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.
of prime positive integers 
such that number 
equals to a square of an integer.
satisfy all conditions below:
for all 
for all 
for all
be real number greater than 
. Let![\[S=\log_a {bc}+\log_b {ca}+\log_c {ab}\]](http://latex.artofproblemsolving.com/5/9/e/59e1a33017116963165f744b49b8515f126c64ef.png)
.
are such that 
, find the minimum value of expression 
for which:
for all 
.
such that 
, 
, and 
are all equal to
, 
be real numbers, such that 
. Prove that 
.
, we have:
(where 
are the digits and 
) we define the \textit{permutation sum} as the value 
For example the permutation sum of 20253 is 
Let 
and 
be two fivedigit integers with the same permutation sum.
.
be the incircle of triangle 
. Line 
is parallel to side 
and tangent to . Line 
is parallel to side 
and tangent to . It turned out that the intersection point of and lies on circumcircle of 
be non-negative real numbers such that 
. Prove that![\[a\sqrt{1+b^{3}}+b\sqrt{1+c^{3}}+c\sqrt{1+a^{3}}\le 5.\]](http://latex.artofproblemsolving.com/f/f/a/ffaeec844a6bdf10a96e8e9d298ef97ac839d713.png)
such that there exists a sequence of positive integers 
, 
,
, 
satisfying: ![\[a_{k+1}=\frac{a_k^2+1}{a_{k-1}+1}-1\]](http://latex.artofproblemsolving.com/7/b/4/7b4ab849cdc409906514e2ebb688802f17df41ea.png)
for every 
with 
.
be an equilateral triangle with circumcenter 
. Let 
and 
be two points on segments 
and 
, respectively, such that 
. If 
is the reflection of with respect to line 
, prove that lines 
and 
are parallel.
and 
be the intersection of 
with 
. Let 
and 
be the reflections of 
and 
wrt 
, it is easy to see that 
is an isosceles trapezoid, 
also belongs to this cyclic since 
, that is, 
is cyclic.
be the midpoint of 
, we know that 
(1), and let 
.
lies on so since 
is altitude of , and 
are isogonal lines wrt 
. (2)
, we see that 
then 
and since 
is a radius 
. Thus, by what was concluded in (2), we have that 
therefore 
is an equilateral triangle.
is the altitude of the equilateral 
then, 
. Let 
be a point on 
such that 
then 
and 
(3)
therefore 
(3), that is 
.
