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of natural numbers such that for every pair of natural numbers 
and 
, the following inequality holds:![\[
\frac{1}{2} < \frac{\gcd(a_r, a_s)}{\gcd(r, s)} < 2
\]](http://latex.artofproblemsolving.com/1/6/7/167679c1707b957d87311298ea5b72347a9bdc45.png)
are given. Prove that one can only use straightedge (which draws a straight line passing two drawn points) and compass (which draws a circle with a chosen radius equal to the distance of two drawn points and centered at a drawn points) to construct a line and two points on it with length 
in only 
steps (Namely, the total number of circles and straight lines drawn is at most .)
inscribed in a circle with center 
. 
is any point not on (O). 
intersect 
at 
. Let 
be the Lemoine points of triangle 
respectively. Prove that 
are collinear.
be a prime number and 
. Prove that![\[\left| p^m - (p - 2)! \right| > p^2.\]](http://latex.artofproblemsolving.com/2/1/c/21ca8bb6e5727d48b18b7ec3b127029c4c97694f.png)
Proposed by Miloš Milićev
with 
, points 
and 
are taken on the rays 
and 
respectively such that 
. The circumcircles of triangles 
and 
intersect at 
and 
. Prove that 
be the circumcenters of the triangles 
and a circle with diameter 
centered at the midpoint of this diagonal of the quadrilateral . Let the lines 
intersect the circle at points 
different from the point 
. The angles 
, 
, 
, 
are all right, because they span the diameter of the circle . The angles 
are all equal to 
:
are all isosceles and consequently, the perpendicular bisectors of the segments 
pass through the points . Let the perpendicular bisectors of the segments meet the circle at points 
. The angles 
are equal, hence, they span the same arc 
of the circle and the points 
are identical. Similarly, the angles 
are equal, hence, they span the same arc 
of the circle and the points 
are identical. Since these angles are all equal to , the points 
are diametrally opposite points of the circle , the points 
are collinear and 
. The perpendicular bisectors 
of the segments 
are parallel, being both perpendicular to the same line 
and they pass through the circumcenters , respectively. Similarly, the perpendicular bisectors 
of the segments 
are parallel, being both perpendicular to the same line 
and they also pass through the circumcenters , respectively. The quadrilateral 
is a parallelogram and the center of the circle is the intersection of its diagonals, being the midpoint of the diagonal 
. Consequently, the centers of the circles 
are collinear. Since these 3 circles intersect at the point , they belong to the same elliptic pencil and they all intersect at one other point, the intersection of the circles 
. Since the angles 
spans the diameter of the circle , it is right and 
.
be the midpoints of 
,
we get 
,
concyclic and we are done.