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 a natural number. Prove that,![\[ \left\lfloor \frac{n}{1} \right\rfloor+ \left\lfloor \frac{n}{2} \right\rfloor + \cdots + \left\lfloor \frac{n}{n} \right\rfloor + \left\lfloor \sqrt{n} \right\rfloor \]](http://latex.artofproblemsolving.com/5/6/0/560a9568b2c09512cac1d317a011931a439e6a02.png)
and the sets 
If 
has exactly three elements, prove that 
cannot be an integer.
such that
holds for all 
.
such that:
with all 
is real number
be a triangle, and let 
, 
, and 
be collinear points such that 
, 
, and 
. Points 
, 
, and 
are the reflections of , , and over 
, 
, and 
, respectively. Prove that if 
is a nondegenerate triangle, then its circumcenter lies on the circumcircle of .
be three positive real numbers satisfying 
Prove that
There is a very elegant proof :-D Could anyone think of it?
such that is a multiple of 
and has exactly positive factors.
opposite the vertex 
be tangent to the side at the point 
. Define the points 
on and 
on analogously, using the excircles opposite 
and 
, respectively. Suppose that the circumcentre of triangle 
lies on the circumcircle of triangle . Prove that triangle is right-angled.
be the circumcircle of triangle , 
is the midpoint of and 
be the second intersection of 
and . Tangent line of at intersects at 
, let 
be a transversal of and 
be intersections of 
with . Let 
be a tangent line of . Prove that 
is cyclic
be positive integers such that 
and 
. Find the maximum possible value of 
.
is a triangle, let 
is orthocenter of 
, let 
is midpoint of 
. Let 
is a line perpendicular with 
at point . Let meet 
at 
respectively. Prove that 
.
be, the orthogonal projections of 
respectively, on the line segment 
and we denote the points 
from 
and 
and then, the points 
and 
are the reflexions of respectively, with respect to 
with 
as their orthocenters respectively and so, their distances from their homologous vertices, are equal.
and the proof is completed.
is a chord of a circle 
and is a point on 
Two distintc chords 
and 
pass through 
and 
cut at 
respectively. Then we have![\[ \frac{1}{PA}-\frac{1}{PX}=\frac{1}{PB}-\frac{1}{PY} \]](http://latex.artofproblemsolving.com/0/f/6/0f6e88f2cc6b36865336290a1ea0bfab7c9359f6.png)
the diametrically oposite point of 
in the circumcircle of 
. 
are collinear since by angle chasing 
is a parallelogram. Also, 
and 
are both cyclic (opposite angles equal 
, sum 
. Since 
this implies is angle bisector of 
then 
.
the diametrically oposite point of in the circumcircle of . are collinear since by angle chasing is a parallelogram. Also, and are both cyclic (opposite angles equal 
, sum . Since 
this implies is angle bisector of then 
.
![$[(x-2)^2+(y-4)^2][ (x-1)^2 + (y-2)^2] = 36$](http://latex.artofproblemsolving.com/b/1/0/b108dc2cac5b7491fb83418eb36c7184f6d5d30c.png)
then
cut 
at 
and 
cut 
at 
.
so the circle from Butterfly Theorem is of center 
.
and 
.
perpendicular to (d), we have that 
.
and 
,
.
