[/quote]
,
has 
as the midpoint of 
and orthocenter 
.
be an arbitrary point on the nine point circle of 
.
intersects 
.
perpendicular to 
intersects 
at 
.
is the midpoint of 
.
that, for all 
and 
,![\[\sum_{i=1}^n \frac{x_i^2}{\sum_{j=1}^i x_j}\geq C \ln n\left(\prod_{i=1}^n x_i\right)^{\frac{1}{n}}\]](http://latex.artofproblemsolving.com/4/8/b/48b9fc4af5a20de7fd9dfc45a74d5aa27e615e46.png)
that satisfy the condition![\[
\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 3.
\]](http://latex.artofproblemsolving.com/2/7/d/27d266570e3a105a4b416bb7c55904ffabf7a07c.png)
Find the maximum value of the expression![\[
P = \dfrac{yz}{\sqrt[3]{3y^2z^2+ 3x^2y^2z^2+ x^2z^2 + x^2y^2}}
+ \frac{xz}{\sqrt[3]{3x^2z^2 + 3x^2y^2z^2 + x^2y^2 + y^2z^2}}
+ \frac{xy}{\sqrt[3]{3x^2y^2 + 3x^2y^2z^2 +y^2z^2 + x^2z^2}}.
\]](http://latex.artofproblemsolving.com/b/a/f/bafc95e59d2f28dbb4e02771db38ce736e3fd45c.png)
.
.
-
numbers 
each equal to either 
or 
.
so that, for any 
so that 
for all 
,
and 
such that
be reals such that 
and 
Find the value of 
players take part. Every two play exactly one match. The schedule is such that no two matches are played at the same time, and each player, after taking part in a match, is free in at least 
next (consecutive) matches. Prove that one of the players who play in the opening match will also play in the closing match.
and 
Find the value of 
,
be the midpoints of arcs 
,
on the circumcircle. Line 
passes through the feet of the perpendiculars from 
and 
.
passes through the feet of the perpendiculars from 
.
denote the intersection of lines 
and 
similarly. Prove that triangle 
and 
are similar to each other.
such that : 
and 
Y by Functional_equation, centslordm, InternetPerson10, Mathmick51, TheStrayCat, megarnie, Aritra12, 554183, adityaguharoy, Aopamy, yshk, deplasmanyollari, eggymath, Kingsbane2139, Rounak_iitr, Funcshun840, cubres
Two squirrels, Bushy and Jumpy, have collected 2021 walnuts for the winter. Jumpy numbers the walnuts from 1 through 2021, and digs 2021 little holes in a circular pattern in the ground around their favourite tree. The next morning Jumpy notices that Bushy had placed one walnut into each hole, but had paid no attention to the numbering. Unhappy, Jumpy decides to reorder the walnuts by performing a sequence of 2021 moves. In the 
-th move, Jumpy swaps the positions of the two walnuts adjacent to walnut .
Prove that there exists a value of such that, on the -th move, Jumpy swaps some walnuts 
and 
such that 
.
-th move, Jumpy swaps the positions of the two walnuts adjacent to walnut .Prove that there exists a value of
such that, on the -th move, Jumpy swaps some walnuts 
and 
such that 
.This post has been edited 2 times. Last edited by popcorn1, Jul 20, 2021, 8:40 PM
,
.
move, color walnut 
brown walnuts between two green walnuts on either side. It's impossible to color all these walnuts green. To see this, note that if 
,
brown walnuts in total. Because this is odd, the groups are of different parities, so one group must have an even number of walnuts. Then keep splitting this group until you get a gap of two brown walnuts, at which point you can't color it.
brown walnuts into two groups. As 
walnuts 
for some 
and 
are the neighbours of 
,
.
,
moves, the number of obedient walnuts with labels greater than 
,
be the mentioned walnuts, in that order. If 
either doesn't change or doesn't matter to us.
consecutive blue points, contradiction.
)
conservative if on the 
and 
with 
,
.
,
is not conservative, since it is swapped after the 
with 
and 
.
.
.
)
remains constant. This can be asserted in a straightforward manner, which I will choose to omit here. This yields that in the end 
.
at the end. This is the contradiction which solves the problem.
.
has to be changed. When we change hole 
must both be unchanged, so we also must change hole 
must be changed. This pattern continues, until we have that hole 
.
.
,
is clear. Now assume that for an even 
'
.
,
is odd, implying that one of 
-
,
,
,
swapped during the process(call this a "forbidden move").
is trivial, so assume that 
work. Then on the first move of that chain, the chain will be split into two parts of length 
,
We want to then show that there is some swap that swaps a number 
and a number 
This is doable iff at every step, between every two consecutive 
walnuts there is an odd number of 
as we can only add a black walnut in between two black walnuts or in between two white walnuts. This is a contradiction as our final state with 
is odd, and after all moves are complete, all acorns are large, so 
.
be 
.
move. If 
,
,
.
,
.
of black walnuts as a maximal chain of consecutive black walnuts (in the circular arrangement).
At every instant 
,
At each move, exactly one new walnut is colored black. Initially, there are no black walnuts. The first black walnut forms a block of size 1, which is odd.
,
,
big walnuts, as well as at most 
walnut which is neither big nor small.
