[/quote]
,
be circumscribed about circle 
,
be the orthocenter of 
.
at 
.
at 
.
be the midpoint of 
,
meet 
.
at 
.
is tangent to 
,
and orthocenter 
be a point on 
.
and 
similarly.
,
.
such that 
.
the point on 
.
is the projection of 
passes through the orthocenter of 
be an acute angled scalene triangle with circumcircle 
.
respectively. Let 
and 
be points on segments 
and 
Prove that 
and 
intersect on the circumcircle of 
.
we consider the last digit different from zero in the decimal expansion of 
.
.
be a quadrilateral such that 
and 
are acute and 
.
,
.
and 
intersect at the point 
.
be a point on the arc 
(not containing 
and 
.
intersect at point 
.
be the reflection of 
and 
intersect at point 
.
,
,
.
and 
.
.
,
exactly once and finally returns with his last jump to 
jumps be a) 
,
?
and 
be points on 
and 
and 
respectively. Prove that the locus of circumcenter of 
is a line.
, such that 
Prove that :
equality cases : ?
such that ![$$\left(a^{3}+b^{3}+c^{3}-3abc\right)^{2}-\left[a^{3}+b^{3}+c^{3}+3abc-ab(a+b)-bc(b+c)-ca(c+a)\right]^{2}\ge 2k\cdot(a-b)^{2}(b-c)^{2}(c-a)^{2}$$](http://latex.artofproblemsolving.com/d/2/4/d24b46e65a587fbd977467b6e4e31838a0eb57aa.png)
forall 
Y by Adventure10, jhu08, megarnie, Mango247, Mango247, EntropiaAwake, EntropiaAwake, EntropiaAwake, cubres
An 
matrix whose entries come from the set 
is called a silver matrix if, for each 
, the 
-th row and the -th column together contain all elements of 
. Show that:
(a) there is no silver matrix for
;
(b) silver matrices exist for infinitely many values of
.
matrix whose entries come from the set 
is called a silver matrix if, for each 
, the 
-th row and the -th column together contain all elements of 
. Show that:(a) there is no silver matrix for
;(b) silver matrices exist for infinitely many values of
.
silver matrices exists for any even 
.
for 
,
there exists a number from 
times, so it can only occur in at most 
columns or rows. So there must be an 
for which it isn't in the 
.
from 
into 
disjoint sets 
of size 
such that each 
with major diagonal filled with 1s and for each pair 
in 
,
.
so what about 
by taking two copies of the original matrix on the diagonal and adding 
by taking the original matrix in the upper left corner, reflecting it over the diagonal and filling the rest appropriately (this reflection works because you can assume that there are only 1's on the diagonal).
and 
you can obtain any even number.
.
which canonically looks like![\[\begin{pmatrix} 1 & 2 & 4 & 5\\3 & 1 & 5 & 4\\6 & 7 & 1 & 2\\7 & 6 & 3 & 1 \end{pmatrix}\]](http://latex.artofproblemsolving.com/3/0/3/303ba582930ca69a8ac9f8265fcdd946c7aee839.png)
Now, you do the reflection over the diagonal:![\[\begin{pmatrix} 1 & 2 & 4 & & & \\3 & 1 & 5 & & & \\6 & 7 & 1 & 2 & & \\ & & 3 & 1 & 5 &4\\& & & 7 & 1 & 2\\ & & & 6 & 3 & 1\end{pmatrix}\]](http://latex.artofproblemsolving.com/5/4/d/54d101c565ab121d3729b0ae2abb5b301087d8ca.png)
(Here of course you need to delete some entries.) And then fill in the gaps:![\[\begin{pmatrix} 1 & 2 & 4 & 8 & 9 & 5 \\3 & 1 & 5 & 9 & 4 & 8 \\6 & 7 & 1 & 2 & 8 & 9\\10 & 11 & 3 & 1 & 5 &4\\11 & 6 & 10 & 7 & 1 & 2\\7 &10 &11 & 6 & 3 & 1\end{pmatrix}\]](http://latex.artofproblemsolving.com/b/4/f/b4ffd8b4c234b0fc0a6eb80e4f7e38341165e489.png)
Of course there some technicalities to describe the general procedure, but hopefully the idea should be clear.
of size 
is defined, using the elements 
,
by setting
if 
if 
if 
if 
.
for all 
.
