be a triangle with circumcenter 
and internal bisector 
.
again at 
and 
cuts 
.
lie on 
.
at 
respectively. Let circumcircle of 
cuts 
.
.
satisfying
Prove
the following equation :
table filled with natural numbers, we say it is a divisor table if:
-
,
-
,
for every 
.
is given. Determine the smallest natural number 
and satisfy ![\[ a^2+b^2+c^2 +abc = 4 . \]](http://latex.artofproblemsolving.com/d/3/a/d3ac01295b077425b5eaae556faa4cd678bce32a.png)
Show that ![\[ 0 \le ab + bc + ca - abc \leq 2. \]](http://latex.artofproblemsolving.com/2/0/9/209b3fcb56c01c335bbcb8d5fb781e28e49a0798.png)
satisfying the following property: If 
and 
are the intersections of line 
with the common external tangent lines of the circumcircles of triangles 
and 
,![\[\left(\frac{PA}{XY}\right)^2+\frac{PB\cdot PC}{AB\cdot AC}=1.\]](http://latex.artofproblemsolving.com/0/8/0/080a689aea002f1b997294cb6aa48d3334881be7.png)
be positive real numbers. Prove that 
.
in the plane and a subset 
of the plane, which are announced to Bob. Next, Bob marks infinitely many points in the plane, designating each a city. He may not place two cities within distance at most one unit of each other, and no three cities he places may be collinear. Finally, roads are constructed between the cities as follows: for each pair 
of cities, they are connected with a road along the line segment 
if and only if the following condition holds:
distinct from 
and 
,
such[/center]
is directly similar to either 
or 
.
is directly similar to 
if there exists a sequence of rotations, translations, and dilations sending 
to 
to 
to 
.
Y by
Given a polynomial 
and a positive integer 
, we denote the -fold composition of by 
. A polynomial with real coefficients is called perfect if for each integer 
there is a positive integer so that 
is an integer. Is it true that for each perfect polynomial , there exists a positive 
so that for each integer there is 
for which is an integer?
and a positive integer 
, we denote the -fold composition of by 
. A polynomial with real coefficients is called perfect if for each integer 
there is a positive integer so that 
is an integer. Is it true that for each perfect polynomial , there exists a positive 
so that for each integer there is 
for which is an integer?This post has been edited 1 time. Last edited by Phorphyrion, Apr 5, 2023, 1:48 PM
and 
is a fixed point.
,
such that the ![\[
P(x) = \frac{4x^3 + 6x^2 + 10x + 3}{2}
\]](http://latex.artofproblemsolving.com/e/9/5/e9576fb4152ac11bea39f043b1f050b0fd81c159.png)
evaluated at ![\[
P\left(\frac{n-1}{2}\right) = \frac{n(n^2 + 7)}{4} - \frac{1}{2}.
\]](http://latex.artofproblemsolving.com/6/f/a/6fa565afb529efed4a90a79c5ddc2977bce75e71.png)
.
must be a half-integer, i.e., 
for some odd integer 
.![\[
2P(n) + 1,\quad 2P(P(n)) + 1,\quad 2P(P(P(n))) + 1, \ldots
\]](http://latex.artofproblemsolving.com/9/b/3/9b3324db81c2cce4cf53ce05ea7fb76515fe42bb.png)
![\[
m \mapsto (m^3 + 7m)/2.
\]](http://latex.artofproblemsolving.com/d/e/e/dee7010b7d067c28abf03240e9b31d2825660684.png)
Hence, the sequence follows the recurrence:![\[
a_{i+1} = (a_i^3 + 7a_i)/2.
\]](http://latex.artofproblemsolving.com/3/f/7/3f76a3ede66c73f59e183a0e70aaa41cc5ba1d3f.png)
![\[
v_2((m^3 + 7m)/2) = v_2(m) - 1,
\]](http://latex.artofproblemsolving.com/3/3/6/3361ac1d7359aa70f7fd71fd34e5a4db106c543d.png)
where 
denotes the 2-adic valuation. That is, each iteration reduces the 2-adic valuation by 1. Therefore, after exactly 
steps, the resulting term in the sequence will be odd. Since the input to 
at that point is an integer, and 
for odd inputs, the composition 
.
.![\[
2P\left(\frac{n-1}{2}\right) + 1 = n(n^2 + 7),
\]](http://latex.artofproblemsolving.com/b/b/9/bb9167c42684c20e5c4941bddc513e1de22a69fc.png)
we note that 
for odd 
is arbitrarily large. It is known that 
is a quadratic residue modulo any power of 2, so there exist such 
