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.
the number of positive integer 
satisfying 
is less than or equal to 
where 
is a constant independent of 
on the board. For each procces, the stutent deletes the number x and replaces it with either 
or 
or 
. Is this possible to make the number 
on the board?
be different real numbers. Prove that![\[\sum_{1 \leqslant i \leqslant n} \prod_{j \neq i} \frac{1-x_{i} x_{j}}{x_{i}-x_{j}}=\left\{\begin{array}{ll}
0, & \text { if } n \text { is even; } \\
1, & \text { if } n \text { is odd. }
\end{array}\right.\]](http://latex.artofproblemsolving.com/6/5/2/6525fc6b451ff5a3c83ff2a0541b063a7cf065c3.png)
lies inside triangle 
such that 
and 
. Point 
is the midpoint of segment 
. Point 
lies on segment 
with 
. Prove that 
.
be a triangle, and let 
be its altitudes. Let 
be its orthocenter, and let 
and 
be the circumcenters of triangles 
and 
. Let 
be the second intersection of the circumcircles of triangles 
and 
. Prove that the lines 
, 
, and 
-median of 
are concurrent.
has orthocenter . Let be a point distinct from the vertices on the circumcircle of . Suppose that circle 
meets 
at 
, and circle 
meets at 
. Prove that as moves on the circumcircle, the reflection of across line 
also moves on a fixed circle.
,
and 
. Interior angle bisector of 
intersects at 
, and 
at 
.Prove that circumcircle of 
passes from circumcenter of 
.
be inscribed in the circle 
. A line through point 
intersects and at points and , respectively. Let 
be the reflection of across the midpoint of , and 
be the reflection of across the midpoint of . Prove that: of triangle across line lies on the circle .
and 
coincide.
, 
be the orthocenter of it and 
be any point on the side 
. The points 
are on the segments 
, respectively, such that the points 
and 
are cyclic. The segments 
and 
intersect at 
is a point on 
such that 
is tangent to the circumcircle of triangle 
at 
and 
intersect at 
. Prove that the points 
and lie on the same line., 
, and 
, are marked on the plane. You have a pencil and a double-edged ruler of width 
. Using them, you may perform the following operations:
and 
are marked, draw the line connecting and .
and 
are drawn, mark the intersection of and .
is drawn, draw a line parallel to that is at distance away from (note that two such lines may be drawn). using these operations.
with centroid . Let 
be the power of w.r.t circumcircle of and 
be the power of w.r.t incircle of . prove that 
.
be a triangle with 
, and let 
be its -excenter. Let be the projection of to . Let 
be the intersection of 
and , and let 
be the points on 
, respectively, such that 
are on a line perpendicular to . Let the circumcircle of 
intersect again at 
. Suppose that the tangent of the circumcircle of at intersects at 
, and the segment 
intersects the circumcircle of at 
. Show that 
.
satisfying the condition: 
is the homothetic center of 
and let be the orthogonal projections of onto 
Then![$[PQR]=[OC'QB']+[OB'PA']+[OA'RC'] \ , \ (1)$](http://latex.artofproblemsolving.com/d/7/f/d7f4973a8794eb0ba44191cc575581ef13d9c384.png)
![$[ABC]=[OBC]+[OCA]+[OAB]=\frac{OX \cdot BC+ OY \cdot CA+OZ \cdot AB}{2} \ , \ (2)$](http://latex.artofproblemsolving.com/8/3/e/83e1840524b1b0cde44241a02bf32cf13a8962f5.png)
![$[OC'QB']=\frac{OX \cdot B'C'}{2} \ , \ [OB'PA']=\frac{OZ \cdot A'B'}{2} \ , \ [OA'RC']=\frac{OY \cdot C'A'}{2}$](http://latex.artofproblemsolving.com/4/b/6/4b688f491fc314171a671bd0ee093bb60f65da22.png)
gives![$[PQR]=\frac{OX \cdot B'C'+OY \cdot C'A'+OZ \cdot A'B'}{2}$](http://latex.artofproblemsolving.com/3/5/e/35ee880490ca295548e3cf0213da978a22bcde1e.png)
denotes the similarity coefficient of 
we have![$[PQR]=\frac{OX \cdot BC+OY \cdot CA+OZ \cdot AB}{2k}. $](http://latex.artofproblemsolving.com/b/a/4/ba475ecae0c4ed73a7d36f15580d88d350bf53df.png)
Together with 
we obtain![${[PQR]=\frac{[ABC]}{k}=k \cdot [A'B'C']}$](http://latex.artofproblemsolving.com/a/6/3/a63c13d02035541a8c61920e75c3431ddb42b990.png)
![$\Longrightarrow \ \frac{[PQR]}{[A'B'C']}=k=\frac{AB}{A'B'}$](http://latex.artofproblemsolving.com/7/7/a/77a0b96d69aacf0c806dc6f5a8522e283a4aed59.png)
be on the sides 
,
be on the segments 
in a way that 
, 
and 
.![\[ \frac{AB}{A'B'}=\frac{S_{PQR}}{S_{A'B'C'}}. \]](http://latex.artofproblemsolving.com/4/2/c/42c7c58a29501eee825a31956e21517c71ce7546.png)
is the surface of the triangle 
.