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 
can divide 
let 
be all distinct prime divisors of . Show that 
is an integer. ( where 
is defined as the number of positive integers 
that are relatively prime to .)
positive real numbers such that 
. Prove that :
be real numbers . Prove that
Where 
of rational numbers such that 
be a non-constant monic polynomial with integer coefficients and let 
be an infinite sequence. Prove that there are infinitely many primes, each of which divides at least one term of the sequence 
.
where they are integers and 
, where 
are in a gp.
be the set of all positive integers. Find all functions 
such that 
for all 
. be real numbers . Prove that
![$$-\frac{1}{18}\leq \frac{ab+a+b+3}{(a^2+3)(b^2+3)}\leq \frac{2+\sqrt[3] 2+\sqrt[3] {4}}{12} $$](http://latex.artofproblemsolving.com/9/9/9/9997397eb38cd4bceec2ff8c85951c78e2797d0f.png)
+1 w
be triangle with circumcenter 
and orthocenter 
. 
intersect again at 
respectively. Lines through parallel to 
intersects 
at 
respectively. Point 
such that 
is a parallelogram. Prove that lines 
and 
are concurrent at a point on 
.
be a triangle with incenter 
, and circumcircle 
. Let the circle with diameter 
intersects at 
and 
. Let 
be the midpoint arc 
and let 
intersect again at a point 
. Let the circumcircle of 
intersect the circumcircle of 
at a point 
. Let the tangent at to intersect 
at a point 
. Prove that 
and 
intersect on .
+1 w
satisfy,
Where {
} is a permutation of . Prove that 
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)
?
such that:
