[/quote]
,
with centroid 
,
.
and 
be the second points of intersection of lines 
and 
with the circumcircle of triangle 
is tangent to the Euler circle of triangle 
q
answer choices. A participant who answers all questions correctly will receive a gift. To ensure that at least one member of my family answers all questions correctly, how many family members need to take the quiz?
,
and let 
be the foot of the
-
.
and 
in line 
and 
respectively. A line through 
while 
and 
and 
are tangent to each other.
are written on a board. 
.
the circumcircle circumscribed to the triangle 
.
be the second intersection point of the circumcircle circumscribed to 
with 
and 
with 
and 
be the intersection points of lines 
and 
with lines 
line passes through the midpoint of 
be six positive real numbers. Prove that
cards on a table, flipped face down. Madina knows that on each card a single number is written and that the numbers are different integers from 
to 
cards, and she is told a number that is written on one of the chosen cards, but not which specific card it is on. After several moves, Madina must determine the written numbers on as many cards as possible. What is the maximum number of cards Madina can ensure to determine?
be nonzero real numbers such that 
.
Show that for any odd integer 
prove that the triangle must have a right angle.
denote the set of natural numbers, and let 
,
,
.
whose product is a perfect cube.
and 
be prime numbers. Prove that if 
is a perfect square, then 
is also a perfect square.
points in 
can be partitioned into two sets whose convex hulls intersect.
and 
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 
