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and integer 
. Denote![\[\lambda = \frac{\sin \frac{m\pi}{p}}{m\sin \frac{\pi}{p}}.\]](http://latex.artofproblemsolving.com/8/3/a/83a38f8f0f9481bdcd07899d1e79c20ff93e0f24.png)
Suppose 
is a subset of 
with 
. For integer 
, denote![\[B_s^{(n)} := \{(a_k)_{k=1}^{n}:a_k\in A, \sum_{k=1}^{n} a_k\equiv s (\mod p)\}.\]](http://latex.artofproblemsolving.com/4/1/7/417e234322f3e87a097d865d42d46b95e2f6589f.png)
Show that for 
large enough, we have![\[\frac{\# B_s^{(n)}}{\# B_t^{(n)}}\leq \frac{1+(p-1)\lambda^n}{1-(p-1)\lambda^n}\]](http://latex.artofproblemsolving.com/4/b/c/4bc2577292e2726f81ae8b97f0f61e0d09d6a6c1.png)
for all integers 
.
and 
be circles with centres 
and 
, respectively, such that the radius of is less than the radius of . Suppose and intersect at two distinct points and 
. Line 
intersects at 
and at 
, so that 
lie on in that order. Let 
be the circumcentre of triangle 
. Line 
meets again at 
and meets again at 
. Let 
be the line tangent to and near than 
through tangent to 
, the incircle 
is tangent to 
, 
at 
, 
, respectively. 
intersects 
at 
and 
. 
, 
intersect 
again at 
, 
, respectively. Let the external angle bisector of 
intersect at 
, the Euler line of at 
.
are concyclic.
is the circumcircle of 
,
is the orthocenter of is an arbitrary point on 
is the reflection point of wrt 
,
meet 
at 
.
is the reflection point of wrt 
. is a point on such that 
is parallel to ,
is a point on such that 
is parallel to . is the circumcenter of 
is a pair of isogonal lines wrt 
and the points
, 
, 
so that 
, 
and 
. Note with 
the midpoints of ![$\left[ GH \right]$](http://latex.artofproblemsolving.com/d/7/5/d7504900544b3abb8a22aace56668cab00ab8a59.png)
, ![$\left[ DI \right]$](http://latex.artofproblemsolving.com/b/b/5/bb5c2d5d026c6c38cc7625b2c98ff297a5dbf8fa.png)
and ![$\left[ EF \right]$](http://latex.artofproblemsolving.com/1/d/d/1dd94e2988cbd7bd9d2390528412b9d599427fc8.png)
and with 
the intersection of the segments 
and .
., 
and 
are concurrent.
through such that is parallel to 
.Let intersect 
at 
and at 
.Obviously 
.
with respect to transversal 
to get 
as desired
and intersect 
and at 
and 
as desired
by law of sines. By trig ceva, we get that the ratios of the respective sines thus equals 
.