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.
, define the sequence 
for 
as
Determine all values of 
such that there exists a number 
such that 
for infinitely many values of 
.
, 
, 
be an infinite sequence of positive integers. Suppose that there is an integer 
such that, for each 
, the number
is an integer. Prove that there is a positive integer 
such that 
for all 
.
that satisfy the equation
has an incenter 
Suppose 
. Let be the midpoint of 
. Suppose that 
. 
meets 
again at point 
. Let points 
and 
be such that is the midpoint of 
and is the midpoint of 
. Point 
lies on the plane such that 
is a parallelogram, and suppose the angle bisectors of 
and 
concur on 
.
and 
meet 
at 
and 
. Prove that 
.
be a triangle. Let 
and 
be the intersections of the 
and 
angle bisectors with the opposite sides. Let 
. Let 
where 
is the major 
arc midpiont.
and are coconic on a conic 
intersects 
again at , not equal to 
or , then prove 
and 
concur on 
be an acute-angled triangle with circumcircle 
and circumcentre 
. Points 
and 
lie on such that 
and 
. Let 
meet 
at , and 
meet at .
and 
have an intersection lie on line 
.
be a convex hexagon with 
and 
. which is equidistant from sides 
and 
.
and 
are the centers of mass of 
and 
, show that 
.
be a point in the plane, distinct from the vertices of 
. Consider 
the reflections of with respect to lines 
and 
, in this order.
are collinear if and only if lies on the circumcircle of . does not lie on the circumcircle of and the centroids of triangles and 
coincide, prove that is equilateral.
be a fixed square and 
a variable point on segment 
. The square 
is constructed such that is on segment 
and is on segment 
. Let be the intersection point of lines 
and 
. Find the locus of as varies along segment .
be a convex quadrilateral, and , 
, and be the midpoints of diagonals and 
, and side , respectively. Also, suppose that 
and that 
, 
. Calculate the perimeter of triangle 
.
be the circumcircle of triangle with center , and the inmixtilinear circle is tangent to 
at 
respectively. is the intersection of 
and 
and is the intersection of 
and . Prove that the isogonal conjugate of lies on the line passing through the midpoint of and .
be an acute triangle with 
, and let 
be the projection from onto . Let be a point on the extension of past such that 
. Let be on the perpendicular bisector of 
such that and are on the same side of and![\[\frac12\angle ALE=1.4\angle ABE+3.4\angle ACE-558^\circ\]](http://latex.artofproblemsolving.com/c/6/3/c63eeb053f1fa8a29bef93516814ae121382219e.png)
Let the reflection of across and be 
and , respectively. Let 
and 
such that 
. Let 
and 
intersect the circumcircles of 
and 
at 
and , respectively. Let 
and 
intersect 
and 
at and . Let intersect at . Prove that 
.
be the set of real numbers. Find all functions 
that satisfy the following condition. Here, 
is the function obtained by composing 
times, that is, 
, 
has no meaning here, so replace it with a given integer 
. Let 
be the assertion
From 
, we can easily observe that 
.
is bijective.
, we have that 
for all reals 
. This implies that is surjective. Now, assume that 
for some 
. Since 
, we have that 
and 
, implying with 
that 
. Hence, is injective, so we can conclude that is bijective.
.
be a nonzero real. With induction, we will prove that 
for any 
. The inductive hypothesis holds when 
. Assume that 
for any 
, for 
. From 
, we have that
Since 
and 
, and , we have that should hold. Therefore, from , we can obtain that 
, or, .
, we have that 
or 
must hold. In both cases, we have that . Hence, if a function 
is defined as 
, we have that 
is the inverse of , and so bijective. Also, 
always holds, and 
. Now, we can simplify as follows.
for any 
.
such that 
. When 
for some 
, from injectivity, we can deduce that 
, so 
for all positive integer . From , we have that 
also holds for all positive integer . From 
, we have that
The first one cannot hold from the reasons mentioned above. Therefore, for any positive integer , we have that 
. This implies that the sequence(orbit)
is an arithmetic progression. Hence, when , 
for any positive integer , so 
, a contradiction. Therefore, must hold for any .
was the inverse of , we can conclude that for all reals , and the identity function clearly satisfies the problem condition.
or 
and 
in which 
,
, so 
holds.![$a_{1}\xrightarrow[]{f}a_{2}\xrightarrow[]{f}a_{3}\cdots a_{101}\xrightarrow[]{f}a_{1}$](http://latex.artofproblemsolving.com/6/5/c/65c7b115f603c715e785d050e482537fff3cb482.png)
![$2a_{1}\xrightarrow[]{f}a_{1}+a_{2}\xrightarrow[]{f}2a_{2}\cdots 2a_{101}\xrightarrow[]{f}a_{101}+a_{1}\xrightarrow[]{f}2a_{1}$](http://latex.artofproblemsolving.com/5/4/7/547bb8b8dd2fbc71e0f92c1fed25a0cc9398cfd0.png)
where c is a constant.
or 
, so by the claim that is bijective,
or 
, so 
or 
for all 
or 
, 
, and 
, so by bijective, 
, contradiction
by bijective, 
, 
such that ![\[f(2x+y) + f(x+f(2y)) = f(x)f(y) - xy\]](http://latex.artofproblemsolving.com/0/6/2/0623b7c328af6d088cfc5dd288afe3e17a94ff29.png)
for all reals 
