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let 
denote the midpoint of side 
. Let the circle passing through point 
and tangent to side at point intersect sides 
and 
at points 
and 
, respectively. Let the line segments 
and 
intersect in point 
. Let 
be the second point of intersection of the circumcircles of triangles 
and 
. Prove that points 
and are collinear.
be an acute triangle and 
it's incenter. Let 
, 
and 
be the projections of onto , and respectively. 
and 
is the projection of onto 
. Let be the middle of the arc not containing and 
the second intersection of 
and the circumcircle of . If is the midpoint of 
, 
, 
and 
, prove that 
is cyclic.
+1 w
players take part. Every two play exactly one match. The schedule is such that no two matches are played at the same time, and each player, after taking part in a match, is free in at least 
next (consecutive) matches. Prove that one of the players who play in the opening match will also play in the closing match.
, point is the midpoint of side and the centers of the - excircles of triangles 
and 
are 
and 
, respectively. The circumcircle of triangle 
intersects line at points 
and . The circumcircle of triangle 
intersects line at points 
and 
. Prove that 
.
and satisfy ![\[ a^2+b^2+c^2 +abc = 4 . \]](http://latex.artofproblemsolving.com/d/3/a/d3ac01295b077425b5eaae556faa4cd678bce32a.png)
Show that ![\[ 0 \le ab + bc + ca - abc \leq 2. \]](http://latex.artofproblemsolving.com/2/0/9/209b3fcb56c01c335bbcb8d5fb781e28e49a0798.png)
be a triangle. Find all points on segment satisfying the following property: If and are the intersections of line 
with the common external tangent lines of the circumcircles of triangles 
and 
, then ![\[\left(\frac{PA}{XY}\right)^2+\frac{PB\cdot PC}{AB\cdot AC}=1.\]](http://latex.artofproblemsolving.com/0/8/0/080a689aea002f1b997294cb6aa48d3334881be7.png)
be positive real numbers. Prove that 
.
and 
in the plane and a subset 
of the plane, which are announced to Bob. Next, Bob marks infinitely many points in the plane, designating each a city. He may not place two cities within distance at most one unit of each other, and no three cities he places may be collinear. Finally, roads are constructed between the cities as follows: for each pair 
of cities, they are connected with a road along the line segment if and only if the following condition holds: distinct from and , there exists 
such[/center]
is directly similar to either or 
.[/center]
is directly similar to 
if there exists a sequence of rotations, translations, and dilations sending 
to , 
to , and 
to 
.
is a right-angled triangle, 
. Draw a circle centered at with radius . Let be a point on the side , and 
is tangent to the circle at . The line through perpendicular to meets line 
at . Line 
meets at point . The line through parallel to 
meets at 
. Prove that 
.
is a right-angled triangle, . Draw a circle centered at with radius . Let be a point on the side , and is tangent to the circle at . The line through perpendicular to meets line at . Line meets at point . The line through parallel to meets at . Prove that .
and be 
, Asume 
be the diameter of 
, denote the intersection of 
be .
, if we can prove 
then since 
,so 
,we can finish the proof .
:
are concyclic ,
(#)
and line 
, by Menelaus Th. , we get :
,
and be " (???) and "
" (???) ..., I will post my solution here. However, I see that we both may have some same ideas:
cut the circle with center , radius again at , what we have to prove is equivalent to prove that 
cut the circle with diameter 
again at , as 
is cyclic, hence 
, thus 
is cyclic, but as 
is isosceles, it is obvious that 
, notice that 
.
we get
, but 
, so 
hence 
and be " (???) and "" (???) ..., I will post my solution here. However, I see that we both may have some same ideas: cut the circle with center , radius again at , what we have to prove is equivalent to prove that cut the circle with diameter again at , as is cyclic, hence , thus is cyclic, but as is isosceles, it is obvious that , notice that . we get, but , so hence 
, 
, 
is also tangent to the circle hence 
. Now,
, we have
which is exactly what we want to prove.
. Notice that 
is cyclic, and furthermore it is the nine-point circle of 
. But is the 
-antipode in 
, hence as is the orthocenter, is the foot of the -altitude, and thus lies on the circle too.
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