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 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 
.
, 
, 
be positive real numbers such that 
. Prove that ![$$\sqrt[3]{(a^2+1)(b^2+1)(c^2+1)} \le \left(\frac{a+b+c}{3}\right)^2 + 1.$$](http://latex.artofproblemsolving.com/2/0/7/207934bca53047858a3820f37705171ff0abfd45.png)
and be positive real numbers satisfying 
and 
Prove that![\[\frac{a}{(b+c-a)^2} + \frac{b}{(c+a-b)^2} + \frac{c}{(a+b-c)^2} \geq \frac{3}{(abc)^2}.\]](http://latex.artofproblemsolving.com/0/a/b/0ab4b4e6c28a1ae67788525642661dbcd04d0bd8.png)
such that ![\[ f(x^2 + yf(z)) = xf(x) + zf(y) , \]](http://latex.artofproblemsolving.com/2/b/a/2baee1d8464992fc38a16561b48b18f4b35807bd.png)
for all 
.
and in 
steps and has to reach again. But in between one of the s steps and can't go . Find probability. For example 
is valid but 
is not valid.
be an acute triangle. Points 
, and 
lie on a line in this order and satisfy 
. Let and 
be the midpoints of 
and 
, respectively. Suppose triangle 
is acute, and let 
be its orthocentre. Points 
and 
lie on lines 
and 
, respectively, such that 
and are concyclic and pairwise different, and 
and are concyclic and pairwise different. Prove that 
and are concyclic.
such for any :
.
satisfy,
Where {
} is a permutation of . Prove that 
be an integer. We are given a balance and 
weights of weight 
. We are to place each of the weights on the balance, one after another, in such a way that the right pan is never heavier than the left pan. At each step we choose one of the weights that has not yet been placed on the balance, and place it on either the left pan or the right pan, until all of the weights have been placed.
to satisfy that the equation 
has at least 1 roots 
of positive integers is called central if for every positive integer , the arithmetic mean of the first 
terms of the sequence is equal to .
, 
, 
, 
of positive integers such that for every central sequence , , 
, , there are infinitely many positive integers with 
.
of the equation![\[n^3=p^2-p-1\]](http://latex.artofproblemsolving.com/f/f/8/ff88093bb0aa93afcfdff3ae25149ec5fbacb61f.png)
is a prime number and is an integer
where 
are the lengths of the altitudes from the three vertices.
is obtuse, then the side opposite of the obtuse angle is the longest one (let's name that side ). Let the other two sides of the triangle be and such that 
. The vertex opposite of the side , i.e. vertex , is in fact such a vertex, as if 
is a cevian drawn from to (with endpoints and 
), then we have 
(based on the fact that in 
, i.e.that 
), 
(based on the fact that 
), as well as 
(based on the fact that 
-- this can be proven the same way was proved--, so we have that 
, which holds because if 
then 
, which contradicts the triangle inequality). is acute, and 
, then vertex is yet again the desired vertex, as if is a cevian from that vertex then (because , which can be shown similarly as it was shown in the previous case), (because ), and finally, in order to prove that , it is sufficient to show that 
, where 
is the altitude from to (and since 
), and if we let 
, then is equivalent to proving 
which is equivalent to 
, i.e. we have to prove that 
, which holds because 
.
?