-
to 
,
times in the matrix. But since there are 
,
,
silver matrix. A 
silver matrix is shown as the base case.
Now, asume that we have found a silver matrix 
where 
for some 
.
.
is
The reason why this works is easy to see once you 
to 
to 
,
-
-
blocks follows by rotating the matrix 
and using the work mentioned above)
for all 
.
,
Note that these two sets have the same number of elements, and the minimum of these is 
and the maximum possible is (by the induction hypothesis) 
.
and 
for any 
,
and 
for all 
.
.
t
t
total rolumns, each of 
must appear at least 
times. So there will be at least a total of 
entries in the matrix. However we only have 
slots for the numbers, and we have a contradiction. Thus there are no silver matrices for odd 
.
,
.
is the dimension of the matrix.
,
.
.
.
matrix into four 
matrices. We call them 
clockwise. Let 
.
.
must be filled in 
and 
.
and 
.
.
,
,
t
Note that if 
,
by the induction hypothesis. The 
and the 
.
.
,
to all elements of matrix 
) and fill in the left large diameter elements with 
and 
respectively.
form. for example :
now, for the for machine a new silver matrix, it's sufficient to convert the silver matrix 
that's clearly gives the silver matrix because matrix 
.
in matrix is row 
are same as 
and they can completely cover numbers 
,
to 
FTSOC, let 
be a 
which is a contradiction since 
is odd. Since 
is odd, there is no such silver matrix.
.
We claim that there exists a silver matrix for all powers of 
.
matrix 
where 
and 
with 
in 
We proceed by induction.
is such a matrix.
We construct the matrix 
Note that 
is a 
Let 
Let 
be the matrix such that 
except we only add 
to the diagonal that spans from upper-left to lower-right.
We see that it will contain row/column 
and 
Since the smallest terms in 
are 
it suffices to show that column 
and by the inductive hypothesis are distinct. The only "overlapping" term is also different because we are subtracting one from the "overlapping" term in 
while there are only 
for 
for each 
.
The construction is described recursively as follows. Let ![\[ M_e' = \left[ \begin{array}{c|c} {\color{red}M_{e-1}} & M_{e-1} + (2^e-1) \\ \hline M_{e-1} + (2^e-1) & {\color{red}M_{e-1}} \\ \end{array} \right]. \]](http://latex.artofproblemsolving.com/9/b/4/9b4bd1d1f0db5477c83338b4e35835fff7a0a611.png)
Then to get from 
to 
'
:
pairs. Let an 
modulo 
,
for some 
.
copies of each element of 
entries in the matrix, which is clearly impossible. Since 
.
for all positive integers 
to 
,
.![\[
\begin{bmatrix}
1 & 2 \\
3 & 1
\end{bmatrix}
\]](http://latex.artofproblemsolving.com/9/c/8/9c89cb379afae7aae3f0560466484913bb8872a4.png)
works for 
as the matrix 
as 
with ![\[
M_{k+1} =
\begin{bmatrix}
M_k & A_k \\
B_k & M_k
\end{bmatrix}
\]](http://latex.artofproblemsolving.com/2/d/d/2dde1a3bbd288c6b417f7506ee2783b912cdd9c0.png)
Now each grass contains all integers 
(since each grass contains a grass of 
and 
as the base case for 
with the labeling describing the location. We define 
and 
.
by the indexes' parity.
works. We wish for every number to be in a structure. We induct on ![\[\begin{tabular}{|c|c|}
\hline
1 & 3 \\
\hline
2 & 1 \\
\hline
\end{tabular}\]](http://latex.artofproblemsolving.com/8/8/0/880173e6e07775dfac6eed298c962e386f996e04.png)
works for 
.
