be positive integers that satisfy the inequality![\[
\frac{a}{b} < \frac{x}{y} < \frac{c}{d}
\]](http://latex.artofproblemsolving.com/8/d/9/8d982d52e655546ac29039fa6e440e68821bfe7f.png)
with 
.
,
.
.
to 
and 
as the reflections of 
be the circumcircle of 
.
with 
with 
is the circumcenter of 
can be inscribed within a circle.
,
,
at 
,
and 
be the midpoints of 
and 
,
and 
lie on 
so that 
and 
.
and 
intersect on the circumcircle of 
.
Y by Rounak_iitr, erringbubble, omgggg, ItsBesi
Let 
be an acute triangle. Let 
be a point on side 
and 
be a point on side 
such that lines 
and 
are parallel. Let 
be an interior point of 
. Suppose rays 
and 
meet side at points 
and 
, respectively, such that both and lie between 
and 
. Suppose that the circumcircles of triangles 
and 
intersect at a point 
. Prove that the points 
, and 
are collinear.
be an acute triangle. Let 
be a point on side 
and 
be a point on side 
such that lines 
and 
are parallel. Let 
be an interior point of 
. Suppose rays 
and 
meet side at points 
and 
, respectively, such that both and lie between 
and 
. Suppose that the circumcircles of triangles 
and 
intersect at a point 
. Prove that the points 
, and 
are collinear.
and 
intersect 
at 
respectively. Since 
and 
we get 
are cyclic. Hence 
which gives that 
as desired
and 
.
is cyclic
,
is cyclic, so we are done by applying the Radical Axes Theorem on 
,
,
.
at 
.
is cyclic.So
is cyclic and with the same way we can found 
is cyclic.Thus, radical axises of 
,
.
which may be obtained by applying DDIT to 
.
intersects 
at 
then 
.
or 
lies on radical axis of 
or 
,
and 
we hav
a
such that 
and it lies on radical axis of 
,
where 
is radical axis.
and 
,
and 
we prove that 
,
and project on 
are pairs under an involution. Thus 
and 
is the radical axis of 
intersects 
intersects 
is cyclic
is cyclic so 
is cyclic
is cyclic and because 
and 
intersect at 
and 
are collinear
is cyclic
and similary for 
and we are done.
and since 
we have that 
so 
and 
meaning it lies on their radical axis i.e 
be 
respectively and 
.
were cyclic, 
would pass through 
.
from 
and 
.
we have 
.
and divide the relations to get:
we get 
.
and divide the relations to get 
.
and 
like we did before and divide the relations to get: 
and 
.
and 
(well known lemma). Finally, we have that: 
is cyclic and hence 
is cyclic.
but 
so 
so 
is cyclic.
cyclic so 
and how 
we have 
so A lies on the radical axis of 
and 
.
and 
.![[asy]
import olympiad;
import geometry;
size(9cm);
// defaultpen(fontsize(10pt));
pair A = (-0.36,0.93);
pair B = (-0.86,-0.51);
pair C = (0.86,-0.51);
pair D = (-0.59, 0.27);
pair E = extension(D,D+(C-B),A,C);
pair X = (-0.13, -0.12);
pair P = extension(D,X,B,C);
pair Q = extension(E,X,B,C);
pair U = intersectionpoints(circumcircle(D,X,E),circumcircle(B,Q,X))[0];
pair V = intersectionpoints(circumcircle(D,X,E), circumcircle(X,P,C))[0];
pair Y = intersectionpoints(circumcircle(X,P,C), circumcircle(X,Q,B))[0];
draw(A--B--C--A, red);
draw(A--X, magenta+dashed);
draw(D--E, red);
draw(D--P, red);
draw(E--Q, red);
draw(circumcircle(P,X,C), orange);
draw(circumcircle(Q,X,B), orange);
draw(circumcircle(D,X,E), deepgreen+dashed);
draw(circumcircle(B,C,U), blue+dashed);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$D$", D, dir(D));
dot("$E$", E, dir(E));
dot("$X$", X, dir(270)*1.8);
dot("$Y$", Y, dir(E)*1.8);
dot("$P$", P, dir(P));
dot("$Q$", Q, dir(Q));
dot("$U$", U, dir(110));
dot("$V$", V, dir(90));
[/asy]](http://latex.artofproblemsolving.com/b/9/4/b945c70388ac9b9439fbccd5cbe75571b0912464.png)
and 
are cyclic quadrilaterals.
.![\[\measuredangle XDE = \measuredangle XPQ = \measuredangle XVC = \measuredangle XVE. \]](http://latex.artofproblemsolving.com/d/2/6/d267e8abd1f9f651c7e167611759e10555eee011.png)
For 
This completes the proof. 
,
and 
concur, it must follow that 
are collinear. 
.
intersect 
intersect 
which means that 
is cyclic. Similarly 
is cyclic. Therefore 
is cyclic. Now observe 
therefore 
is cyclic and since 
,
,
and 
,
which implies the points 
gives the points 
and 
,
which means points 
as desired.
.
and we get 
;
;
are pairs under some involution. Projecting this onto 
;
;
are pairs under some involution.
and we are done.
.
,
.
to get 
is cyclic.
,
concur at some point. But we know that 
and 
concur at 
again at 
and 
are cyclic. Thus, 
is cyclic, so by Reim's theorem again, 
,
and 
concyclicities then the statement is equivalent to show that 
.
.
=
and hence 
and 
and hence the conclusion 
.
,
.
is cyclic.
(1)
,
.
,
.
, 
and 
Points 
and 
Points 
and 
Points 
and 
Points 
and 
but 
Points 
,
and 
,
.
is in the radical axis of 
Then, by the ratio lemma at 
in 
in 
but, by the paralelism and from the the fact that 
what makes us get that 
intersect 
intersect 
at 
.
is the radical axis of the two circles, and 
is cyclic. We have that 
so 
is cyclic. But 
so 
is cyclic. So the entire pentagon 
is cyclic. Now 
is cyclic if 
or if 
. But 
and 
so 
,
,
.
.
are concyclic, and finish by PoP and radax.
Then 
Similarly 
so pentagon 
is cyclic. Now by homothety 
is cyclic, and we finish by Radical Axis. QED
and 
circumcircles of 
and 
,
different of 
different of 
and 
are cyclic.
is cyclic, then 
.
cyclic, we have 
.
then 
.
that means 
then 
that means 
is cyclic.
and 
,
means 
is cylic. Then 
where we obtain 
