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 an integer, 
a real number, and 
be positive real numbers such that 
. Prove that:![\[
\frac{1 + x_1^k}{1 + x_2} + \frac{1 + x_2^k}{1 + x_3} + \cdots + \frac{1 + x_n^k}{1 + x_1} \geq n.
\]](http://latex.artofproblemsolving.com/0/8/d/08de0c74a4e36b50a64d17875d3fd93eeb5b52de.png)
+1 w
with 
. Let 
be the midpoint of side 
, and let 
and 
be points on segments 
and 
, respectively, such that 
. Let 
be the circumcircle of 
, and 
the circumcircle of 
. The common tangent 
to and , which lies closer to point 
, touches and at points 
and 
, respectively. Let the line 
intersect again at 
, and the line 
intersect again at 
. Prove that the circumcircle of triangle 
is tangent to both and .
be positive real numbers. Prove that there exists a cyclic permutation 
of 
such that for all positive real numbers 
the following holds:![\[
\frac{a}{x\,a + y\,b + z\,c}
\;+\;
\frac{b}{x\,b + y\,c + z\,a}
\;+\;
\frac{c}{x\,c + y\,a + z\,b}
\;\ge\;
\frac{3}{x + y + z}.
\]](http://latex.artofproblemsolving.com/1/5/2/15261b3f712c2cc8ffbeb92a9711e06ad0992b5d.png)
islands of the Arctic Ocean. Every bear sometimes swims from one island to another. It turned out that every bear made at least one swim in a year, but no two bears made equal swams. At the same time, exactly one swim was made between each two islands 
and 
: either from to or from to . Prove that there were no bears on some island at the beginning and at the end of the year.
and 
such that 
.
be positive integers such that 
, 
, and 
are perfect squares. Prove that 
is also a perfect square.
be a convex polyhedron with the following properties: has exactly 
edges. differ by at most 
. with 
colors such that for each color 
and any two distinct vertices 
, there exists a path from 
to 
all of whose edges have color . for which such a polyhedron exists.
such that
divides 
for every 
, and
we have![\[
f\bigl(f(a)+kb\bigr)\;=\;f\bigl(a + k\,f(b)\bigr).
\]](http://latex.artofproblemsolving.com/c/f/f/cff41352422c8765978b8fe6dc966ba934a9df49.png)
be an acute scalene triangle. Denote by 
the circle with diameter 
, and let 
be the contact points of the tangents from to , chosen so that and 
lie on opposite sides of 
and 
and 
lie on opposite sides of 
. Similarly, let 
be the circle with diameter 
, with tangents from touching at 
, and 
the circle with diameter , with tangents from touching at 
.
are concurrent.
were selected 
pairwise distinct points, abscissas of which are integer numbers from the segment ![$[0; 100000]$](http://latex.artofproblemsolving.com/0/5/e/05e1d2855ef6a6dfaf9b24f2713455ab1017a30d.png)
. Prove that it is possible to choose six different selected points , , , 
, 
, 
such that areas of triangles 
and 
are equals.
baskets, each containing at least one candy. The total number of candies is 
. Asad and Sevinch make moves alternatingly, with Asad going first. On one move, Asad takes all the candies from 
consecutive non-empty baskets, while Sevinch takes all the candies from a single non-empty basket that has at least one empty neighboring basket. Prove that Asad can take overall at least 
candies, regardless of the initial distribution of candies and Sevinch's actions.
for which the following statement holds: there exists at least one triple of integers such that
and all triples of real numbers, satisfying the equations, are such that 
are integers. be a triangle. On side take points 
and 
such that 
take points 
and 
such that 
and on side take points 
and 
such that 
Let 
and 
are concurrent.
be a convex quadrilateral with ![\[\angle ADC = 90^\circ, \ \ \angle BCD = \angle ABC > 90^\circ, \mbox{ and } AB = 2CD.\]](http://latex.artofproblemsolving.com/b/4/1/b416d5c0b29c3a69a3e87cd22b6000aa8e70d456.png)
The line through 
, parallel to 
, intersects the external angle bisector of 
at point 
. Prove that the angles 
, 
, 
, 
, 
can be divided into two groups, so that the angles in each group have a sum of 
.
kawaii if it satisfies 
(
is the number of positive factors of , while 
is the number of integers not more than that are relatively prime with ). Find all that is kawaii.
works, now we show other 
,
.
be the two smallest prime factors of 
![\[\frac{\varphi(n)}n = \prod_{\text{prime }p\mid n} \left(1-\frac1p\right)\le \left(1-\frac1{q_1}\right)\left(1-\frac1{q_2}\right)=1+\frac1{q_1q_2}-\left(\frac1{q_1}+\frac1{q_2}\right){\overset{\text{AM-GM}}{\le}} \left(1 - \frac1{\sqrt{q_1q_2}}\right)^2\le\left(1-\frac1{\sqrt{n}}\right)^2\]](http://latex.artofproblemsolving.com/7/3/2/7322fddf32a558fede16a67bdb13ca22ab4850e6.png)
contradiction. Otherwise, 
,
,
which gives 
.