constructions on the diagonal of \textbf{centers}. Then, we replace each number in the 
occupies the row 
column 
then this number is in exactly two groups. Suppose this number is poor. Then every occurence of this number is in two groups. Since we need a total of 
occurences of each poor number, which is absurd as ![\[\begin{pmatrix} 1 & 4 & 2 & 7\\5 & 1 & 6 & 3\\3 & 7 & 1 & 5\\6 & 2 & 4 & 1 \end{pmatrix}\]](http://latex.artofproblemsolving.com/3/4/3/3431349e16306ff8f41c2109de8cbbd8df10eb0c.png)
Suppose 
.
table with entries as ordered pairs 
such that each entry is distinct, and in any row, for any two entries, we have four distinct numbers; i.e. 
must be in different rows. It is possible to construct such a table using a greedy algorithm. Then, each entry in the matrix on 
as the set of entries in either the 
copies of every element in order to be in ![\[(4k+1)(k+1)>(2k+1)^2,\]](http://latex.artofproblemsolving.com/b/c/e/bcee8b3c67781aac3f5444421dc30baea8e98122.png)
![$$
M_k' = \left[ \begin{array}{c|c} M_{k-1} & M_{k-1} + (2^k-1) \\ \hline M_{k-1} + (2^k-1) & M_{k-1} \\ \end{array} \right].
$$](http://latex.artofproblemsolving.com/1/a/4/1a41439e191686b0cc13a5eabc0f24854def4b04.png)
to 
diagonal points. Now, we claim that a silver matrix occurs for all 
We will prove this by induction. This is clearly true for 
If a silver matrix exists for 
let quadrant 
be the same silver matrix. Let qudrant 
be 
We use a bounding argument for odd 
.
appears in at most two 
sets, so we require at least 
of them. Thus the total number of integers needed is at least![\[(2n-1)\left(\frac{n+1}{2}\right) = \frac{2n^2+n-1}{2},\]](http://latex.artofproblemsolving.com/b/3/d/b3d08b731d9e3a596c985ca79a0aeb2de5b1c014.png)
,
We construct matrices for 
satisfies the given conditions to show ![\[A_0 = [1], \quad A_1 = \begin{bmatrix} 1 & 3 \\ 2 & 1 \end{bmatrix}, \quad A_{i+1} = \begin{bmatrix} A_i & B_i \\ C_i & A_i \end{bmatrix},\]](http://latex.artofproblemsolving.com/1/d/7/1d76d2ba5a2c8d905bb664622fd4dd053d6fedd4.png)
represents the matrix 
with each cell increased by 
and 
represents the matrix 
with 
,
distinct elements.
matrix, there are only 
distinct elements of 
for some positive integer 
.
This arrangement ensures that each row contains all elements of 
This arrangement ensures that each column also contains all elements of 
list all the entries of the column 
the list is exactly the elements of the set 
i.e on the diagonal. So in total there are exatcly 
and that is the contradiction.
let ![$A_1=[1]$](http://latex.artofproblemsolving.com/c/4/d/c4d72d9152ecde1c636d7bd6fd4e5499f900784c.png)
and 
Where 
and 
.
be the multiset of elements in the 
be the multiset of elements in the 
be the element in the ![\[ 2 \times M = \sum_{1 \le i \le n} (r_i + c_i) = n \times \{1, 2, \dots, 2n - 1 \} + \{d_1, d_2, \dots, d_n \}. \]](http://latex.artofproblemsolving.com/b/7/8/b78d015eac19543a3b4a6713d8dfde7187a5237f.png)
Clearly the frequency of any integer 
is odd in 
,
,
where 
case is obvious. Now assume a silver matrix 
,
be an arbitrary matrix with rows being cyclic shifts of 
.![\[ M_{k + 1} = \begin{bmatrix}
A_k & M_k \\
M_k & B_k
\end{bmatrix} \]](http://latex.artofproblemsolving.com/6/7/2/6720e89109b0ac1a89e7ce9e235ff8e0c2a06938.png)
is a silver matrix, completing the induction.
.
times in the matrix when 
for 
Let 
where the long diagonal going down and to the right in 
is adjusted to only contain 
,
.
,
,
.
(for the areas "marked" with a 
to 
