[/quote]
,
such that 
is a perfect square.
such that, for every positive integer 
the integer
is a multiple of 
(Note that 
denotes the greatest integer less than or equal to 
For example, 
and 
)
If the value of 
where m and n are co-prime positive integers then find the value of 
If we set 
then it's the unique polynomial such that for all integers 
In general, for each positive integer k,there is a unique polynomial 
such that 
Find the value of 
for 
and 
if 
then find the coefficient of 
in 
such that:
be a prime and 
a nonempty set. Suppose that for each element 
there exist elements 
,
such that 
is divisible by 
,
denotes the cardinality of 
.
,
,
is divided into cyclic quadrilaterals with pairwise disjoint interiors. None of the vertices of the cyclic quadrilaterals in the decomposition is an interior point of a side of any cyclic quadrilateral in the decomposition or of a side of the quadrilateral 
.
is an integer. Is it true that for each perfect polynomial 
so that for each integer 
for which 
with the property that 
is divisible by 
.
Y by ehuseyinyigit, Rounak_iitr
Triangle 
is inscribed in circle 
. Let 
denote its incenter and 
its 
-excenter. Let 
denote the midpoint of arc 
. Line 
meets a second time at 
. The perpendicular to 
at meets sides 
and 
at 
and 
respectively. The circumcircle of 
meets 
a second time at 
, and the circumcircle of 
meets 
a second time at 
. Prove that 
passes through the antipodal to on .
is inscribed in circle 
. Let 
denote its incenter and 
its 
-excenter. Let 
denote the midpoint of arc 
. Line 
meets a second time at 
. The perpendicular to 
at meets sides 
and 
at 
and 
respectively. The circumcircle of 
meets 
a second time at 
, and the circumcircle of 
meets 
a second time at 
. Prove that 
passes through the antipodal to on .
,
midpoint of minor arc 
and let 
,![\[\angle UTB=90-\frac{\angle A}{2}=\angle AFE=\angle AEF=\angle UTC \implies TBFU, TCEU \; \text{cyclic}\]](http://latex.artofproblemsolving.com/7/d/6/7d610eb6140fd31f4ac8d216e9dcb44c93c83d0e.png)
Now we also have 
so 
is cyclic, so by PoP:![\[I_AB \cdot I_AP=I_AU \cdot I_AT=I_AC \cdot I_AQ=I_AI \cdot I_AM\]](http://latex.artofproblemsolving.com/4/6/3/4630d2ffb195ff12ea73c8b274de33bca14a6d35.png)
Which imply that 
are both cyclic and also since 
we notice that this means 
therefore not only 
are colinear, in fact they lie on the perpendicular bisector of 
.
.
and 
.
in 
inversion lies on 
.
meet again in point 
and 
.
cyclic, and also 
cyclic by angle chase. So the 
,
m
and 
.
and 
.![\[\measuredangle BTX = \measuredangle BTN = \measuredangle BCN = \measuredangle AFE = \measuredangle AFX \]](http://latex.artofproblemsolving.com/2/6/4/26464ce4620f43f5594757059b6563c4e8e18a26.png)
Thus, 
must be cyclic. Similarly, 
is also cyclic, and we have our claim.
, 
and 
are concurrent at 
is cyclic. Next, we can observe the following.
and 
are parallel.
from which the claim follows. Next, we have the following rather convenient cyclic quadrilaterals.
and 
are cyclic.
it is clear that 
.
.
which implies that 
and 
so points 
,
be the reflection of 
about 
intersects 
.
be the antipodal point of 
.
as the midpoint of segment 
.
and 
are similar.
.
,
and 
are similar. Therefore:![\[ \cfrac{BT}{TJ} = \cfrac{BT}{TI_a} = \cfrac{BI_a}{CI_a} \]](http://latex.artofproblemsolving.com/a/e/5/ae5862303ad751437155ce4dd212785cf875dcd9.png)
Then, by an angle chase: 
.
.
and 
are similar.
and 
.
.
.
.
,![\[ \angle BFT = \angle BZJ = 180^{\circ} - \angle BP'T \]](http://latex.artofproblemsolving.com/1/b/8/1b812074679162d49f600dd0877ff0fcfa84cfee.png)
So 
are concyclic, thus 
.
.
.
so we should have that 
are collinear, as desired.
be the interssection point of 
and 
its obvious from Reim's theorem that 
is cyclic, same goes for 
.
is the radical axis of these two circles. we get 
cyclic. let 
be the intersection of 
and 
since 
we get that 
.
so 
is cyclic. from PoP, we get 
cyclic then by reim's theorem we get 
cyclic. now if we let 
be the second intersection of 
so 
Antipode.
