[/quote]
,
satisfying 
.![\[
0 \le (x^2 + 1)(y^2 + 1) + 4(x - 1)(y - 1) \le 164.
\]](http://latex.artofproblemsolving.com/4/2/d/42d0b88388462f16dd99d928b94024e689ba7c2e.png)
and 
,
as follows:
where 
are all the positive divisors of 
.
Find all positive integers 
.
Find all positive integers 
.
be a convex quadrilateral. Suppose that the circles with diameters 
and 
intersect at points 
.
and 
.
,
,
attaining all positive values and positive numbers 
are given. Numbers 
satisfy
for every 
.
.
?
and 
points on a plane is 
,
such that: in every set of 
points of a plane, no three of which are collinear, consisting of 
,
be the set of all positive integers not exceeding ![\[
P_n(x) = \sum_{k \in S_n} x^{k - 1}
\]](http://latex.artofproblemsolving.com/3/a/c/3ac60cb70baf4dfcf297e9af87480a8ee34ba70c.png)
and a polynomial 
with integer coefficients such that![\[
P_n(x) = (x^{r_n} + 1) Q_n(x).
\]](http://latex.artofproblemsolving.com/7/0/e/70e19a9d3e0a7e97abc7aa08d3f23126c14ea7d0.png)
is irreducible in ![$\mathbb{Z}[x]$](http://latex.artofproblemsolving.com/4/b/7/4b7d768fffbb7a1a6163af392ed348f2d49d8783.png)
.
Y by Adventure10, Mango247
An acute triangle 
is inscribed in a circle 
. Let 
be the point of diametrically opposite 
. Determine all points 
of the arc 
(which does not contain the vertex 
) of such that the quadrilateral 
and the triangle have equal areas .
is inscribed in a circle 
. Let 
be the point of diametrically opposite 
. Determine all points 
of the arc 
(which does not contain the vertex 
) of such that the quadrilateral 
and the triangle have equal areas .This post has been edited 2 times. Last edited by parmenides51, Jul 23, 2021, 3:56 PM
Reason: title
Reason: title
.![$[ABC]=\frac{1}{2}\times AC\times AB\sin\theta$](http://latex.artofproblemsolving.com/3/b/b/3bb8da5b4286fa5b6a5b96b4059ec174d4bc9ba0.png)
and ![$[CXD]=\frac{1}{2}\times CX\times XB\sin(180^\circ-\theta)=\frac{1}{2}\times XC\times XB\sin\theta$](http://latex.artofproblemsolving.com/5/8/1/5810298f06cb3392a98802b8df582c7de093c305.png)
and ![$[CBD]=\frac{1}{2}\times BD\times DC\sin\theta$](http://latex.artofproblemsolving.com/8/c/e/8ce077e3f6d7ce76cb6fce051ff121028cb37223.png)
.![$[BXC]+[CBD]=\frac{1}{2}\sin\theta(XC.XB+BD.DC)$](http://latex.artofproblemsolving.com/a/f/b/afb32d8eedbd7dc408a5fae4a939547b22fdf267.png)
.
we are done.
