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 all the divisors of 
. It is known, that 
are all divisors for some 
(except 
). Find all such 
be a sequence of positive real numbers that satisfies 
and 
for every natural number . Prove that 
for every natural number .
be positive real numbers such that : 
. Prove that : 
be the set of positive integers. Find all functions 
such that![\[ m^2 + f(n) \mid mf(m) +n \]](http://latex.artofproblemsolving.com/2/f/4/2f409d1de993f1af8fd839bb8e9f87a57e1b8608.png)
and .
(ans- 
)
satisfying 
Prove that 
When does equality hold?
red balls and 
white balls. Audi took two balls at once from inside the bag. The chance of taking two balls of the same color is ...
unit. The area of the hexagon is ...
and are the roots of the cubic equation 
. The value of 
is ...
so that 
and 
is ...
, 
, 
, 
has an average of 
and a median of 
. The largest number of such data is ...
are integers greater than 
which are four consecutive quarters of an arithmetic row with 
. If 
and 
are squares of two consecutive natural numbers, then the smallest value of 
is ...
, with 
, 
and 
. The points 
and 
lies on the line segment 
. with 
and 
. The measure of the angle 
is ...
meet 
for each natural number . The value of 
is ....
provided that the difference of any two numbers at least 
is ...
which satisfies 
are as many as ... with 
and 
. Point lies on the side so that 
. Suppose 
is a point on the side extension 
so that 
is perpendicular to . The point lies on the ray such that 
and 
. The large angle 
is ...
consists of integers with the following properties: For every three different members of there are two of them whose sum is a member of . The largest value of is ....
with 
positive reals is ....
with 
is ....
square units. Two plots are said to be neighbors if they both have one side in common. Initially, there are a total of 
coins on the chessboard where each coin is only loaded exactly on one square and each square can contain coins or blanks. At each turn. You must select exactly one plot that holds the minimum number of coins in the number of neighbors of the plot and then you must give exactly one coin to each neighbor of the selected plot. The game ends if you are no longer able to select squares with the intended conditions. The smallest number of so that the game never ends for any initial square selection is ....
be an isosceles triangle such that 
. Let be a point on the side such that 
and a point on side 
such that 
. If 
, calculate 
.
are not congruent, but 
.
reflection of 
about 
, see that 
( 1 ). From angle bisector theorem we get 
( 2 ), but from Thales 
( 3 ); with (1) and (3), (2) becomes 
, so 
, or is tangent to the circle 
, i.e. 
( 4 ) thus, if 
, then 
. With (4) we get 
, or 
, i.e. 
.
and 
.
?
Write only the answers to the questions given.
to be more exact:
in years 2002-08 time was 90' for part A and 120' for part B
.
so that 
.
, 
.
.
![[asy]
import markers;
import geometry;
pair A=0, B=(6cos(2pi/10),6sin(2pi/10)),
C=(12cos(2pi/10),0),D=(3,3tan(2pi/10)),E=(3,0);
label("$A$",A,SW);
label("$B$",B,N);
label("$C$",C,SE);
label("$D$",D,NW);
label("$E$",E,S);
draw(A--D,blue);
draw(B--D);
draw(A--E,red);
draw(B--C,red);
draw(C--D,blue);
draw(C--E);
draw(D--E);
markangle(Label("$\theta$"),D,C,A,blue);
markangle(Label("$\theta$"),B,C,D,blue);
markangle(Label("$2\theta$"),C,A,B,blue);
markangle(Label("$3\theta$"),C,D,B,blue);
markangle(Label("$\pi-4\theta$"),A,B,C,blue);
perpendicular(E,NW,size=5mm,blue);
[/asy]](http://latex.artofproblemsolving.com/9/d/c/9dc6e9df8e8228f5a97fe496efd233233c8065aa.png)
