[/quote]
,
be an acute triangle. Points 
,
lie on a line in this order and satisfy 
.
and 
be the midpoints of 
and 
,
is acute, and let 
be its orthocentre. Points 
and 
lie on lines 
and 
,
and 
and 
and 
of positive integers is called central if for every positive integer 
terms of the sequence is equal to 
,
,
,
of positive integers such that for every central sequence 
,
,
,
.
be all the positive integers smaller than 
such that
for all 
.
and 
be the orthogonal projections of 
onto the internal bisectors from 
and 
is parallel to 
.
be a positiv integer. Ana chooses the positive integers 
and for any non-empty subset 
she computes the sum ![\[s_A=\sum_{k
\in A}a_k.\]](http://latex.artofproblemsolving.com/7/8/a/78afe6d1036a006b2252023d8f9619285fe0c6d5.png)
She orders these sums 
Prove that there exists a subset 
with 
elements such that, regardless of the integers 
with 
such that for any set of consecutive integers taken from the given set of integers, there will always be a colour colouring exactly one of them? That is, for all integers 
such that 
,
.
is a permutation of the elements of the set 
}
and 
be different prime numbers. We are given an infinite decreasing arithmetic progression in which each of the numbers 
and 
occurs. Show that the numbers 
![\[
\text{Given that } 0 < \theta < 90^\circ,\ \text{solve the equation: } \sin(\theta - 60^\circ)\sin\theta + \sin(54^\circ - \theta)\sin 54^\circ = 0
\]](http://latex.artofproblemsolving.com/b/2/4/b24276c1fbddbcc97e50cef52db03ea9b264d7ee.png)
![\[
\text{What is the value of } \theta\ (\text{in degrees})\ \text{that satisfies the equation?}
\]](http://latex.artofproblemsolving.com/0/0/f/00fbcd5839fedced0e5636c1def5360d65fc0444.png)
and circumcenter 
.
.
be the incenter of triangle 
be the midpoint of the arc 
,
be a point on the arc 
of 
.
be the circumcenter of triangle 
.
and 
meet on 
.
respectively.
and 
as the reflections of 
.
with 
with 
can be inscribed within a circle.
where the 
are positive integers such that each 
and none of two consecutive terms when summed are equal to 
.
Y by megarnie, HWenslawski, Mogmog8, centslordm, OlympusHero, Zyz1010, 554183, The_Turtle, Pluto1708, Tastymooncake2, ItsBesi
Let 
be a fixed positive integer and consider an 
grid of real numbers. Determine the greatest possible number of cells 
in the grid such that the entry in is both strictly greater than the average of 's column and strictly less than the average of 's row.
Proposed by Holden Mui
be a fixed positive integer and consider an 
grid of real numbers. Determine the greatest possible number of cells 
in the grid such that the entry in is both strictly greater than the average of 's column and strictly less than the average of 's row.Proposed by Holden Mui
This post has been edited 2 times. Last edited by MathLuis, Nov 4, 2021, 11:10 PM
.
,
of the cells are bad, so at most 
:
If there ar
fat squares in a different column and row. Select one fat square and rearrange it such that the fat square is on the bottom left of the 
.
,
if you consider the column while 
if you consider the row. This implies that 
unique dumb square that corresponds to it as long as it is not covered by a "fat" square.
and 
then the number of dumb squares 
,
:
no. of dumb squares 
If we have 
f
dumb square for 
c
Number of 
-
by 
'
and the top right grid with 
,![[asy]
int n = 5;
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= n; ++j) {
if (i < n && j < n || i == n && j == n) {
label("0", (i, j));
} else if (i == n) {
label("$1$", (i, j));
} else {
label("$-1$", (i, j));
}
}
}
[/asy]](http://latex.artofproblemsolving.com/e/a/2/ea24b6ce58c9b671d61f3e616f404cb5d3c768bb.png)
![\[n^2 - (2n-1) = (n-1)^2.\]](http://latex.artofproblemsolving.com/b/c/b/bcb2d1d37d4213a0fb3ab5b8f675a44a09aeab41.png)
Call a cell "good" if it satisfies the conditions and "bad" if not. Assign coordinates to our cells with 
the bottom left corner and 
the top right.
and the down-most row be 
We see that all the zeros are good as 
and 
Also note that all the ones and negative ones are bad. 
coordinate 
with ![$k\in[1,n]$](http://latex.artofproblemsolving.com/c/6/d/c6d3158118e6d7d94f118de715bde448b455da44.png)
be called the 
and 
and all cells with 
FTSOC assume at least one cell in the 
Since all other cells in the column are good, they are greater than the average. But then all the cells are greater than the average, so a contradiction.
Then, at least one other cell is less than or equal to 
so a contradiction since all other cells must be good.
Then, at least one other cell is less than 
and 
Since 
it is greater than 
it is less than 
Thus, 
Similarly, 
from 
good cells. 
the average of the row. Let 
.
.
,
,![\[X>M>Y>N>X\]](http://latex.artofproblemsolving.com/7/1/5/715bd5eab43ddb78708764da1163f07edc3fe5e5.png)
contradiction.
,
squares are good.
except for the rightmost cell. We label all other cells 
of these cells minus the number of cells that are the smallest in its column and largest in its row. None of these can be a value of 
.
,
which easily generalizes:![[asy]
unitsize(14);
int n = 4;
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= n; ++j) {
if (i < n && j < n || i == n && j == n) {
label("1", (i, j));
} else if (i == n) {
label("$2$", (i, j));
} else {
label("$0$", (i, j));
}
}
}
[/asy]](http://latex.artofproblemsolving.com/2/5/9/2596e2b06833c8f77425a5edbcc47ecb13a1b742.png)
.
