and a positive integer 
,
.
there is a positive integer 
is an integer. Is it true that for each perfect polynomial 
so that for each integer 
for which 
with the property that 
is divisible by 
.
Y by Amir Hossein, nguyendangkhoa17112003, Adventure10, Mango247, ItsBesi, Rounak_iitr
The incircle of triangle 
touches the sides 
, 
, 
at 
respectively. 
is a point inside triangle of such that the incircle of triangle 
touches at 
, and touches 
and 
at 
and 
respectively.
Show that
are concyclic.
touches the sides 
, 
, 
at 
respectively. 
is a point inside triangle of such that the incircle of triangle 
touches at 
, and touches 
and 
at 
and 
respectively.Show that
are concyclic.
and 
,
,
are concurrent,and let 
,
,
,
,
are concyclic.
.
.
,
(I mean polar wrt the incircle). Since 
and 
belongs to the polar of 
.
,
.
and 
to find that 
and 
coincide (they must both be equal to the harmonic conjugate of 
,
.
and 
.
,
is on the circle with center 
and radius ![$[BD]$](http://latex.artofproblemsolving.com/3/2/6/3261e689901bce018ecdef47b9bc60a78ead3746.png)
.
.
and 
.
.
are concyclic.
; 
; 
are concurrent .
.
.
(or 
,
)
,
and 
are concurrent), and 
is such that 
is harmonic (because 
,
and 
concur), than 
.
are concurrent, you can use Menelaus' Theorem on 
,
.
gives
gives
we have
and we are done!!
and 
.
are tangent to their incircle and 
tangent to theirs. Then 
and 
.
so 
is cyclic as desired. 
is isosceles with 
!
.
are concurrent, and so 
.
are concurrent, and thus 
.
are concurrent.
and we are done.
and 
,
is a bisector of 
Because of the concurency of the lines 
,
which means that the cross ratio 
,
is harmonic bundle. A similar argument shows that, 
is harmonic bundle, where 
is the intersection of 
and 
Now, it is easy to see that, 
Therefore, we have
where 
denote the power of point 
with respect to the circle 
This prove that the points 
,
are the angle bisectors of 
,
,
,
.
is cyclic iff bisectors of 
is circumscribed, what can be easily proven by calculating it's side lengths.
.
:
and 
.
for some real constants 
,
,
and so 
.
,
,
still holds, along with the relative positions of 
.
and 
intersecting on 
and 
,
by the circle centered at 
passing through 
)
satisfy the relation that 
is tangent to 
at 
,
,
are concylic.
centered at 
and a circle 
centered at 
.
,
and 
both become lines perpendicular to 
,
becomes a rectangle. In particular, it is cyclic, so 
is cyclic.
and 
.
.
.
concur at the Gregonne Point of 
.
.
.
concurs at a point 
.
and another circle centered at 
.
![[asy]
size(11cm);
defaultpen(fontsize(10pt));
pair A = dir(140), B = dir(210), C = dir(330), I = incenter(A,B,C), D = foot(I,B,C), E = foot(I,C,A), F = foot(I,A,B), P = extension(B,C,E,F), I1 = I+dir(I--D)*abs(I-D)*0.35;
path p = circle(I1,abs(D-I1));
pair Z = intersectionpoints(circle(B,abs(B-D)),p)[1], Y = intersectionpoints(circle(C,abs(C-D)),p)[0], X = extension(B,Z,C,Y);
dot("$A$", A, dir(120));
dot("$B$", B, dir(270));
dot("$C$", C, dir(330));
dot("$D$", D, dir(270));
dot("$E$", E, dir(15));
dot("$F$", F, dir(135));
dot("$P (Q)$", P, dir(210));
dot("$X$", X, dir(100));
dot("$Y$", Y, dir(260));
dot("$Z$", Z, dir(315));
draw(C--B--A--C, linewidth(1.2));
draw(B--X--C);
draw(Y--P--E^^P--B, dashed);
draw(incircle(A,B,C)^^p);
draw(circumcircle(E,F,Y), dotted);
[/asy]](http://latex.artofproblemsolving.com/a/9/4/a94da0dc9eabe4e02f53d769c83b8def501552df.png)
then 
and 
.
.
,
We know that 
,
,
,
,
implying 
by Power of a Point, and the result follows. 
concur at the Gergonne Point of 
concur at the Gergonne Point of 
or 
concur(say at a point 
,
,
and 
,
hence 
,
are concurrent 
are concurrent 
are concyclic.
and 
at 
and 
respectively. Since 
Hence, 
.
.
,
.
.
.
so 
are concyclic, as desired.
are concurrent at some point 
which clearly finishes. 
and 
.
the lines 
are concurrent at 
.
And 
are different
and 
meet 
i
but since 
and 
,
.
concur.
by Power of a Point, which implies the result. 
