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.
be finite.
satisfying the following condition for all real numbers 
:![\[
f(x - 1 - f(x)) = f(x) - x - 1.
\]](http://latex.artofproblemsolving.com/3/a/5/3a51479b61a5c025ed81f539e964ce96b5857f0f.png)
of positive integers such that ![\[\sqrt[3]{7x^2-13xy+7y^2}=|x-y|+1.\]](http://latex.artofproblemsolving.com/e/c/7/ec778edf97100f3b19350d8a817317ebf1f391bd.png)
such that 
is the square of an integer.
be a positive integer. Find the number of permutations 
, 
, 
of the
, 
, 
, satisfying
.
be a right angled triangle with 
.Let 
be the midpoint of 
, and 
be an arbitrary point on 
. The reflection of 
over 
intersects lines 
and at 
and 
, respectively. The circumcircles of 
and intersect again at 
. Prove that the center of the circumcircle of 
lies on 
.
and be on segment of an acute triangle such that 
and 
. Let and 
be the points on 
and 
, respectively, such that is the midpoint of and is the midpoint of 
. Prove that the intersection of 
and 
is on the circumference of triangle . with 
and 
, let be the midpoint of 
. Choose point on the extension of 
past 
and point on segment 
such that lies on 
. Let 
be on the opposite side of 
from 
such that 
and 
. Let 
intersect the circumcircle of 
again at 
, and let 
intersect the circumcircle of 
again at 
. Prove that , 
, and 
are collinear. with the following property: the 
positive divisors of have a permutation 
such that for 
, the number 
is a perfect square.
such that there exists a sequence of positive real numbers 
, 
, with the property that for every it holds that:![\[
a_1 + \cdots + a_{n+1} < \alpha \cdot a_n.
\]](http://latex.artofproblemsolving.com/9/b/7/9b73deb46032844b324288b02470ff0623ac0ff1.png)
Proposed by Pavle Martinović
and 
such that![\[
\frac{1}{\gcd(x, y)} + \frac{3}{xy} + \frac{y}{\operatorname{lcm}(x, y)} = y,
\]](http://latex.artofproblemsolving.com/c/8/2/c828f866a97dbb38fe5b6d7b8d190b3684365938.png)
where 
denotes the greatest common divisor of and , and 
denotes their least common multiple.
that satisfies: for all different 
, we have 
?
, let 
be the point on the segment such that 
. Suppose that the points 
, and the centroids of triangles 
and 
lie on a circle. Prove that 
.
at origin and on the 
axis. be 
and be 
, 
be 
, D be 
are 
means
-
hence the - coordinate of must be 
- coordinate of the circumcentre of 
, I got two relations,
be the centroids of 
, 
respectively. Note that as 
, we have 
. Since 
is a cyclic quadrilateral, it must be isosceles, so 
. As 
and 
concur at the midpoint 
of 
, it must hold true that 
.
It is given that , so through some manipulation 
. Thus the roots of 
and 
concur, whence 
. However, if 
and 
, then 
, contradiction (hmm maybe this is what TUAM meant?). Therefore as desired. 
, 
be respectively .
and 
cut at 
and resp.
. hence , 
is parallel to is a cyclic trapezoid . hence , it is isosceles . hence , 
[and ] cut at [it's mid pt.]
. Now , using Appolnius theorem gives , 
[as 
] 
+
= +
, apply stewart's theorem and then law of cosines to get 
= 
. This proves the result.
. Let the centroids be 
and be the midpoint of .
), we get 
and so, by appolonius theorem, we get 
.
then we are already done. So assume both of these lengths are distinct. For the sake of contradiction, assume that 
. Call the sum condition in the problem to be 
. be the midpoint of 
and let 
and 
be the centroid of and 
respectively. Since medians divide each other in the ratio of 
we know that![\[ \frac{ZX_1}{ZB} = \frac{ZX_2}{ZC} = \frac{1}{3}\]](http://latex.artofproblemsolving.com/0/2/7/027794e1b8964bf20c7acec46b09350cc0861257.png)
which gives 
. But since 
is cyclic, we get 
. Invoking ``Appolonius Theorem'' in and , we get that
Divide 
from to get 
which violates the triangle inequality which means the assumption is false and thus we are done. 
)
, therefore 
. Hence 
. is given by 
and the centroid of is given by 
has equation of the form 
. For both centroids, 
and 
, therefore 
must be the same.
should be equal if we interchange 
. This reduces immediately to 
, which implies 
so 
.
as 
. It is trivial that 
As 
are concylic by power of point we get,
,
,
and 
where 
denotes semi-perimeter of the triangle . Applying Apollonius Theorem in triangle and and equating the median lengths we get:
Clearly the later violates triangle inequality, hence 
is an isosceles triangle as desired. (QED) 
. Thus, is a isosceles trapezoid which implies . Therefore the medians passes through and are equal. By Median Theorem
and the result follows easily.
.
. and , we have 
. But note that 
, contradicting the triangle inequality. So we have 
. 
satisfying
Prove
and 
meet at 
is cyclic, 
.
is one-third 
and 
is one-third 
.
.
touches to the side 
.
.
we get that 
.
let Midpoint of 
midpoints of 
and 
perpendicular 
lies on 
and everything follows
and solving with AB+BD=AC+CD, we get AB=CD