and 
for 
.
.
for which 
is divisible by 
.
is divisible by 
satisfy,
Where 
}
be a fixed square and 
a variable point on segment 
.
is constructed such that 
is on segment 
and 
is on segment 
.
be the intersection point of lines 
and 
.
be a function taking in positive integers and outputting nonnegative integers, defined as follows:
is the number of positive integers 
with 
such that the equation 
has an integer solution 
.
such tha
Adit Aggarwal, India.
Y by Davi-8191, Adventure10, Mango247
Given is a trapezoid 
where 
and 
are parallel, and 
are clockwise in this order. Let 
be the circle with center 
passing through 
, 
be the circle with center 
passing through 
. The intersection of line 
and is 
. Denote by 
the circle with diameter 
, and let and meet at 
. and meet at 
. If the circumcircle of triangle 
and meet at 
, prove that 
are collinear.
where 
and 
are parallel, and 
are clockwise in this order. Let 
be the circle with center 
passing through 
, 
be the circle with center 
passing through 
. The intersection of line 
and is 
. Denote by 
the circle with diameter 
, and let and meet at 
. and meet at 
. If the circumcircle of triangle 
and meet at 
, prove that 
are collinear.
be the intersection of 
and 
are conyclic, which means 
and 
as the intersection of 
and 
,
are cyclic.
and we also have 
so we have 
giving 
,
be the point of intersection of 
and 
.
.
to 
.
to 
to ![\[\frac{OB}{OD}=\frac{OP}{OQ}\]](http://latex.artofproblemsolving.com/4/5/d/45d669a6d4320e17f1558e14fd5f7bd758292776.png)
which proves our claim.
and 
and 
.
.![\[\frac{OB'}{OD'}=\frac{OB}{OD}\]](http://latex.artofproblemsolving.com/1/2/1/12168853d87349c5ed5ec83341f14b2475c804c1.png)
by the homothety and![\[OD'\cdot OD''=OD\cdot OQ'\]](http://latex.artofproblemsolving.com/3/2/6/32640244b02c6d740f38ff71c75a8b0cc148b0fa.png)
so 
.
stays constant, we have that 
and 
are swapped, which means 
,
and we are done.
