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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 
of radius 
composed with reflection in bisector of 
maps this "problem" to the same problem in a similar triangle; the closer intersection is mapped to the farther one in the new figure. Using this similarity we get the isogonality.
by factor 
. Then reflect everything wrt angle bisector of . Finally invert with power 
. It can be seen that after this transformation 
are interchanged with one another. So 
has gone to 
and we are done.
with radius 
where is the foot from to . This inversion maps the circle with diameter to that with diameter 
. Now, their intersections are mapped to each other! (Since they are inverses) Since the intersection points are not fixed (check!) we get that 
are isogonal.
with radius 
and reflect all trough the angle bisector of angle 
. We will prove that and swap. Let and 
be the projections of and 
onto and 
. The circle pass trough 
, and while pass trough 
, and . By simple segment calculating, we get that: 
thus, our inversion and reflection makes:
as we wanted.
, and are collinear?
, 
and 
(where and are circles with diameter and ,respectively) and concur on 
and since this intersection point is 
.But how it can help? 
be the respective midpoints of 
. Then it's clear that 
, so the internal angle bisector of 
also bisects , this yields![\[\frac{NL}{LM}=\frac{CN}{NM}=(\bigstar)\]](http://latex.artofproblemsolving.com/4/0/f/40f87eddfeed2c10b75982c935732ca85485c89e.png)
Now let the external angle bisector of meet 
at 
,
bisects 
bisects 
are concyclic on a circle with diameter 
.
lie on the -Appolonian cirle of 
are concyclic.
bisects 
.
, as desired. 
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 
.
The circles with diameters 
