special sets
by ChubbyTomato426, Mar 23, 2025, 3:58 PM
Let 
be a positive integer. A subset 
with four distinct elements is special if there exists a rearrangement 
of 
such that 
. Prove that the set 
cannot be partitioned into special disjoint sets.
be a positive integer. A subset 
with four distinct elements is special if there exists a rearrangement 
of 
such that 
. Prove that the set 
cannot be partitioned into special disjoint sets.This post has been edited 1 time. Last edited by ChubbyTomato426, an hour ago
satisfying 
and their sum is an even number, consider the quadratic polynomial:![\[
P(x) = x^2 - (m^2 - m + 1)x + (m^2 - n^2 - m)(n^2 + 1).
\]](http://latex.artofproblemsolving.com/d/5/1/d51aaddfddfe06cc91abfdd3476a155770a8a5fd.png)
are positive integers but are not perfect squares.
.
the number of changes of signs in the
Then prove that
.
greater than 
,
,
and ![$P_1(x),P_2(x), \ldots P_n(x) \in \mathbb{Z}[x]$](http://latex.artofproblemsolving.com/c/6/4/c64176e52d94032e571b4579e35f9e6adfde675a.png)
such that 
.
such that every equation: 
has no real roots
,
and 
e
are tangent to 
at 
and 
respectively. 
intersects line 
.
intersects 
,
intersects 
.
intersects 
.
.
such that 
is a perfect square but none of the integers 
are perfect squares.
(first player) and 
,
chips from the remaining pile of chips, where 
. No one can skip a turn. The player who at some point is unable to make a move (cannot remove chips from the pile) loses the game. Who among the two players can force a win on this game?
be an acute triangle with orthocenter 
be a point on the side 
and 
.
and 
are the feet of the altitudes from 
is the circumcircle of 
,
be the point on 
is a diameter of 
the circumcircle of triangle 
,
be the point such that 
is a diameter of 
and 
all distinct points on opposite sides; for Pascal, no two opposite sides parallel)
,
on 
is above, 
is below 
is a plane, etc.
,
,
,
,
are coplanar, so 
and 
intersect at a point on the line equal to the intersection of planes 
and 
,
and 
,
,
,
(possibly degenerate, but then the result's obvious) intersects plane 
,
,
,
be positive rational numbers and let 
be integers greater than 1. If ![$\sum_{i=1}^{n}\sqrt[k_i]{a_i}\in\mathbb{Q}$](http://latex.artofproblemsolving.com/6/d/3/6d3268640d63424770f2abb2e6c42657d054ace0.png)
,![$\sqrt[k_i]{a_i}\in\mathbb{Q}$](http://latex.artofproblemsolving.com/1/e/9/1e98b3af4db054d32f00d529032658c15875251a.png)
for all 
.
and go by contradiction. For motivation we look at the 
case. WLOG 
are ``minimal'' in the sense that 
for 
.
for all 
is the minimal polynomial of 
over the rationals, so 
.
;
and we can easily get a contradiction by considering the ![$[x^{k_1-1}]$](http://latex.artofproblemsolving.com/a/1/3/a131a3600dec6f0f8fc7d7d771d8e0224b7b78cc.png)
coefficients.![\[x^{k_1}-a_1\mid \prod_{\omega_i^{k_i}=1,2\le i\le n}(S-x-\sum_{i=1}^{n}\omega_i a_i^{1/k_i})\in\mathbb{Q}[x],\]](http://latex.artofproblemsolving.com/1/c/e/1ce1cb394f58859b98d50fe5b410b813d4c9a5a1.png)
t
and (any) nontrivial 
t
.![\[ 0=\Re(S-S) = \Re(a_1^{1/k_1}(1-\omega_1))+\sum_{i=2}^{n}\Re((1-\omega_i)a_i^{1/k_i}) > \sum_{i=2}^{n}\Re((1-\omega_i)a_i^{1/k_i}) \ge 0,\]](http://latex.artofproblemsolving.com/2/3/9/2399ccddd04a1f97962cffb620d26fa61541fa66.png)
are all positive, but then 
would be in 
.
square board contains either 
or 
.
there are two; we restrict our attention to 
.
.
of each entry so we're working in 
instead; we're choosing some of the 
entires to be 1. Viewing this as the equation 
(where boundary cases are 0 for convenience), we can set up a matrix equation 
where 
iff entries 
(there are 
total) are neighboring, and 
has a 1 for each entry 
that 
(whence 
must be the all-zero vector and there's exactly one solution)---note that for 
(as we're working 
that contributes 1 to the sum (i.e. doesn't vanish), draw an arrow from 
for each 
board into disjoint directed cycles of size 1 or 
for some 
(arrows connect neighboring elements---these directed cycles correspond to permutation cycles). These cycles are nontrivially ``reversible'' iff 
,
)
,
,
tiles. Reflecting over the horizontal axis we're left with horizontally symmetric tilings; now reflecting over the vertical axis we're left with tilings that are both horizontally and vertically symmetric.
where 
,
,
is Fibonacci, as all dominoes intersecting one of two axes must lie within the axes (by symmetry). By induction we get 
.
)
boards using the same recursive linear algebra ideas, but when both of 
are even it seems hard to reduce the problem.
be two positive integers. A graph 
is 
-
such that 
and 
are 
-
to a vertex colored 
in 
,
,
and 
,
for 
and 
and define 
by the first and second coordinates, respectively---the induced subgraphs 
can easily be split in the desired form.) 
and 
are both bipartite.
without inducing an odd cycle. Now assume for the sake of contradiction that 
)
,
.
and 
(indices modulo 
)
are connected in 
for all 
.
is connected in 
to at least one vertex of 
be positive real numbers so that 
.![\[ \left( a - 1 + \frac 1b \right) \left( b - 1 + \frac 1c \right) \left( c - 1 + \frac 1a \right) \leq 1.
\]](http://latex.artofproblemsolving.com/7/6/e/76e794a62090c8c4b5f3a70ad4d036669418ca69.png)
July 2014
June 2014