AoPS Academy
.
+1 w
such that ![\[ \forall x, y \in \mathbb{R}^+ \ , \ \ f(x+f(xy))=f(x)+xf(y).\]](http://latex.artofproblemsolving.com/f/2/7/f27575aa4092155d285af0eccebe3865043011bc.png)
be a triangle with incentre 
and circumcircle 
. Let 
and 
be the second intersection points of with 
and 
, respectively. The chord 
meets 
at a point 
, and 
at a point 
. Let 
be the intersection point of the line through parallel to 
and the line through parallel to 
. Suppose that the tangents to at 
and 
meet at a point 
. Prove that the three lines 
and 
are either parallel or concurrent.
be the point such that 
is a parallelogram . Hence 
and 
Hence 
is a parallelogram.
. But 
. Hence 
is cyclic.
. 
. Then points 
are concyclic. Using 
is a parallelogram we get
Similarly 
. Hence,
![[asy]
size(200);
pair B=dir(-160),C=dir(-20),E=dir(70),F=dir(135),A=extension(B,F,C,E),P=extension(B,E,C,F),Q=B+C-P;
draw(unitcircle,cyan);
fill(P--E--F--P--cycle,purple+grey+grey);
fill(B--Q--C--B--cycle,purple+grey+grey);
dot("$A$",A,dir(A));
dot("$B$",B,dir(B));
dot("$C$",C,dir(C));
dot("$E$",E,dir(E));
dot("$F$",F,dir(F));
dot("$P$",P,dir(-90));
dot("$Q$",Q,dir(Q));
draw(A--B--C--A,royalblue);
draw(B--E^^C--F,red);
draw(P--A--Q,green);
draw(B--Q--C^^E--F,brown);
[/asy]](http://latex.artofproblemsolving.com/c/3/a/c3a5e6bde8cb758fbb2214e5c8fa14940fbd8308.png)
denote homothety at with scale 
followed by reflection in internal angle bisector of 
. Note 
and 
. Thus 
is a point such that 
Hence 
. As also consists of reflection in internal angle bisector of , so 
follows. 
and 
. Since these are parallel pairs of lines, 
are the points at infinity for those pencils of parallel lines respectively. Note that 
by 
and 
. Thus, 
is isogonal to 
wrt. 
. Hence by DDIT on complete quadrilateral 
, there exists a projective involution swapping 
, 
, 
. This is taking the isogonal, and so 
are isogonal. 
Putting them together, we get![\[\frac{\sin(\angle QAC)}{\sin(\angle BAQ)} = \frac{CQ}{BQ}= \frac{BP}{PC} = \frac{\sin(\angle BAP)}{\sin(\angle PAC)}\]](http://latex.artofproblemsolving.com/2/1/9/219c9710c194c69c02102a61f129a2c27089f106.png)
and since 
, we have that 
and we're done. .
. Note that 
. Also note that 
. Both angle equalities are there since is a parallelogram. So, consider homothety 
under follow by reflection along the angle bisector of with scale 
. Note that sends to 
and to . Then sends to 
such that 
. So, it also implies that the line 
is a reflection of the line 
along the angle bisector of , which implies . So, we're done.
be the isogonal conjugate of 
Consider the lines through parallel to 
These lines intersect 
at points 
with between and 
and they intersect at points 
with between and 
Since 
we have that 
is cyclic. Also, since 
we have 
is cyclic, and similarly 
is cyclic, so 
is cyclic, and by Reim's theorem we have 
parallel. Now, take a homothety at sending to 
It is not hard to see that this takes to 
so 
are collinear, and we have 
as desired. 
be the isogonal conjugate of in , the angle condition implies that is on the perpendicular bisector of . Use barycentric coordinates with respect to and let 
. Then 
and 
. We have
where the last equality holds because is on the perpendicular bisector of . Hence are collinear, so 
. Done.
and 
. Then 
is cyclic and is a parallelogram, so![\[\angle PXY = \angle BCP = \angle CBQ, \quad \angle PXY = \angle CBP = \angle BCQ.\]](http://latex.artofproblemsolving.com/c/4/8/c48139ea59c1b9b90c2540bb70d5bc11e2b8d5eb.png)
as well as 
, so the quadrilaterals 
and 
are also similar with opposite orientation, which implies the desired. 
be the reflection of across the midpoint of 
. It is clear that 
is a parallelogram. Then, notice that 
, and 
. Hence, is a parallelogram, meaning that 
.![\[\angle ABQ = \angle PBQ+\angle ABP =\angle PCQ+\angle PCA = \angle ACQ.\]](http://latex.artofproblemsolving.com/3/1/f/31f97b897eac7a55612a5c680e47def8dfe46b2c.png)
, so is cyclic. Thus,![\[\angle BAP = \angle ABR = \angle AQR = \angle CAQ. \ \square\]](http://latex.artofproblemsolving.com/1/6/a/16a8865ddddc47799abcab7efec03a458234fb5f.png)
be the point such that 
is a parallelogram. First note that 
since is a parallelogram as well. Additionally, this gives 
so 
is also a parallelogram. Then 
so 
is cyclic.
and 
so 
Finally, 
as desired.
, 
, and .
is a parallelogram since 
and 
. Similarly, 
is a parallelogram. Then we have 
so 
is cyclic. Then 
so we are done.
, 
, 
, 
. The Sine Law in 
and 
gives 
, i.e. 
. Analogously 
. However, 
and 
from the parallelogram , thus 
. Hence 
, equivalent to 
, i.e. 
, as desired.
from point .
, and 
, so we get that , 
, and are involutions.
, so and are both involutions around the angle bisector of . So it follows that and are isogonal as desired.
over the midpoint of be 
We see that 
is a parralelogram. We show 
is cyclic because then 
Let 
Then 
Then note that 
Therefore 
as desired
until is the orthocenter of , then 
is the antipode so it's obvious.
to be parralelogram and trivial by similarity of 
and .
, 
being isogonal. From here Isogonal Line Lemma (or DDIT if you wish) gives the required.
, then 
is cyclic and 
.
and its reflection 
the 
angle bisector. is sent to and to and as this transformation preserves similarity, it sends to , hence and are isogonal.
and 
to intersect and at 
respectively. Then, is cyclic and thus we have . Furthermore, note that 
(with similar logic following for 
and 
) so in fact, 
and are homothetic to each other. Then, since they are oppositely oriented but share , it thus follows that they must be reflections about the angle bisector and thus as desired.
.
is a parallelogram. Prove that 
.
,
,
,
and comparing them we get that
.
and 
then the equation reduces to 
be its reflection over 
and 
are obviously parallel. 
are antiparallel wrt 
since 
form a parallelogram, so done.![[asy]
unitsize(1cm);
pair A, B, C, P, Q, X;
A=(2,3sqrt(5));
B=(0,0);
C=(8,0);
P=(2,4sqrt(5)/5);
Q=B+C-P;
X=A+P-B;
draw(A--B--C--A--P--B--Q--C--P--A--X--P--A--Q--C--X);
draw(circumcircle(A,C,P));
label("$A$", A, NW);
label("$B$", B, W);
label("$C$", C, SE);
label("$P$", P, dir(-105));
label("$Q$", Q, S);
label("$X$", X, N);
[/asy]](http://latex.artofproblemsolving.com/7/0/4/704b5f7f0625f0924ab5d52d0476cee1554d07cc.png)
be the reflection of 
and 
,
is a paralellogram. Therefore, 
,
is cyclic. Therefore, 
.
and 
to meet 
;
and that 
.
