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.
satisfying 
, find the biggest value the following equation could acquire:
, let 
be the circumcentre, 
be the orthocentre, and 
be the centroid.
satisfy the following condition:
is constant.
be a point outside circle 
centered at . Tangents from to touch at 
. Line 
intersects bigger 
arc at 
.The line drawn from parallel to 
intersects 
at 
. Ray 
intersects small 
arc at 
. Prove that the circumcircle of 
is tangent to .
be the set of real numbers. Let 
be a function such that ![\[f(x+y)f(x-y)\geqslant f(x)^2-f(y)^2\]](http://latex.artofproblemsolving.com/5/d/3/5d3476fd9195d085d0030d6036a9fde1cf6756a7.png)
for every 
. Assume that the inequality is strict for some 
.
for every 
or 
for every .
square is filled with numbers 
.The numbers inside four 
squares is calculated,and arranged in an increasing order. Is it possible to obtain the following sequences as a result of this operation?
amount of balloons, where 
. Then, on his first turn, Bob takes 
amount of ballons where 
. After first turn, Alice and Bob alternately takes as many balloons as his/her partner has. Is it possible for Bob to take amount of balloons at first, such that after a finite amount of turns, one of them have a number of balloons that is a multiple of 
?
such that the number 
is
consecutive even numbers, such that the square of their product is divisible by the sum of their squares.
to 
(including ). He writes some of the numbers on the numerator, and writes 
(product) between each number. Then he writes the rest of the numbers in the denominator and also writes between each number. There is at least one number both in numerator and denominator. The teacher ensures that the fraction is equal to the smallest possible integer possible.
prime number, there exists only one positive number that makes the equation 
a perfect square of a positive integer
such that there exists a prime number 
, such that 
is a power of 
. is a number of the form 
where 
is a positive integer.
determine the minimum and maximum value of![\[\frac{\sum_{i,j=1}^n|a_i+a_j|}{\sum_{i=1}^n|a_i|}.\]](http://latex.artofproblemsolving.com/6/f/2/6f2427edf9a5e8e8f2e12276d0a841765e8322f1.png)
is a triangle
are points in , such that 
is a point at the line such that 
is perpendiculat to 
is perpendicular to 
.
)
.
is a triangle are points in , such that is a point at the line such that is perpendiculat to is perpendicular to .).
is a triangle are points in , such that is a point at the line such that is perpendiculat to is perpendicular to ..
. But 
(since 
)
.
(or anything of such kind). Then this problem becomes rather nice.
, 
.
be 
. Then 
and 
, 
.
.
.
.
.
. But 
. Hence 
.
. But (since ).(or anything of such kind). Then this problem becomes rather nice., . be . Then and , . .
is the bisector of the angle 
, i.e. 
.
, then 
). The problem is the picture: the angle C is very obtuse!
whit 
obtuse. Clearly 
because 
and is the midpoint of the hypotenuse 
at the right triangle 
. Let's call 
to be the 
. 
. As 
the 
is isosceles whit 
. The 
is isosceles whit 
and 
because 
. Let 
be the point of intersection of whit . As 
and then 
, therefore 
. And as 
it is clear that 
is the bisector external of the 
. Now, as is the bisector external of the we have for the Angle Bisector Theorem that 
. Let 
be the midpoint of , it is clear that 
is the bisector of because 
, we see that the triangles 
, 
and 
are congruent because 
and 
then cleary 
, and as 
then 
., and are collinear points on the sides , 
and 
of the 
, then for the Menelaus's Theorem we have that 
, then 
, 
.![[asy]
usepackage("tikz");
label("\begin{tikzpicture}
\coordinate[label=below:$A$] (A) at (-4,0);
\coordinate[label=below:$E$] (E) at (0,0);
\coordinate[label=below:$F$] (F) at (4,0);
\coordinate[label=below:$B$] (B) at (8,0);
\coordinate[label=above left:$D$] (D) at (.8,2.4);
\coordinate[label=above right:$C$] (C) at (3.2,1.6);
\coordinate[label=above:$K$] (K) at (2.4,3.2);
\draw[thick] (A)--(K)--(F)--cycle;
\draw[thick,blue] (E)--(D)--(B)--cycle;
\draw[thick,red] (D)--(F);
\draw[thick,purple] (E)--(K);
\foreach \s in {A,B,C,D,E,F,K}\filldraw (\s) circle (1.5pt);
\end{tikzpicture}");
[/asy]](http://latex.artofproblemsolving.com/f/4/0/f40ff1a0ea2e86d0463ca5c934be72c2e37797a8.png)
and 
. Because 
, then and are circumcenters of 
and 
, respectively. We have ![\[\angle EFD = 2\angle EBD = 2\angle BDF = 2x\implies \angle DFC=x=\angle CDF\implies CF=DC.\]](http://latex.artofproblemsolving.com/0/e/3/0e32bd04099a6d605941c145cce3f6f6c608e0f3.png)
Let 
. From LoS of 
, we have 
. BBy LoS of 
, we have 
. By simple angle chasing, we have 
and 
. From LoS of 
, we have![\[\frac{4a\cos^2 (x)}{\cos (3x)} = \frac{6a\cos (x)}{\cos (2x)} \implies 2\cos (2x)\cos(x)=3\cos (3x)\implies 4\cos^3(x) - 2\cos (x) = 12\cos^3(x) - 9\cos (x)\]](http://latex.artofproblemsolving.com/e/1/3/e133bf2700347f69b54fa61bdbce84259c32f3cc.png)
and we have 
. Therefore,![\[\frac{DB}{DC} = \frac{4a\cos^2(x)}{a} = 4\cos^2(x)=\boxed{\frac{7}{2}}.\]](http://latex.artofproblemsolving.com/8/c/7/8c7556f8196009b50f1cc692aa2d346efc81c080.png)
.
