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be a positive integer. For a set of points in the plane 
, we call 
distinct points 
connected if the line 
contains exactly 
points from .
such that there exists a set of points in the plane with the property that any point 
is connected with exactly 
other points from .
be a point in the plane, distinct from the vertices of 
. Consider 
the reflections of with respect to lines 
and 
, in this order.
are collinear if and only if lies on the circumcircle of . does not lie on the circumcircle of and the centroids of triangles and 
coincide, prove that is equilateral.
then prove \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{2}\cdot\frac{abc}{ab+bc+ca}\ge \frac{5}{3}.$$
and 
Prove that
Let and 
Prove that
Let and 
Prove that
Let and 
Prove that
is the incentre and point 
is defined such that 
and 
is the perpendicular bisector of 
let 
and 
be the circles passing through 
and 
respectively such that 
is tangent to and define 
to be a point such that it lies on both the circles and prove that 
and 
are perpendicular.
weights, each weighing 
and another 
weights with 
each. Write a algorithm that determines the ways in which a scale can be balanced with a weight 
on the left pan, and display the number of possible solutions. (The weights can be placed on both pans and the program starts with the numbers 
. What will be displayed after three successive runs: 5,2,5,1,4 | 5,2,5,1,11 | 5,2,5,1,20?
with orthocenter 
. Extend 
to meet at 
. The angle bisector of 
meets the midpoint of 
. If 
, compute the area of 
.
with side length of 
. The point 
is selected on the diagonal 
such that 
. Denote the circumcenters of 
as 
. Find the length of 
be a triangle, (
). The line 
passing trough the vertices 
and orthogonal to the angle bisector 
of 
, meets and the median 
of the side at points and 
, respectively. Prove that segment 
bisects the segment 
.
be a triangle, (). The line passing trough the vertices and orthogonal to the angle bisector of , meets and the median of the side at points and , respectively. Prove that segment bisects the segment .
intersect at . It is clear that 
, thus is the midpoint of 
. is the midpoint of , we get 
.
.
. intersect at . Since is a midline of 
, we have 
is a midline of 
, and this implies that 
, which in turn implies that bisects 
.
and 
. Since 
and 
is the angle bisector of 
, we have that 
is isosceles, i.e. 
. Also, 
is the midpoint of 
. Hence, 
is the midline for 
, i.e. 
, from where it is clear that 
is the midpoint of 
.
. It is sufficient to show that![\[
\frac{BG}{GD} = \frac{CE}{ED}.
\]](http://latex.artofproblemsolving.com/0/a/8/0a8d82d51c1332e1f9d955307f2b98c62481f0a4.png)
From and 
, we have![\[
\frac{BG}{GD} = \frac{BK}{DF} = \frac{2BK}{AK} \tag{1}.
\]](http://latex.artofproblemsolving.com/9/9/1/99106be2266fdfcf22f5a7ecd1a9056609cb5aee.png)
Also,![\[
\frac{CE}{DE} = \frac{CD - DE}{DE} = \frac{CD}{DE} - 1 = \frac{AD}{DE} - 1 = \frac{AE - DE}{DE} - 1 = \frac{AE}{DE} - 2.
\]](http://latex.artofproblemsolving.com/1/b/b/1bb87694ed76019680ef01486dec765c89682a84.png)
![\[
= \frac{AB}{DF} - 2 = \frac{AK + BK}{\frac{AK}{2}} - 2 = 2 + \frac{2BK}{AK} - 2 = \frac{2BK}{AK}. \tag{2}
\]](http://latex.artofproblemsolving.com/8/8/6/886723577cd4ce2e320706eb8d909097f466b11d.png)
From (1) and (2), we have![\[
\frac{BG}{GD} = \frac{CE}{ED},
\]](http://latex.artofproblemsolving.com/d/9/4/d94000488a007d705c3e3aa87485afeed3a034a7.png)
so . Since is the midpoint of , it follows that the segment bisects the segment 
.
and 
Prove that
such that ![\[f(2x+y) + f(x+f(2y)) = f(x)f(y) - xy\]](http://latex.artofproblemsolving.com/0/6/2/0623b7c328af6d088cfc5dd288afe3e17a94ff29.png)
for all reals 
and 
.
