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 
.
be an acute-angled triangle. The extension of the altitude 
of 
beyond intersects 
at . Let is the midpoint of side 
. The circles and 
intersect again at 
. Let be the projection of onto 
. Prove that 
.
over and let it be 
, from spiral sim and gliding principle we have 
and in general from this and miquelpoint we get 
tangent to 
and from Shooting Lemma we get 
colinear now simply finish by noting that is circumcenter of 
thus we are done 
.
and 
then is the midpoint of 
, and we finish by power of a point.
be the reflection of over , and let 
.
is the antipode of 
.
is right, is the circumcenter, so 
. Then we have 
. Therefore 
, so is the antipode of ., , and are collinear.
, 
, and 
be on the unit circle. We have 
, 
, and 
. Scale everything up by a factor of 
.![\[\begin{vmatrix} ab+ac-bc & \frac{b+c-a}{abc} & 1 \\ -ab & -\frac 1{ab} & 1 \\ \frac c2(a-b) & \frac{1}{2abc}(b-a) & 1 \end{vmatrix} = 0.\]](http://latex.artofproblemsolving.com/7/b/3/7b3195bf773c9745373321f2c1c528617bda5160.png)
Scaling up the second column and third row, it suffices to show that ![\[\begin{vmatrix} ab+ac-bc & b+c-a & 1 \\ -ab & -c & 1 \\ c(a-b) & b-a & 2 \end{vmatrix} = 0.\]](http://latex.artofproblemsolving.com/0/3/8/038b25ac3bf4e72035b2bcd8e9fc6a278ca2f029.png)
I don't feel like LaTeXing this but it's a relatively quick computation.
. We have 
. Since 
, 
is cyclic. Then 
, so as desired.
