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.
such that 
is the square of an integer.
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)
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 
?
such that 
. The lines 
and intersect at a point and the line passing through which is parallel to , intersects 
at . Prove that
and parallel to intersects and 
at 
and 
, respectively. We have
Also, 
.
.
is 
.
. We have
Also, 
. Hence, we conclude that the angle bisector of is .
and 
. Note that by the given angle condition, 
is isosceles with 
. By angle chasing, it suffices to prove 
, or that 
.
and 
, we have 
, this gives 
, which implies the result. 
, foot of the altitude from to be 
, the perpendicular at to meet 
at , foot of the altitude from to be , 
, 
meet 
at 
respectively. is the angle bisector of 
![\[(R,A;B,F)=(TC,TA;TB,TF)=(C,G;B,TF\cap BC)=(FC,FG;FB,FT)=(K,G;A,T)=(C,G;H,L)\]](http://latex.artofproblemsolving.com/a/5/c/a5c2884b0bc82141aaf4d009631dca8cf204614f.png)
![\[\frac{GH}{GL}=\frac{AH}{TL}=\frac{AB}{TP}=\frac{GB}{GP}=\frac{GM}{GN}=\frac{DM}{DN}=\frac{CH}{CL}\]](http://latex.artofproblemsolving.com/f/7/b/f7b3f4c9655d05a98a3e6a39b66a028e2fa62b2e.png)
As desired.
fixed. Animate over 
. Denote 
as reflection of 
to 
.![\[f:A\rightarrow AC\rightarrow R(AC,BC)\rightarrow R(AC,BC)\cap BQ\]](http://latex.artofproblemsolving.com/8/4/5/84560e0cca6bcbd3effff4a7d1b08a21c1d377fc.png)
![\[g: A\rightarrow AD\rightarrow AD\cap BC=T\rightarrow T(QB)_{\infty}\cap BD=P\rightarrow PC\cap QB\]](http://latex.artofproblemsolving.com/9/8/0/9801aae710aba9cd1dd54164b94777e46a64b68c.png)
has degree .
.
to intersect at .
. to intersect at 
and the parallel from to intersect at .
are same at 
points as desired.
satisfying
Prove
