AoPS Academy
lying outside the circle 
, two tangents are drawn touching at points 
and 
. A point 
is chosen on the segment 
. Let points 
and 
be the midpoints of segments 
and 
respectively. The circumcircle of triangle 
intersects again at point 
(
). Prove that the line 
passes through the centroid of triangle 
.
has circumcircle 
, drop the perpendicular line from 
to 
and meet at point 
, similarly, altitude from 
to 
meets at 
. Prove that if 
be the circumcircle of a triangle 
. Denote by and the midpoints of the sides 
and , respectively, and denote by 
the midpoint of the arc of not containing . The circumcircles of the triangles 
and 
intersect the perpendicular bisectors of and at points 
and 
, respectively; assume that and lie inside the triangle . The lines and 
intersect at . Prove that 
.
be the set of positive rational numbers. Construct a function 
such that![\[ f(xf(y)) = \frac {f(x)}{y}
\]](http://latex.artofproblemsolving.com/5/9/1/59172e31cf6b7d885295f6e1daf6891a71739247.png)
, 
in .
bisects 
. Denote 
and 
.![\[(PQ; EF) = -1.\]](http://latex.artofproblemsolving.com/a/7/7/a778b533bd051f336bd505b11b4a0b07dafde6e2.png)
Combining this with 
, we use the Apollonius Circle Lemma to deduce the desired.![[asy]
size(225); defaultpen(linewidth(0.4)+fontsize(10));
pair A, B, C, H, D, E, F, P, Q;
A = dir(110);
B = dir(220);
C = dir(320);
D = foot(A, B, C);
H = .5A + .5D;
E = extension(A, C, B, H);
F = extension(A, B, C, H);
P = extension(E, F, B, C);
Q = extension(E, F, A, D);
filldraw(E--P--D--cycle, palegreen);
draw(A--B--C--A--D);
draw(F--D--E--P--B);
draw(B--E^^C--F);
dot("$A$", A, N);
dot("$B$", B, S);
dot("$C$", C, S);
dot("$P$", P, SW);
dot("$Q$", Q, dir(60)*1.5);
dot("$H$", H, dir(0));
dot("$E$", E, NE);
dot("$F$", F, NW);
dot("$D$", D, S);
[/asy]](http://latex.artofproblemsolving.com/0/1/b/01b15151e074158d0cfed459dc83a90462df7134.png)
and 
. Using the Ceva-Menelaus Configuration, it then follows that 
. Projecting through , we thus find that
Since 
and 
, the Angle Bisector Configuration then implies that 
bisects 
, so we therefore conclude that 
, as desired. 
and 
. We first use a lemma:
be collinear points in that order and let 
be any point not on this line. Then any of the following two conditions imply the third:
is harmonic;
;
bisects 
.
we have 
, and further that 
. Hence 
bisects , and we are done. 
and let 
. Then by right angles and bisectors lemma it suffices to show that 
. However, projecting we find,
as desired.
but this is true since 
are concurrent cevians in 
and 
. Then by EGMO 
, we have 
. And then by 
we have 
harmonic and 
, so 
bisects .
, 
, 
, 
, 
.
and 
.
and 
are opposites so 
so , as desired.
.
as the image of a point , and 
as the image of a line 
. lies on , 
and 
are parallel lines.
, so 
and 
are perpendicular to 
.
is a rectangle.
.
as desired.
. If 
the statement is obviously true.
to meet at ; it's well-known that 
. Then, we cite EGMO lemma 9.18. Because and 
, then we have that bisects , as desired. 
and 
. From EGMO 9.11,![\[(X,D;B,C) \overset{A}{=} (X,Y;F,E) = -1.\]](http://latex.artofproblemsolving.com/f/5/5/f55404d99ef4c0e5b595c7f9dc36d391a1aa8808.png)
, EGMO 9.18 finishes. 
which finishes from right angles and bisectors
be any interior point on 
,
at 
respectively. Prove that 
and by ceva's theorem, 
which gives 
and since both are acute, 
.
,
,
are Cevians of 
.
perpendicular to 
.
.
,
,
,
.
,
,
.
and 
,
respectively. This gives 
and 
.
,
to get 
.
,
,
.
,
.
,
.
,
and 
.
and 
,
until 
. Can someone please explain to me.
.
.
.
being a pencil we actually get that 
.
and 
,
or just 
Therefore by right angles and bisectors we conclude 
and 
Its clear that (perspectivity at 
From this, 
bisects 
(since 
and follows from law of sines), as desired.