[/quote]
,
be a point in the plane of triangle 
such that the segments 
,
,
are the sides of an obtuse triangle. Assume that in this triangle the obtuse angle opposes the side congruent to 
is acute.
and 
be two distinct rays not lying on the same line, and let 
be a circle with center 
that is tangent to ray 
and ray 
.
be a point on segment 
.
be the intersection of lines 
and 
be the intersection of line 
is tangent to 
lie on the plane such that 
is a parallelogram. 
is the orthocenter of 
intersects 
at 
and 
intersects 
at 
The circle with diameter 
intersects the circumcircle of 
again at 
Prove that: The circumcircle of 
passes through the reflection point of 
,
denote the incenter and 
.
perpendicular to 
intersects the tangent from 
.
and 
.
.
such that for any positive integer number 
and any arrangement of the numbers from 1 to 
on a circle, there exists a pair of adjacent numbers whose ratio (when dividing the larger number by the smaller one) is less than 
from the nonnegative integers to the integers satisfying 
and ![\[ f(mn) = f(m) + f(n) + f(m)f(n) \]](http://latex.artofproblemsolving.com/e/f/f/eff10a32cffaec742403ebfc46ff09bc8ef7b7c5.png)
for all nonnegative integers 
and 
![\[\sum_{i=1}^n \sum_{j=1}^n \left\lfloor \frac{ij}{n+1} \right\rfloor=\frac{n^2(n-1)}{4}\]](http://latex.artofproblemsolving.com/e/7/f/e7fb8c20a43535b5aaed5ab254f1ef043263c62b.png)
true?
be the binary representation of 
.
,
,
,
equals 
.
,
is chosen on the side 
.
and 
,
and 
are drawn, respectively. The point 
is the circumcenter of 
.
and 
lies on the line 
.
such that![\[
4k^3 + 4k + 1
\]](http://latex.artofproblemsolving.com/f/3/5/f355e456cd47ecfdb13597d52fec3d848b67c594.png)
numbers, none of which appears more than 
times. Prove that they can be divided into 
groups of 
be a cyclic quadrilateral. 
be the projections of 
,
is a kite.
.
and 
,
,
.
such moves the blackboard will contain a single number. What is its maximum possible value? (Reminder that 
is the maximum integer not exceeding 
.
Y by Jc426, Adventure10, Mango247, and 1 other user
A 
table consists of 
unit cells. A block is a 
square consisting of 
unit cells of the table. A set 
of 
blocks covers the table (i.e. each cell of the table is covered by some block of ) but no 
blocks of cover the table. Find the largest possible value of .
table consists of 
unit cells. A block is a 
square consisting of 
unit cells of the table. A set 
of 
blocks covers the table (i.e. each cell of the table is covered by some block of ) but no 
blocks of cover the table. Find the largest possible value of .
.
(X's are centers of blocks):
(every 3x1 rectangle has one non-marked point). If a 3x3 square has 6 points, then it must be this or its rotation (use brute force, considering that first configuration):
,
be the maximum number of matchings that can begotten 
grid of cells (with corresponding 
grid of interior vertices). Clearly 
since there are no vertices to be matched. We now show that 
.
,
.
cells on each edge by chaotic_iak's forbidden 1 by 3 pattern, but you need to use two corners to get everything to fit so 2 must be subtracted.
of vertices with corresponding 
grid of interior cells, which satisfies all the same conditions as the original problem. Swap the role of vertices and cells and we have an instance for 
,
maximum matchings that can be put on it. This completes the proof of the recursion.
instead of 
for the bound.
; 39 looks to be possible in the original version of the problem. The search has not been exhaustive, so further improvements might be possible.
in 
parts: central square 
and a "frame" around it. It's not difficult to establish that there are no more than 
blocks, which are entirely situated inside the "frame". Also note that if a block is situated (maybe, not entirely) in square 
then its unique cell is situated there too.
blocks, which are situated (maybe, not entirely) in central square 
.
unique cells, which are covered by different blocks. 
There exists a square 
of square 
There doesn't exist a square 
.
combinatorial problem. Was convinced that 
is the answer, when an extra block can be squeezed.
--- at least 1 out of 4 of the cells it covers which is only covered by that block. So, with each block's presence, we draw an arrow from its center to the 
lines: 
for 
and 
for 
,
(0,0), and the highest and rightmost square is the 
square.
