[/quote]
,
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 
.
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 
.
is the midpoint of segment 
.
lies on segment 
with 
.
.
be its altitudes. Let 
and 
be the circumcenters of triangles 
and 
.
be the second intersection of the circumcircles of triangles 
and 
.
,
,
are concurrent.
meets 
,
meets 
.
also moves on a fixed circle.
,
.
intersects 
,
at 
.
passes from circumcenter of 
.
.
and 
coincide.
,
be the orthocenter of it and 
be any point on the side 
.
are on the segments 
,
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 
.
and 
.
and 
are marked, draw the line connecting 
and 
are drawn, mark the intersection of 
is drawn, draw a line parallel to 
be the power of 
be the power of 
.
,
be its 
be the intersection of 
and 
be the points on 
,
are on a line perpendicular to 
intersect 
.
intersects the circumcircle of 
.
.
satisfying the condition: 
Y by Adventure10
Let 
be a center of a circle 
, 
a diameter of it and 
any point in different than 
and 
. Let 
be the intersection of the perpendicular drawn from to 
. On the line 
is marked the point 
, so that 
is half of 
and does not belong to the segment . By we draw the parallel to that cuts the line in 
. We call 
the point of intersection of the lines 
and 
. Prove that , and are collinear.
be a center of a circle 
, 
a diameter of it and 
any point in different than 
and 
. Let 
be the intersection of the perpendicular drawn from to 
. On the line 
is marked the point 
, so that 
is half of 
and does not belong to the segment . By we draw the parallel to that cuts the line in 
. We call 
the point of intersection of the lines 
and 
. Prove that , and are collinear.This post has been edited 3 times. Last edited by NiltonCesar, May 11, 2018, 5:40 AM
with transversal 
,
.
.
,
,
and 
.
.
we have that:
we get that 
are colinear we have that:
solving for 
we get that 
we easily get 
,
are colinear.
.
