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 and 
is a sequence of real numbers for which 
and:
Prove that all elements of the sequence are positive.
is made up of 64 small unit cubes. Exactly 16 of these small cubes must be colored red, subject to the following condition:
, 
, and 
cubes, there must be exactly one red cube.
from the set of positive integers to the set of positive integers such that, for all positive integers 
and 
, there exists a non-degenerate triangle with sides of lengths![\[ a, f(b) \text{ and } f(b + f(a) - 1).\]](http://latex.artofproblemsolving.com/4/a/b/4ab97da27e05d0f9e7762f74d0d0bd2dc8b9842f.png)
,such that the inequality 
holds for all positive reals 
and for every positive real 
, there exist positive reals 
, such that the equality holds.
, let 
be the real root of equation 
. Prove that
;
.
be an integer, 
a finite set of integers (not necessarily positive) and 
subsets of . Suppose that, for every 
, the sum of the elements of 
is 
. Prove that contains at least 
elements.
Prove that
Where 
Prove that
Let 
Prove that
Let 
Prove that
be an acute triangle. Define 
as its incenter. Let 
and 
be the incircle's tangent points to 
and 
, respectively. Let 
be the midpoint of 
. Let 
be the intersection point of a perpendicular line passing through to 
. Line 
intersects the circumcircle of 
at 
. The circumcircle of 
intersects line 
at 
. Prove that quadrilateral 
is cyclic.
for which there exists a sequence of distinct positive integers 
, 
, 
, 
satisfying ![\[s_1^{s_2}=s_2^{s_3}=\cdots=s_{n-1}^{s_n}.\]](http://latex.artofproblemsolving.com/a/f/5/af594ff07aa2f89cb70ae669263558049556cf0e.png)
be a positive integer, and let be a subset of 
. An -partition of into 
parts is a representation of n as a sum 
, where the parts 
belong to and are not necessarily distinct. The number of different parts in such a partition is the number of (distinct) elements in the set 
.-partition of into parts is optimal if there is no -partition of into 
parts with 
. Prove that any optimal -partition of contains at most ![$\sqrt[3]{6n}$](http://latex.artofproblemsolving.com/2/0/1/201489b1f95ad200036698a606f712da7bb465b9.png)
different parts.
other knights where friendship is mutual. Prove that King Arthur can place some of his knights around a round table in such a way that every knight is friends with the 
people adjacent to him and between them there are at least 
friendships of knights that are not adjacent to each other.
cities in a country, where 
is an integer. Some pairs of cities are connected by direct (two-way) flights. For two cities and 
we define:
A 
between and as a sequence of distinct cities 
, 
, such that there are direct flights between 
and 
for every 
;
A 
between and as a path between and such that no other path between and has more cities;
A 
between and as a path between and such that no other path between and has fewer cities. and in the country, there exist a long path and a short path between them that have no cities in common (except and ). Let 
be the total number of pairs of cities in the country that are connected by direct flights. In terms of , find all possible values 
, 
.
.
are fixed positive integers.
are fixed positive real numbers.
be the set of all functions 
with the following properties: is defined for all real numbers and takes only real values.
the following equality holds: 
such that
,
.
be a parallelogram and point be the midpoint of ![$[AB]$](http://latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)
so that the quadrilateral 
is cyclic. If 
is the point of intersection of the lines 
and 
, and 
, then prove that the ray 
is the angle bisector of 
if and only if 
.
be a parallelogram and point be the midpoint of so that the quadrilateral is cyclic. If is the point of intersection of the lines and , and , then prove that the ray is the angle bisector of if and only if .
so 
is an isosceles trapezoid with 
. Hence triangle 
is isosceles and bisector of angle is perpendicular to 
. By ASA 
thus 
. Let 
be the midpoint of 
. We have 
is bisector iff 
