is written on the board. A move consists of adding either 
or 
to the number written on the board, but only if the chosen number is coprime with the current number (for example, if the current number is 
,
of positive integers is called central if for every positive integer 
, the arithmetic mean of the first 
terms of the sequence is equal to 
of positive integers such that for every central sequence 
there are infinitely many positive integers 
.
be a circle passing through 
intersecting 
and 
.
be the intersection of 
and 
and 
with the tangent to 
at 
.
be the second intersection of 
and 
and 
Y by Maths01, AlastorMoody, Jc426, chessgocube, Adventure10, megarnie, Mango247, aidan0626, ItsBesi, buddyram, Rounak_iitr, Yiyj1, and 1 other user
1. Let 
be an acute-angled triangle with 
. The circle with diameter 
intersects the sides 
and 
at 
and 
respectively. Denote by 
the midpoint of the side . The bisectors of the angles 
and 
intersect at 
. Prove that the circumcircles of the triangles 
and 
have a common point lying on the side .
be an acute-angled triangle with 
. The circle with diameter 
intersects the sides 
and 
at 
and 
respectively. Denote by 
the midpoint of the side . The bisectors of the angles 
and 
intersect at 
. Prove that the circumcircles of the triangles 
and 
have a common point lying on the side .Attachments:
.
an
.
and 
However 
,
and the quadrilateral 
is cyclic.
,
.
,
is cyclic.
and common point of the two circumcircles 
lying on BC is on the bisector of 
(All my team too 
this problem was proposed by two Romanians, can you guess who? 
:D
:P
with center 
.
,
.
,
on the side 
the circumcircle of 
the circumcircle of 
and 
.
and 
and 
are homologous (because 
)![\[\frac{AR}{AT}= \frac{AM}{AC}.\quad(1)\]](http://latex.artofproblemsolving.com/7/9/e/79e9fccbaf19f9a82ec9f19db7b28e4f4ceb197c.png)
![\[\frac{AX}{AC}=\frac{AB}{AT} .\qquad (2)\]](http://latex.artofproblemsolving.com/6/3/7/6373eb035b60f88c12c89d0510fcf41f020a39b2.png)
)![\[ AX\times AR=AB\times AM\]](http://latex.artofproblemsolving.com/3/0/c/30c4424075b56f9338167102b1faf5843fbe3709.png)
and then 
.
intersect 
.
(This is called miquel point) . Hence, 
is collinear.
be the intersection of lines 
and 
and 
are cyclic, leading to the desired result.
and line
.
,
is isosceles,
and 
,
.![\[ \angle BMR = \angle BMO + \angle OMR = \angle OBM + \angle BAK = \angle KBA + \angle BAK = \angle BKA = \angle BKR. \]](http://latex.artofproblemsolving.com/3/7/1/3713d256f95008228dc40230d1d5a93dba0d79e1.png)
Thus, we are done.
in 
.
,
.
,
,
is cyclic, the finish is then obvious from miquel.
,
and 
and 
must concur on 
,
.
(it is indeed the midpoint of 
)
and 
intersect on segment 
so 
,
.
and the perpendicular bisector of 
to be the point at which one of the circle hits 
is cyclic. This reduces to proving 
is cyclic. 
and 
are congruent, so 
.
and 
are tangent to 
The intersection of the angle bisector of 
Then ![\[\angle RPC=180-\angle RPB=\angle BMR=180-\angle AMR=\angle ANR=180-\angle RNC\]](http://latex.artofproblemsolving.com/a/9/f/a9f2eb811f2cdcb5bbce3edf050722e1138838b8.png)
so 
cyclic and we are done.
Therefore by the Law of Sines 
Hence either 
or 
If the former holds, we have that 
so 
and thus 
a contradiction.
is a cyclic quadrilateral, and the rest follows by Miquel's Theorem. QED
,![\[\angle MRN = \frac{180^\circ - 2\angle BAC}{2} + 90^\circ = 180^\circ - \angle BAC,\]](http://latex.artofproblemsolving.com/8/7/4/8744a535d4a52c6e265383bb1e743a545377de7f.png)
so 
is cyclic. But 
and 
are congruent, so 
is cyclic, we have![\[\angle MBR' = \angle MNA = \angle MRA,\]](http://latex.artofproblemsolving.com/6/5/e/65e4fe8bd828d0ed85006a23b37985f3e218cfdf.png)
so 
is cyclic. And so is 
.
where 
are tangents to 
is the incentre of 
bisects 
QED
