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.
, 
, 
, the inequality![\[ \frac {1}{a^3 + b^3 + abc} + \frac {1}{b^3 + c^3 + abc} + \frac {1}{c^3 + a^3 + abc} \leq \frac {1}{abc}
\]](http://latex.artofproblemsolving.com/f/a/9/fa964e29de91f0e2f3cb32e335a52d4521bc38fd.png)
is a rational number.
and other permutations.
,
, 
are perpendicular foot from 
to sides 
, prove that 
bisect segment
.
. Prajit then tries to place n points such that no four points are concyclic and no 
points are collinear in Euclidean plane, such that Shining cannot find a group of three points such that their circumcircle contains none of the other remaining points. Is he always able to do so?
be an infinite sequence of positive integers. Prove that there exists a unique integer 
such that![\[a_n < \frac{a_0+a_1+a_2+\cdots+a_n}{n} \leq a_{n+1}.\]](http://latex.artofproblemsolving.com/9/4/f/94fc5a51b7e68588123c5b527fe75183bc4c4937.png)
such that
.
denote the number of positive integers less than 
that are relatively prime to . Prove that there exists a positive integer 
for which the equation 
has at least 
solutions in .
with real coefficients for which there exists a polynomial with real coefficients such that for all real 
, 
of points in the plane is balanced if, for any two different points 
and 
in , there is a point 
in such that 
. We say that is centre-free if for any three different points , and in , there is no points in such that 
.
, there exists a balanced set consisting of points. for which there exists a balanced centre-free set consisting of points.
inside the rectangle such that each of the distances from to the 4 corners of the rectangle is an integer.
inside the rectangle such that each of the distances from to the 4 corners of the rectangle is an integer.
exists.
, 
, 
, and 
where 
and 
are positive odd integers.
, and 
be the distances from to 
, 
, 
, and 
respectively. So 
and 
and 
. Then 
and 
, so 
and 
are perfect squares.
is an integer, which implies 
is rational with a denominator not divisible by 
(when in lowest terms). Similarly, 
, and are also rational with denominators not divisible by (when in lowest terms).
is the denominator of and 
, while 
is the denominator of 
and , we scale the rectangle through 
by 
(note this cannot be divisible by ).
denote the point 
after scaling and 
denote the distance 
after scaling.
, 
, 
, and 
, implying that 
, and 
are positive integers. Also, 
, 
, 
, and 
must be positive integers.
and 
are odd (if 
), or both are 
(if 
). and are both odd, then one of 
or 
and one of 
or 
must be odd. Thus, one of 
, 
, 
, or 
must be , which cannot be a perfect square, so one of , , , or cannot be an integer, contradiction. and are both , then at least one of or is even (and ). and are , then since 
were multiplied by a number , we must have that 
are all odd. Hence, , , , and are all , which cannot be perfect squares, so , , , and cannot be integers, contradiction. or is even, WLOG assume is odd while is even. So 
and 
must be even while 
and 
must be odd. But 
, so one of or must be 
, while the other is . So one of 
or 
must be 
, while the other is 
. Meanwhile, 
and 
are 
. Thus, one of , , , or must be 
, which cannot be a perfect square, so one of , , , or cannot be an integer, contradiction. does not exist.
. Let there be points, namely, 
on 
respectively. Since the cases are symmetric, 
: 
. Finally, connect 
to 
, to 
, to 
and to . is not an integer.
. By Pythagoras Theorem, 
yields no integer solution. 
, and since is not integer, it implies that either 
or both are not integers. 
and 
Prove that
.
and 
Prove that
![$$ a+b+kc^2\geq\frac{3\sqrt[3]{k}}{4}$$](http://latex.artofproblemsolving.com/5/5/b/55be0378562556d1aa4400bef11d1510ff722854.png)
Where 
,