We have the following Claims;
,
,
and 
grid, where its block centers must lie on its horizontal midline. Then, there exists either 
or 
,
is the maximum number of blocks which can be placed in a 
grid. There's a catch here: that block centers may lie in the grid's boundary (so squares can lie in a 
grid, but only the 
We proceed chunk by chunk:
squares in a 
grid which has at least 3 redeeming cells (since each block is matched with a redeeming cell.) It's easy to see that this is impossible.
and at most 
Since each block has a unique redeeming cell, say a block belongs to a grid if its redeeming cell is located inside that grid. Dissect the 
grids. Then, a ![[asy]usepackage("tikz");label("\begin{tikzpicture}[scale=0.7]
\draw[blue,thick](0,0)--(0,10)--(10,10)--(10,0)--cycle; \draw[blue,thick](2,2)--(2,8)--(8,8)--(8,2)--cycle; \draw[blue,thick](2,5)--(5,5)--(5,2); \draw[blue,thick](3,2)--(3,5); \draw[blue,thick](4,2)--(4,5); \draw[blue,thick](2,3)--(5,3); \draw[blue,thick](2,4)--(5,4); \draw[green,thick,->] (2,2)--(2.5,2.5); \draw[green,thick,->](4,2)--(3.5,2.5); \draw[green,thick,->](5,5)--(4.5,4.5); \draw[red,dashed,->,very thick](3,4)--(3.5,4.5);\draw[red,dashed,->,very thick](3,4)--(3.5-1,4.5);\draw[red,dashed,->,very thick](3,4)--(3.5,4.5-1);\draw[red,dashed,->,very thick](3,4)--(3.5-1,4.5-1); \draw[green,thick,->](5,3)--(4.5,3.5);
\end{tikzpicture}");[/asy]](http://latex.artofproblemsolving.com/b/9/9/b99976d5303d548068035167d2da02371bb45e55.png)
The center cell (the 
square, diagram-wise) must be covered by a block belonging to this grid, because the blocks belonging to other 
or 
.
,
and 
and 
All four 
This is just trial-and-error, as for even three of the grids to behave this way, the lower left grid must be oriented in a specific way (i.e. the block which covers the 
can be obtained almost effortlessly:![[asy]usepackage("tikz");label("\begin{tikzpicture}[scale=1]
\fill[red!10!white](2,3)--(2,4)--(3,4)--(3,3)--cycle; \fill[red!10!white,shift={(1,-1)}](2,3)--(2,4)--(3,4)--(3,3)--cycle;
\foreach \i in {0,1,2,3,4,5,6} \draw[blue,thick](\i,0)--(\i,6); \foreach \j in {0,1,2,3,4,5,6} \draw[blue,thick](0,\j)--(6,\j);
\draw[green,thick,shift={(0,1)},->](0,0)--(0.5,0.5); \draw[green,thick,shift={(0,3)},rotate=-90,->](0,0)--(0.5,0.5); \draw[green,thick,shift={(0,6)},rotate=-90,->](0,0)--(0.5,0.5); \draw[green,thick,shift={(1,0)},rotate=0,->](0,0)--(0.5,0.5); \draw[green,thick,shift={(2,2)},rotate=180,->](0,0)--(0.5,0.5);\draw[green,thick,shift={(1,4)},rotate=0,->](0,0)--(0.5,0.5); \draw[green,thick,shift={(2,6)},rotate=180,->](0,0)--(0.5,0.5);\draw[green,thick,shift={(3,0)},rotate=90,->](0,0)--(0.5,0.5); \draw[green,thick,shift={(2,3)},rotate=0,->](0,0)--(0.5,0.5);\draw[green,thick,shift={(3,5)},rotate=180,->](0,0)--(0.5,0.5);\draw[green,thick,shift={(3,2)},rotate=0,->](0,0)--(0.5,0.5);\draw[green,thick,shift={(4,1)},rotate=0,->](0,0)--(0.5,0.5);\draw[green,thick,shift={(5,3)},rotate=180,->](0,0)--(0.5,0.5);\draw[green,thick,shift={(4,4)},rotate=0,->](0,0)--(0.5,0.5);\draw[green,thick,shift={(4,6)},rotate=-90,->](0,0)--(0.5,0.5);\draw[green,thick,shift={(6,0)},rotate=90,->](0,0)--(0.5,0.5);\draw[green,thick,shift={(6,2)},rotate=180,->](0,0)--(0.5,0.5);\draw[green,thick,shift={(6,4)},rotate=90,->](0,0)--(0.5,0.5);\draw[green,thick,shift={(6,6)},rotate=180,->](0,0)--(0.5,0.5);
\end{tikzpicture}");[/asy]](http://latex.artofproblemsolving.com/c/4/1/c413f99bd64011a2e283aa075da3dbc362e3b944.png)
and 
are required to curb the middle intersection of the blocks. This is a valid move since the redeeming cells still follow the same formation. 
to the ultimate configuration.
s
is replaced with 
,
or 
,
and 
have two blocks which can cover them, and the square 
cannot be covered by any of those blocks --- which results in a total of exactly nine blocks.
