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.
such that 
divides 
for all positive integers 
and 
with 
.
, are there infinitely many positive integers 
such that 
is a perfect square? (Here 
denotes the integer part of the real number 
?
be an acute triangle with 
, Let 
be the intouch triangle with 
,
,
,, let 
be the intersecttion of the perpendicular from 
to 
with 
, and 
.
and 
are concylic
be a triangle with circumcircle 
, circumcenter 
, and orthocenter 
. Assume that 
and that 
. Let 
and 
be the midpoints of sides and 
, respectively, and let 
and 
be the feet of the altitudes from 
and 
in 
, respectively. Let 
be the intersection of line 
with the tangent line to at 
. Let 
be the intersection point, other than , of with the circumcircle of 
. Let 
be the intersection of lines 
and . Prove that 
. be the foot of perpendicular from to the Euler line (the line passing through the circumcentre and the orthocentre) of an acute scalene triangle . A circle 
with centre 
passes through and , and it intersects sides and at and 
respectively. Let be the foot of altitude from to 
, and let be the midpoint of . Prove that the circumcentre of triangle 
is equidistant from and .
is a cyclic quadrilateral inscribed in a circle of radius one unit. If 
, prove that is a square.
, 
, 
, 
satisfy 
and 
. Prove that there exists a permutation 
of 
such that 
.
be a cyclic quadrilateral. A circle centered at passes through and and meets lines 
and again at points and (distinct from 
). Let denote the orthocenter of triangle 
Prove that if lines 
are concurrent, then triangle and 
are similar.
such that 
, where 
are non-negative reals such that 
.
.
a tangencial quadrilateral that is not a rhombus, let 
be lengths of tangents from 
to the incircle of the quadrilateral which center is 
. Let 
be the midpoints of 
resp. Prove that![\[ \frac{MI}{IN}=\frac{a+c}{b+d} \]](http://latex.artofproblemsolving.com/e/7/0/e709acb8b17c0466a385ba8e7a89bb1bf16c72af.png)
is given. It turns out that for every positive integer number 
, for which the numbers 
and 
are positive, they have the same number of divisors. Is it necessarily true that 
?
. Prove that if![\[\dfrac {1} {0^2+1}+\dfrac{1}{1^2+1}+\cdots+\dfrac{1}{(p-1)^2+1}=\dfrac{m} {n},\]](http://latex.artofproblemsolving.com/d/5/f/d5f55cc86a95d51109ace2dc3e737bddecc144b0.png)
(and the fraction 
is in reduced terms) then 
.
and 
(it is essential that 
is a quadratic residue modulo 
such that 
.
,![\[(p-1)! = \left (\prod_{\ell=1}^{(p-1)/2} \ell\right )\left (\prod_{\ell=1}^{(p-1)/2} (p-\ell)\right ) \equiv (-1)^{\frac {p-1} {2}}(((p-1)/2)!)^2 \pmod{p},\]](http://latex.artofproblemsolving.com/7/1/7/7174f97c5aeae035b4d7b4995a6f0c1fe9fe5b38.png)
.
we will have 
(although knowing the exact value is irrelevant in the sequel).
for 
(and it divides 
and 
)
;
yields 
,
.![\[E=\sum_{\ell\neq \pm \textrm{i}} \dfrac {1} {\ell^2 + 1} = \sum_{\ell\neq \pm \textrm{i}} \dfrac {1} {(\ell - \textrm{i})(\ell + \textrm{i})} = \dfrac {1} {2\textrm{i}} \sum_{\ell\neq \pm \textrm{i}} \left (\dfrac {1} {\ell - \textrm{i}} - \dfrac {1} {\ell + \textrm{i}}\right ).\]](http://latex.artofproblemsolving.com/1/d/f/1df3809beb7b98cf17abb6cd0497275473cd9d19.png)
.
,
,