,
find the maximum number of cells that are painted such that each fantastic set has at least one cell choosed. (Of course the answer is 
,
cells.
cells for 
,
fantastic special sets.
.
,
,
cells, then are at least 
,
.
and 
the number of cells in the first column and in the first row respectively. We paint 
cells, then in the 
(that no contain the first column and row) are 
painted cells, by hyphotesis, there are at least 
fantastic special sets. Let 
.
(Consider the cell not painted on the top left). And see that there are at least 
and 
rows and columns respectively that are specials for 
fantastic special sets
.
painted cells, and for 
,
and 
are marked. This implies 
,
is the average of all numbers in the set 
have already been constructed above.
Let 
but 
Thus we have a contradiction, and thus we are done with proving the claim.
so an upper bound for the number of cells that is asked for the answer is 
be the row-bad square of row 
for each 
.
.
(value 
(value 
and 
the value in 
.
.
which means 
,
bad cells, so there are at most 
grid of 
distinct bad squares, as needed.
and 
be arbitrary squares in the columns and rows respectively, which are greater than the average of the row and less than the average of the column, respectively. The key in my above solution is to instead make these the maximum and minimum, removing the necessity of everything else in the row/column being good. With this old definition, I made an inequality chain similar to my solution above's, where instead of the maximum and minimum giving the inequalities, it was the fact that one is greater and one is less than the average in the row/column. Again, while the claim is technically true for this definition, it is useless, since 
does not imply 
denote the set of squares with the desired property. Wlog slightly tweak the value on each cell so that they are all distinct. We let 
are not in 
,
,
,
.
and 
,
and 
.
.
.
when looking at rows, but 
when looking at columns, contradiction.
.
cells.
.
.
.
,
.
,
.
and 
.
,
and for the last two cells, one from column 
and 
,
,
blue squares, so there are most 
non-blue squares. Given this, the key claim is the following:
,
blue cells, and in the remaining 
rows there are at most 
,
which being a upwards-opening parabola in 
is maximized when 
,
,
,
cells in total as well. Thus Hall's condition holds so a perfect matching exists. 
denote the column paired with row 
the (blue) intersection between 
denote the average of row or column 
Summing over all 
where 
denote the sets of rows and columns respectively. But this is absurd, as the two quantities should be equal by double counting. Hence 
is impossible, as desired. 
Thus 
be the average of the 
be the average of the 
cell satisfying the condition. Draw a bipartite graph with vertex 
.
,
,
,
and everything else 
.
and 
.
and 
.
.
,
,
.
,
bad cells, so at most 
which is achived by similar constructions described in the above solutions (say, make a 
then we have in total at most 
cells and we are done. Similarly, we take off the cases where each column and each group has at most 
cells.
cells. Then let 
If 
is not an empty cell, we can remove 
so that we find a row, line or group 
cells in 
,
This obviously works when 
portion of the grid satisfies the conditions.
simultaneously, which is obviously impossible. 
as required.
a contradiction. 
which is an example for a 
grid, but we can extend this by letting the upper left 
where 
.
is 
be the numbers in a row, 
be the numbers in a column, and let 
for all 
,
that also counter for 
,
,
,
.
(obviously in different rows and columns) which satisfies this condition. Then the cell in the same column as 
![\[\begin{matrix}
0 & 0 & \cdots & 0 & -1 \\
0 & 0 & \cdots & 0 & -1 \\
\vdots & \vdots & \cdots & \vdots & \vdots \\
0 & 0 & \cdots & 0 & -1 \\
1 & 1 & \cdots & 1 & 0
\end{matrix}\]](http://latex.artofproblemsolving.com/2/a/a/2aa4793073bd2b280c06033d5e79b03eebea7785.png)
at the intersection of row 
and 
.
.
from left to right.
from top to bottom.
.
.
.
.
and if we take 
.
.
Note 
.
Similarly, 
which motivates the construction. Consider 
(just because it can be anything really). Let every other entry in the n-th column be 
,
.
.
.
![\[\begin{matrix}
-1 & -1 & \cdots & -1 & 0 \\
0 & 0 & \cdots & 0 & 1 \\
\vdots & \vdots & \cdots & \vdots & \vdots \\
0 & 0 & \cdots & 0 & 1 \\
0 & 0 & \cdots & 0 & 1
\end{matrix}\]](http://latex.artofproblemsolving.com/3/c/d/3cd754841498f5aa313e936117785c188e9afac5.png)
.
.
and 
.
which implies, maximum number of good cells 
,
is a rectangle – we have 
and 
simultaneously, which is a contradiction.
cells which do.
the rightmost column except the top-right square a really big number (call it 
Call this other entry 
![[asy]
size(5cm); // Default size based on your preference
unitsize(1cm);
for (int i = 0; i < 4; ++i) {
for (int j = 1; j < 5; ++j) {
draw(shift(i, j) * unitsquare);
label("0", (i+0.5, j+0.5), fontsize(10pt));
}
}
for (int i = 4; i < 5; ++i) {
for (int j = 1; j < 5; ++j) {
draw(shift(i, j) * unitsquare);
label("1", (i+0.5, j+0.5), fontsize(10pt));
}
}
for (int i = 0; i < 4; ++i) {
for (int j = 0; j < 1; ++j) {
draw(shift(i, j) * unitsquare);
label("-1", (i+0.5, j+0.5), fontsize(10pt));
}
}
draw(shift(4, 0) * unitsquare);
label("1434", (4+0.5, 0.5), fontsize(10pt));
[/asy]](http://latex.artofproblemsolving.com/4/d/7/4d7e4e575996faa0af4568d82d86c79e0f1d31a1.png)
numbers (cells) that don't satisfy the problem's conditions.
;
,
,
,
,
.
and 
.
