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: 
on s.t. 
. Now let 
. The line 
passes through (because of well-known projective properties of the complete quadrilateral). Let 
. We have 
, and since is the midpoint of 
, we find 
, so is the midpoint of . Furthermore, we have 
, which is fixed, so 
is constant, meaning that lies on a circle obtained from by a homothety of centers and (one of them is the external center, the other one is the internal center).
, with directed segments. Let p be a number such that 
, and consider the circle P(D, p).
. On the other hand, let the parallel to the diamater MN through the point D meet the line BM at V. Then, Thales yields 
. Since , we have 
, or, equivalently, 
. Since MA = AN, this implies DU = DV. This implies that the points U and V coincide, i. e. the parallel to the diameter MN through the point D meets the lines CN and BM at the same point. This point, of course, must then coincide with the point of intersection K of the lines CN and BM. Thus, K = U = V. Therefore, the equation becomes 
. On the other hand, we know that 
; thus, 
. Now, AN = r; hence, DK = p, and it follows that the point K lies on the circle P(D, p), completing our proof., it follows that the points B and C are the two centers of similitude of the circles T and P.]
