with the following property:
number 
.
and 
need not be different.
and 
Find the minimum 
satisfying![\[f(x+f(y))=x+f(f(y))\]](http://latex.artofproblemsolving.com/8/d/a/8daa6bcb3155768ee1ba4bd22beb53a85a6e3923.png)
and 
,
.
be positive odd integers. Prove that 
doesn't divide 
.
Y by Adventure10, Mango247
Let 
be a cyclic quadrilateral. Let 
be the midpoint of the arc 
of its circumcircle which does not contain 
or 
. Let the lines 
and 
meet at 
and the lines 
and 
meet at 
. Prove that the lines 
and are parallel.
See here: http://www.bmoc.maths.org/home/bmo1-2015.pdf
be a cyclic quadrilateral. Let 
be the midpoint of the arc 
of its circumcircle which does not contain 
or 
. Let the lines 
and 
meet at 
and the lines 
and 
meet at 
. Prove that the lines 
and are parallel.See here: http://www.bmoc.maths.org/home/bmo1-2015.pdf
is parallel to 
,
) we get 
.
is an isosceles triangle ( 
) 
and 
are equal then so are the chord lengths 
.
,
is cyclic. Now 
,
and 
have their usual meanings, but noting 
.![\[ p = \frac{da+dc-ac-acd}{d-ac} \text{ and } q = \frac{c+acd-ad-cd}{ac-d} \]](http://latex.artofproblemsolving.com/6/9/6/69622af313eb190927cb3e0feed9676247d3f935.png)
so ![\[ p-q = \frac{c-ac}{d-ac} \]](http://latex.artofproblemsolving.com/0/5/2/052013904dbec7dc106ef0e655c63c5048d28565.png)
which is easily verified to be pure imaginary.
is similar to 
,
/
=
/
.
and 
are similar 
/
.
=
using that 
and 
are bisectors and 
then 
is cyclic.Let 
points of intersections of 
and 
and 
thus 
is cyclic therefore by angle chase or Reim's theorem the result follows
= arc 
=
=
=
is cyclic
=
=
,
