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.
, an arbitrary point 
is chosen on the side 
. In triangles 
and 
, the angle bisectors 
and 
are drawn, respectively. The point 
is the circumcenter of 
. Prove that the second intersection point of the circumcircles of triangles 
and 
lies on the line 
.
be a point in the plane of triangle 
such that the segments 
, 
, and 
are the sides of an obtuse triangle. Assume that in this triangle the obtuse angle opposes the side congruent to . Prove that 
is acute.
and 
be two distinct rays not lying on the same line, and let 
be a circle with center 
that is tangent to ray at 
and ray at 
. Let 
be a point on segment 
. The line through parallel to intersects line at . Let 
be the intersection of lines 
and , and let 
be the intersection of line and the line through parallel to . Prove that line 
is tangent to .
be an acute triangle with point 
lie on the plane such that 
is a parallelogram. 
is the orthocenter of 
intersects 
at 
and 
intersects 
at 
The circle with diameter 
intersects the circumcircle of 
again at 
Prove that: The circumcircle of 
passes through the reflection point of over 
with 
, let 
denote the incenter and the midpoint of side 
. The line through 
perpendicular to 
intersects the tangent from to the incircle (different from line ) at a point > Show that is tangent to the circumcircle of triangle 
.
and 
. At each step, it is allowed to add to one of the polynomials and to multiply another one by the polynomial 
. Can the polynomials become equal at some point?
such that for any positive integer number 
and any arrangement of the numbers from 1 to 
on a circle, there exists a pair of adjacent numbers whose ratio (when dividing the larger number by the smaller one) is less than .
from the nonnegative integers to the integers satisfying 
and ![\[ f(mn) = f(m) + f(n) + f(m)f(n) \]](http://latex.artofproblemsolving.com/e/f/f/eff10a32cffaec742403ebfc46ff09bc8ef7b7c5.png)
for all nonnegative integers 
and 
.
make the equation ![\[\sum_{i=1}^n \sum_{j=1}^n \left\lfloor \frac{ij}{n+1} \right\rfloor=\frac{n^2(n-1)}{4}\]](http://latex.artofproblemsolving.com/e/7/f/e7fb8c20a43535b5aaed5ab254f1ef043263c62b.png)
true?
be the binary representation of 
. Prove that for any positive integer , at least one of the digits 
, 
, 
, 
equals 
.
positive integer numbers is equal to 
. The product of their factorials is a perfect tenth power of some positive integer. Prove that all numbers are equal to each other.
. If not all the numbers are equal, then 
. So, the product of the factorials is divisible by 
, and therefore must also be divisible by 
. Since all numbers are less than 
, 
for all 
. Hence, 
for all , which means 
. However, 
, contradiction!
,
and 
.
since 
.
,
are all 
,
,
.
,
So 
and so 
for all 
and 
and as 
so 
and with same logic 
and with same logic 
and with same logic 
which is contradiction