 
, 
is the incenter, 
is the intersection of -excircle and . Point 
lies on the external angle bisector of 
such that and lieas on the same side of the line 
and 
. Point lies on such that 
. Circle 
is tangent to 
and at 
, circle 
is tangent to and at 
and both circles pass through the inside of triangle . if is the Midpoint od the arc , which does not contain , prove that lies on the radical axis of and . be a triangle with incircle 
, which touches sides 
and at points and , respectively. Choose points and on the line 
such that 
and 
. Let be the intersection of lines 
and 
. Define and 
as the intersections of 
and 
with lines 
and 
, respectively. Assume that 
is the intersection of 
and . Prove that 
.
be the altitudes in acute triangle 
, and let 
be an arbitrary point. Let 
be the feet of the perpendiculars from to the lines 
. Prove that 
are concurrent.
be the altitudes in acute triangle , and let be an arbitrary point. Let be the feet of the perpendiculars from to the lines . Prove that are concurrent.
be the altitudes in acute triangle , and let be an arbitrary point. Let be the feet of the perpendiculars from to the lines . Prove that are concurrent.
reflected about 
, and analogous results hold for the other two sides. Since the altitudes of ABC are the angle bisectors of its orthic triangle, the images of MN, PQ, and RS concur at the isogonal conjugate of X in the orthic triangle. Thus MN, PQ, and RS themselves must also concur.
reflected about , and analogous results hold for the other two sides. Since the altitudes of ABC are the angle bisectors of its orthic triangle, the images of MN, PQ, and RS concur at the isogonal conjugate of X in the orthic triangle. Thus MN, PQ, and RS themselves must also concur.
.
is given and let be, an arbitrary fixed point, and let 
be, it’s cevian triangle. be, an arbitrary also point and 
the traces on sidelines 
from the lines through and parallel to 
respectively.
be, the traces on segment lines from the lines through and parallel to the sidelines respectively.
are concurrent at one point so be it, 
as the midpoint of the segment between the point and the 
conjugate of 
In my drawing 
– We denote the intersection points of sidelines of 
from the segment lines 
as follows:
the intersection points of 
respectively, from the segment line 
the intersection points of respectively, from the segment line 
the intersection points of respectively, from the segment line 
the intersection point of 
the intersection point of 
the intersection point of 
pass through the points respectively.
– PROPOSITION 1. - A triangle is given and let be, an arbitrary point inwardly to it. We denote as the intersection points of respectively, from the line through and parallel to 
Also we denote as the intersection points of from the lines through and parallel to respectively. Prove that the lines 
are concurrent.
- PROOF. - We denote as the intersection point of 
and as 
the one of from the segment line 
because of 
and 
and so, the points 
are collinear. Hence the lines are concurrent and the proof of proposition 1, is completed.
– In our configuration now, we conclude that the line passes through the point 
intersection point of 
and similarly the lines pass through the points respectively.
and the point inwardly to it, we conclude that the lines are concurrent. Hence, the line 
passes through the point and similarly the lines pass through the points respectively.
– We denote as 
the intersection points of the line 
trom the segment lines 
respectively and as 
the intersection point of 
from the line 
We will prove that the points 
are collinear.
and 
where 
is the intersection point of 
and then we have that
congruent double ratios of these bundles of lines 
intersection of 
from the line 
intersection of from the line 
intersect the pencil 
and so, we have that 
we conclude that the lines 
are concurrent. So, the line 
passes through the vertex of as the intersection point of 
Hence, the points are collinear.
– We will prove now, that the lines are concurrent. Let be, the intersection point of 
and it is enough to prove that this point lies on 
because of it has already been proved that the line passes through the point We will prove that the pencils 
have congruent double ratios. That is, we will prove that
and 
They have the sideline of as their common ray and the points as the intersection point of 
as the intersection point of 
and as the intersection point of 
lie on the same line 
So, we have that
and because of 
is the common ray of pencils 
and we conclude that the points 
as the intersection point of 
as the intersection point of 
and as the intersection point of 
are collinear. Hence, the lines are concurrent at one point, so be it 
– We will prove now, that the concurrency point of the lines is the midpoint of the segment between the point and the conjugate of 
be, the intersection point of 
So, from the parallelogram 
we have that 
Hence, the line through vertex of and parallel to 
intersects the line 
at a point so be it 
such that 
of and parallel to 
intersects the line 
at the same point because of 
from the parallelogram 
of and parallel to 
respectively, intersect each other at points 
lie on the lines 
respectively. The triangle 
is the anticevian triangle of with respect to the point 
– We will prove now, that 
In triangle 
and from 
we have that
because of 
and 
and so, we conclude that the line 
connecting the vertices 
of the triangles respectively, intersects the segment line 
see the figure t=126138(c) as parallel to the sidelines of respectively, intersect them at pairs of points 
and 
and such that the intersection points 
lie on the segment lines respectively. of and parallel to 
respectively, intersect each other at the points also lie on the segment lines respectively. as the Cevian triangle of with respect to the point Hence the triangle is the Anticevian triangle of wrt 
are concurrent at one point here the point and also the segment lines 
are concurrent at one point here the point , based on the well known proposition 2, we conclude that the segment lines 
are concurrent at one point, so be it 
which in a such our configuration, is called the conjugate of with respect to the triangle 
see the proof 1 that the segment lines pass through the points as the midpoints of the segments 
respectively.
and 
and 
we conclude that the segment lines intersect the segment 
at the same point, so be it as it’s midpoint. That is the segment lines are concurrent at the point as the midpoint of the segment between and as the conjugate of with respect to the triangle and the proof is completed.
This proof is dedicated to Vladimir Zajic. is given and let 
be, the Cevian triangle of an arbitrary point in the plane. If are three points on the sidelines 
of respectively, such that the segment lines 
to be concurrent at one point, prove that the segment lines 
are also concurrent at one point.
. and so, we conclude that the line connecting the vertices of the triangles respectively, intersects the segment line at a point so be it such that 
because of 
are parallel to respectively and intersect the segment line at the sane point 
So, the triangles 
are perspective with perspector the point lies on the segment line and such that 
is the anticevian triangle of with respect to the point since is the cevian triangle of 
wrt we conclude that the point the concurrency point of is the midpoint of the segment between the point and the point as the conjugate of and the proof is completed. This proof is dedicated to Babis Stergiu.
and 
.
