AoPS Academy
be three non-negative real numbers satisfying 
such that any real numbers 
and 
satisfy
Proposed by EmilXM
board has some of its cells coloured red, blue, or green. Each cell is coloured with at most one colour, and some cells may remain uncoloured. Additionally, there is at least one cell of each colour. Two coloured cells are said to be friends if they have different colours and lie in the same row or in the same column. The following conditions are satisfied:
be an acute-angled triangle with 
and let 
be the centre of its circumcircle 
. Let 
be a point on the line segment 
such that 
. Let 
be the second point of intersection of and the line 
. If 
, 
and 
are the midpoints of the line segments 
, 
and 
, respectively, show that the points , and are collinear.
be the intersection of 
and . It is well-known that in configurations like this one, the midpoint of arc 
on 
is the center of the spiral similarity mapping to 
, and that the line 
is parallel to the -angle bisector by this spiral similarity. Next, see that the -angle bisector is parallel to the internal angle bisector of 
and we are done.
the midpoint of the line segment 
. Let 
. Notice that 
, and 
.
is a parallelogram. This gives 
is point 
.
is perpendicular to line which connects middles of sides of line segments 
and 
.
![[asy]
size(200);
defaultpen(fontsize(8pt));
pair A = (0,0);
pair B = (1,2);
pair C = (4,2);
pair D = (3,0);
pair P = rotate(15,C)*B;
pair M = midpoint(A--P);
pair N = midpoint(C--D);
pair Pp = extension(B,P,M,N);
pair I = midpoint(P--B);
draw(A--B--C--D--cycle^^C--P, lightblue);
draw(M--N, orange);
draw(B--P, magenta);
draw(P--Pp, dashed+magenta);
draw(A--P, palered);
draw(M--I--C, dashed+purple);
dot("$A$", A, dir(-100));
dot("$B$", B, dir(120));
dot("$C$", C, dir(40));
dot("$D$", D, dir(-30));
dot("$P$", P, dir(135));
dot("$M$", M, dir(140));
dot("$N$", N, dir(N));
dot("$I$", I, dir(I));
[/asy]](http://latex.artofproblemsolving.com/f/a/1/fa100a270c6133ab0a4e56e01a6efe6494e1067b.png)
be the midpoint of 
is a parallelogram.
and 
since 
is the midline of 
.
.
.
,
is an isoceles triangle, we have 
.
,
,
,
.
,
,
,
.
.