be positive reals with 
.![\[\sum_{i=1}^n \frac{x_i}{2-x_i} \ge \frac{n}{2n-1}\]](http://latex.artofproblemsolving.com/f/6/5/f65f6fce4465125dfacf002f0659c898df7ae150.png)
to the cross product 
reveals that a negative determinant [of 
] corresponds to a matrix which reverses the direction of the cross product of two vectors.[/quote]
and 
,
and 
although its supposed to be a quick comparison.
![$ a,b \in [0 ,1] . $](http://latex.artofproblemsolving.com/9/1/9/91980d5ca5d97fe4b61797e6fbb378211674d9cd.png)
Prove that
be a function defined in any real numbers 
with 
Prove that on the 
plane, the line 
passes through the fixed point which isn't on the 
axis in regardless of the value of 
and 
.
Y by shinichiman, Ankoganit, vjdjmathaddict, SAUDITYA, 62861, sunfishho, megarnie, Adventure10, Mango247
nic
y is drawing kappas in the cells of a square grid. However, he does not want to draw kappas in three consecutive cells (horizontally, vertically, or diagonally). Find all real numbers 
such that for every positive integer 
nicy can label at least 
cells of an 
square.
Proposed by Mihir Singhal and Michael Kural
y is drawing kappas in the cells of a square grid. However, he does not want to draw kappas in three consecutive cells (horizontally, vertically, or diagonally). Find all real numbers 
such that for every positive integer 
nicy can label at least 
cells of an 
square.Proposed by Mihir Singhal and Michael Kural
.
.
kappa ?
x
so all 
'
doesn't make much sense to me. 
is the answer, but I can't prove that 
is possible as pointed out below.
kappas in a 
ways of putting 
kappa's (say 21) atleast 3 squares must have 
kappa's and atleast one has 
kappa's
for 
and 
and other boards containing 2 kappas exactly.
.
,
,
,
.![[asy] size(9cm);
int N = 12; for (int i=0; i<=N; ++i) { draw((2*i,0)--(2*i,2*N), grey); draw((0,2*i)--(2*N,2*i), grey); }
pen efill = opacity(0.2)+lightcyan; pen eborder = heavycyan; pen ifill = opacity(0.8)+grey+blue; pen iborder = black;
picture pic1; filldraw(pic1, (1,-1)--(3,1)--(1,3)--(-1,1)--cycle, efill, eborder); filldraw(pic1, scale(2)*unitsquare, ifill, iborder); add(shift(4,18)*pic1); picture pic2; filldraw(pic2, (1,-1)--(3,1)--(3,3)--(1,5)--(-1,3)--(-1,1)--cycle, efill, eborder); filldraw(pic2, box((0,0),(2,4)), ifill, iborder); add(shift(12,18)*pic2); picture pic3; filldraw(pic3, box((-1,-1),(5,5)), efill, eborder); filldraw(pic3, scale(4)*unitsquare, ifill, iborder); add(shift(16,12)*pic3); picture pic4; filldraw(pic4, (-1,-1)--(3,-1)--(7,3)--(7,5)--(3,5)--(-1,1)--cycle, efill, eborder); filldraw(pic4, (0,0)--(4,0)--(4,2)--(6,2)--(6,4)--(2,4)--(2,2)--(0,2)--cycle, ifill, iborder); add(shift(10,4)*pic4); picture pic5; filldraw(pic5, (1,-1)--(5,-1)--(5,1)--(1,5)--(-1,5)--(-1,1)--cycle, efill, eborder); filldraw(pic5, (0,0)--(4,0)--(4,2)--(2,2)--(2,4)--(0,4)--cycle, ifill, iborder); add(shift(4,10)*pic5); [/asy]](http://latex.artofproblemsolving.com/3/a/6/3a6e5ea87f9388062cf0e4818bc40c0b0e3614ca.png)
times the original area). Thus, Nicky can cover at most 
of the squares with kappas.
-
kappas, 
.
is a small error due to the edges, and as 
then we must have 
by counting the squares where a kappa may not appear in row 1 due to two-in-a-row in rows 2 and 3, and noting that a kappa is blocked from a cell by at most one or two two-in-a-rows, never three. The argument doesn't generalize to other distributions very easily though, and might require some more ingenious counting. I'm afraid I haven't found a way to resolve this 
and the bound is of the order 
so if makes no difference. But why wasn't it mentioned in the official solution. I think it's a non-trivial observation.
.
and we are done.
for bounding instead of 
-- this did not end well. Another larger issue was the lack of seeing the area argument ;-;
can be done.
(and the bottom right square is 
)
or less squares surrounding it and let 
be the grid for a 
config. Define 
for a square 
to be the number of kappas in the surrounding ring. Let 
be the indicator function for the presence of a Kappa.![\[ \sum_{s \in G} R(s) = \sum_{s \in G} 8 \kappa(s) + O(n) \]](http://latex.artofproblemsolving.com/c/4/f/c4fd8583e74c7cb7817007f22f65f840171a230a.png)
so it follows that ![\[ \frac{\sum_{s \in G} R(s)}{8n^2} = \sum{s \in G}{n^2} + O\left(\frac1n\right) \]](http://latex.artofproblemsolving.com/e/4/9/e4909f02ecdd3d275690ecebe9be58865c270dbe.png)
It thus remains to show that 
.
,
.
,
such that their average is at most 
.
.
-
,
,
and there's an 
