AoPS Academy
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 triangle
are points in , such that 
is a point at the line such that 
is perpendiculat to 
is perpendicular to 
.
)
.
is a triangle are points in , such that is a point at the line such that is perpendiculat to is perpendicular to .).
is a triangle are points in , such that is a point at the line such that is perpendiculat to is perpendicular to ..
. But 
(since 
)
.
(or anything of such kind). Then this problem becomes rather nice.
, 
.
be 
. Then 
and 
, 
.
.
.
.
.
. But 
. Hence 
.
. But (since ).(or anything of such kind). Then this problem becomes rather nice., . be . Then and , . .
is the bisector of the angle 
, i.e. 
.
, then 
). The problem is the picture: the angle C is very obtuse!
whit 
obtuse. Clearly 
because 
and is the midpoint of the hypotenuse 
at the right triangle 
. Let's call 
to be the 
. 
. As 
the 
is isosceles whit 
. The 
is isosceles whit 
and 
because 
. Let be the point of intersection of whit . As 
and then 
, therefore 
. And as 
it is clear that 
is the bisector external of the 
. Now, as is the bisector external of the we have for the Angle Bisector Theorem that 
. Let be the midpoint of , it is clear that 
is the bisector of because 
, we see that the triangles 
, 
and 
are congruent because 
and 
then cleary 
, and as 
then 
., and are collinear points on the sides , 
and 
of the 
, then for the Menelaus's Theorem we have that 
, then 
, 
.![[asy]
usepackage("tikz");
label("\begin{tikzpicture}
\coordinate[label=below:$A$] (A) at (-4,0);
\coordinate[label=below:$E$] (E) at (0,0);
\coordinate[label=below:$F$] (F) at (4,0);
\coordinate[label=below:$B$] (B) at (8,0);
\coordinate[label=above left:$D$] (D) at (.8,2.4);
\coordinate[label=above right:$C$] (C) at (3.2,1.6);
\coordinate[label=above:$K$] (K) at (2.4,3.2);
\draw[thick] (A)--(K)--(F)--cycle;
\draw[thick,blue] (E)--(D)--(B)--cycle;
\draw[thick,red] (D)--(F);
\draw[thick,purple] (E)--(K);
\foreach \s in {A,B,C,D,E,F,K}\filldraw (\s) circle (1.5pt);
\end{tikzpicture}");
[/asy]](http://latex.artofproblemsolving.com/f/4/0/f40ff1a0ea2e86d0463ca5c934be72c2e37797a8.png)
and 
. Because 
, then and are circumcenters of 
and 
, respectively. We have ![\[\angle EFD = 2\angle EBD = 2\angle BDF = 2x\implies \angle DFC=x=\angle CDF\implies CF=DC.\]](http://latex.artofproblemsolving.com/0/e/3/0e32bd04099a6d605941c145cce3f6f6c608e0f3.png)
Let 
. From LoS of 
, we have 
. BBy LoS of 
, we have 
. By simple angle chasing, we have 
and 
. From LoS of 
, we have![\[\frac{4a\cos^2 (x)}{\cos (3x)} = \frac{6a\cos (x)}{\cos (2x)} \implies 2\cos (2x)\cos(x)=3\cos (3x)\implies 4\cos^3(x) - 2\cos (x) = 12\cos^3(x) - 9\cos (x)\]](http://latex.artofproblemsolving.com/e/1/3/e133bf2700347f69b54fa61bdbce84259c32f3cc.png)
and we have 
. Therefore,![\[\frac{DB}{DC} = \frac{4a\cos^2(x)}{a} = 4\cos^2(x)=\boxed{\frac{7}{2}}.\]](http://latex.artofproblemsolving.com/8/c/7/8c7556f8196009b50f1cc692aa2d346efc81c080.png)
