[/quote]
,
be an integer, 
a real number, and 
be positive real numbers such that 
.![\[
\frac{1 + x_1^k}{1 + x_2} + \frac{1 + x_2^k}{1 + x_3} + \cdots + \frac{1 + x_n^k}{1 + x_1} \geq n.
\]](http://latex.artofproblemsolving.com/0/8/d/08de0c74a4e36b50a64d17875d3fd93eeb5b52de.png)
with 
.
be the midpoint of side 
,
and 
be points on segments 
and 
,
.
be the circumcircle of 
,
the circumcircle of 
.
to 
,
and 
,
intersect 
,
intersect 
.
is tangent to both 
be positive real numbers. Prove that there exists a cyclic permutation 
of 
such that for all positive real numbers 
the following holds:![\[
\frac{a}{x\,a + y\,b + z\,c}
\;+\;
\frac{b}{x\,b + y\,c + z\,a}
\;+\;
\frac{c}{x\,c + y\,a + z\,b}
\;\ge\;
\frac{3}{x + y + z}.
\]](http://latex.artofproblemsolving.com/1/5/2/15261b3f712c2cc8ffbeb92a9711e06ad0992b5d.png)
islands of the Arctic Ocean. Every bear sometimes swims from one island to another. It turned out that every bear made at least one swim in a year, but no two bears made equal swams. At the same time, exactly one swim was made between each two islands 
:
and 
such that 
.
be positive integers such that 
,
,
are perfect squares. Prove that 
is also a perfect square.
be a convex polyhedron with the following properties:
edges.
.
colors such that for each color 
and any two distinct vertices 
,
to 
all of whose edges have color 
such that
divides 
for every 
,
we have![\[
f\bigl(f(a)+kb\bigr)\;=\;f\bigl(a + k\,f(b)\bigr).
\]](http://latex.artofproblemsolving.com/c/f/f/cff41352422c8765978b8fe6dc966ba934a9df49.png)
be an acute scalene triangle. Denote by 
the circle with diameter 
,
be the contact points of the tangents from 
lie on opposite sides of 
and 
and 
lie on opposite sides of 
.
be the circle with diameter 
,
,
the circle with diameter 
.
are concurrent.
were selected 
pairwise distinct points, abscissas of which are integer numbers from the segment ![$[0; 100000]$](http://latex.artofproblemsolving.com/0/5/e/05e1d2855ef6a6dfaf9b24f2713455ab1017a30d.png)
.
,
,
such that areas of triangles 
and 
are equals.
baskets, each containing at least one candy. The total number of candies is 
.
consecutive non-empty baskets, while Sevinch takes all the candies from a single non-empty basket that has at least one empty neighboring basket. Prove that Asad can take overall at least 
candies, regardless of the initial distribution of candies and Sevinch's actions.
for which the following statement holds: there exists at least one triple 
and all triples 
are integers.
and 
such that 
and 
such that 
and on side 
and 
such that 
Let 
and 
are concurrent.
be a convex quadrilateral with ![\[\angle ADC = 90^\circ, \ \ \angle BCD = \angle ABC > 90^\circ, \mbox{ and } AB = 2CD.\]](http://latex.artofproblemsolving.com/b/4/1/b416d5c0b29c3a69a3e87cd22b6000aa8e70d456.png)
The line through 
,
,
at point 
.
,
,
,
,
can be divided into two groups, so that the angles in each group have a sum of 
.
Y by
Let 
and 
be positive integers with 
. In a group of people, each one or always
speak the truth or always lie. Arnaldo can ask questions for any of these people
provided these questions are of the type: “In set
, what is the parity of people who speak to
true? ”, where is a subset of size 
of the set of people. The answer can only
be “
” or “
”.
a) For which values of and is it possible to determine which people speak the truth and
which people always lie?
b) What is the minimum number of questions required to determine which people
speak the truth and which people always lie, when that number is finite?
and 
be positive integers with 
. In a group of people, each one or alwaysspeak the truth or always lie. Arnaldo can ask questions for any of these people
provided these questions are of the type: “In set
, what is the parity of people who speak totrue? ”, where
is a subset of size 
of the set of people. The answer can onlybe “
” or “
”.a) For which values of
and is it possible to determine which people speak the truth andwhich people always lie?
b) What is the minimum number of questions required to determine which people
speak the truth and which people always lie, when that number is finite?
This post has been edited 14 times. Last edited by betongblander, Jun 7, 2021, 8:08 PM
,
for truth and 
for liar.
about the set ![\[ (p+1) + \sum_{a \in A} \pmod 2. \]](http://latex.artofproblemsolving.com/a/a/0/aa06fdf63273fd2e1568549f5e92de569dc526b2.png)
So in other words, a query amounts to sampling a set of either 
elements 
)
elements 
)
changes no responses.
for 
By summing, one gets the value of 
,
for 
.
for 
.
for 
.
possible final answers for Arnaldo, and each query has two possible responses.
questions, by Pidgeonhole there exists some string of answers that corresponds to more than one possible configuration, hence he cannot determine the configuration, which shows the lower bound.
is even, these two numbers are either both odd or both even. If they are both odd, then the people that said odd are truth-tellers and the people that said even are liars, and vice versa for the both even case. In both cases, he can determine the type of each person in 
people to determine their type, hence done.
the process is impossible. For even 
'
'
people remaining, from which 
queries is sufficient to determine who are liars and truthtellers--by taking a set of 
people reply ``odd" and 
people reply ``even". Then 
possibilities for the distribution of liars.
for each 
answer sequences. Since 
there exists atleast two possible states for some answer sequence for 
denote our required (with 
meaning it ain't possible). Then ![\[\boxed{f(n,k) = \begin{cases} \infty \text{ if $k$ is odd} \\ n \text{ if $k$ is even} \end{cases}}\]](http://latex.artofproblemsolving.com/8/8/d/88db4b91b2b8d6bc02978e7bf4c06c986688c2dd.png)
To see why 
questions atleast.
