be positive real numbers such that 
.
,
the number of positive divisors of 
is Burapha integer if it satisfy the following condition
is an odd integer.
holds for every positive divisor 
of 
be real numbers such that 
for all 
.
.
.
and 
meet at 
and 
denote the circumcircle and the circumcenter of triangle 
.
and 
denote the circumcircle and circumcenter of triangle 
.
meets 
and 
(other than 
and 
)
and 
be the midpoints of minor arcs 
(not including 
(not including 
.
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Consider the convex quadrilateral 
. The point 
is in the interior of . The following ratio equalities hold:
![\[\angle PAD:\angle PBA:\angle DPA=1:2:3=\angle CBP:\angle BAP:\angle BPC\]](//latex.artofproblemsolving.com/9/3/8/9388a91325842f1ec67a4693f0f10af46ed66c6d.png)
Prove that the following three lines meet in a point: the internal bisectors of angles 
and 
and the perpendicular bisector of segment 
.
Proposed by Dominik Burek, Poland
. The point 
is in the interior of . The following ratio equalities hold:![\[\angle PAD:\angle PBA:\angle DPA=1:2:3=\angle CBP:\angle BAP:\angle BPC\]](http://latex.artofproblemsolving.com/9/3/8/9388a91325842f1ec67a4693f0f10af46ed66c6d.png)
Prove that the following three lines meet in a point: the internal bisectors of angles 
and 
and the perpendicular bisector of segment 
.Proposed by Dominik Burek, Poland
This post has been edited 2 times. Last edited by djmathman, Oct 6, 2020, 5:51 PM
be circumcenter of 
.
are cyclic.Hence 
.Thus 
is angle bisector of 
and similarly 
is angle bisector of 
and we are done
and 
so that 
and 
are both isosceles. The given angle conditions imply 
and 
,
and 
.
and 
of 
,
.
intersect 
at 
and 
respectively. It can be seen by angle chase that 
so the angle bisectors in the problem are just bisectors of 
and 
.
)
Now using the angle conditions note that 
is cyclic, so an angle chase shows 
.
.
bad.
.
and 
are cyclic quads. By Fact 5, 
at 
,
and 
,
is isosceles and since 
is a chord, the angle bisector of 
is the perpendicular bisector of 
is the perpendicular bisector of 
and intersects this center of the circle 
and 
.
.
.
,
.
.
,
and 
.
meet 
and 
,
,
bisects 
.
.
,
is just the perpendicular bisector of 
.
.
,
,
,
.
tells us 
,
and 
in the order they appear; and let the bisectors of 
and 
meet 
and 
,
.
,
,
.
,![\[\angle PDA = 180-4\alpha = 180 - 2\angle ABP = 180-\angle AOP.\]](http://latex.artofproblemsolving.com/f/1/e/f1e80e78d04b2a6ba87ef68dddd215c716ef1507.png)
Similarly, 
.
and 
are cyclic.
,
is the angle bisector of 
is 
We claim that this works. Notice that 
and 
are cyclic. Thus, we are done.
.
,
.
so 
so 
,
.
,
with 
and 
.
and 
so 
.
.
,
,
and 
are cyclic. The intersection of the internal bisectors of 
We get 
Thus, 
Clearly, 
which finishes.
,
and 
are concyclic.
and 
,
and in triangle 
the angle chase follows similarly, so 
and segment 
,
,
,
,![\[
\angle PAK=\frac{1}{2} \angle PAB=\angle CBP=\angle KBP.
\]](http://latex.artofproblemsolving.com/b/1/5/b159733e4e44657a65ee7d4c39b16f6a4462063a.png)
,![\[
\angle CKP=\angle KBP+\angle KPB = 2\angle KBP=\angle BPC -\angle BPK=\angle CPK.
\]](http://latex.artofproblemsolving.com/d/6/a/d6abb8a7c0e069687ac3ec9a31fd73b070612662.png)
similarly 
.
,
are concurrent, which is obvious by the claim.