bounding, seeing that the blocks with center 
will cover up sets of 2 squares of 
and 
block covers two squares from 
and two squares from 
,
together. Equivalently, any redeeming cell 
block because their redeeming cells seem economically efficient, and I shaped the other blocks to have an (even,even) center, but there's immediately a problem:![[asy]usepackage("tikz");label("\begin{tikzpicture}[scale=0.9]
\fill[gray!30!white,shift={(-2,-3)}](2,3)--(2,4)--(3,4)--(3,3)--cycle; \fill[gray!30!white,shift={(-2,2)}](2,3)--(2,4)--(3,4)--(3,3)--cycle; \fill[red!10!white,shift={(-1,-2)}](2,3)--(2,4)--(3,4)--(3,3)--cycle;
\foreach \i in {0,1,2,3} \draw[blue,thick](\i,0)--(\i,6); \foreach \j in {0,1,2,3,4,5,6} \draw[blue,thick](0,\j)--(3,\j);
\draw[green,thick,shift={(0,0)},->](0,0)--(0.5,0.5); \draw[green,thick,shift={(0,2)},rotate=-90,->](0,0)--(0.5,0.5); \draw[green,thick,shift={(0,4)},rotate=0,->](0,0)--(0.5,0.5); \draw[green,thick,shift={(0,6)},rotate=-90,->](0,0)--(0.5,0.5); \draw[green,thick,shift={(2,0)},rotate=90,->](0,0)--(0.5,0.5); \draw[green,thick,shift={(1,3)},rotate=-90,->](0,0)--(0.5,0.5);\draw[green,thick,shift={(2,4)},rotate=90,->](0,0)--(0.5,0.5); \draw[green,thick,shift={(2,6)},rotate=180,->](0,0)--(0.5,0.5);\draw[green,thick,shift={(3,2)},rotate=180,->](0,0)--(0.5,0.5);
\draw[red,semithick,shift={(1,1)}] (-0.13,-0.13)--(0.13,0.13); \draw[red,semithick,shift={(1,1)}] (0.13,-0.13)--(-0.13,0.13); \draw[red,semithick,shift={(1,2)}] (-0.13,-0.13)--(0.13,0.13); \draw[red,semithick,shift={(1,2)}] (0.13,-0.13)--(-0.13,0.13); \draw[red,semithick,shift={(2,1)}] (-0.13,-0.13)--(0.13,0.13); \draw[red,semithick,shift={(2,1)}] (0.13,-0.13)--(-0.13,0.13); \draw[red,semithick,shift={(2,2)}] (-0.13,-0.13)--(0.13,0.13); \draw[red,semithick,shift={(2,2)}] (0.13,-0.13)--(-0.13,0.13);
\end{tikzpicture}
\begin{tikzpicture}
\node at (0,0) {}; \node at (0,2.6) {$\quad \Rightarrow \quad$};
\end{tikzpicture}
\begin{tikzpicture}[scale=0.9]
\fill[gray!30!white,shift={(-2,-3)}](2,3)--(2,4)--(3,4)--(3,3)--cycle; \fill[gray!30!white,shift={(-2,2)}](2,3)--(2,4)--(3,4)--(3,3)--cycle; \fill[red!10!white,shift={(-1,-2)}](2,3)--(2,4)--(3,4)--(3,3)--cycle;
\foreach \i in {0,1,2,3} \draw[blue,thick](\i,0)--(\i,6); \foreach \j in {0,1,2,3,4,5,6} \draw[blue,thick](0,\j)--(3,\j);
\draw[green,thick,shift={(0,0)},->](0,0)--(0.5,0.5); \draw[green,thick,shift={(0,2)},rotate=-90,->](0,0)--(0.5,0.5); \draw[green,thick,shift={(0,4)},rotate=0,->](0,0)--(0.5,0.5); \draw[green,thick,shift={(0,6)},rotate=-90,->](0,0)--(0.5,0.5); \draw[red,thick,shift={(2,1)},rotate=90,->](0,0)--(0.5,0.5); \draw[green,thick,shift={(1,3)},rotate=-90,->](0,0)--(0.5,0.5);\draw[green,thick,shift={(2,4)},rotate=90,->](0,0)--(0.5,0.5); \draw[green,thick,shift={(2,6)},rotate=180,->](0,0)--(0.5,0.5);\draw[red,thick,shift={(3,3)},rotate=180,->](0,0)--(0.5,0.5);
\end{tikzpicture}");[/asy]](http://latex.artofproblemsolving.com/3/4/b/34b3c19329ea5b0e974cde3e72a8575099ab256e.png)
,
grid with 
blocks. Then, another problem surfaced up when I attempted to extend this configuration to a 
.
--- making the board a bit heavier on other blocks which are supposed to 
there.![[asy]usepackage("tikz");label("\begin{tikzpicture}[scale=0.9]
\foreach \i in {0,1,2,3,4,5,6} \draw[blue,thick](0,\i)--(3,\i); \foreach \i in {0,3,4,5,6} \draw[blue,thick](\i,0)--(\i,6); \draw[blue,thick](3,0)--(6,0); \draw[blue,thick](3,6)--(6,6);
\foreach \i/\j/\k in {0/1/0,0/3/-90,0/4/0,0/6/-90,1/0/0,2/2/180,2/3/90,2/5/180,3/0/90,3/2/-90,3/3/0,3/5/-90,4/6/-90,5/1/90,5/3/180,5/4/90,6/6/180}
\draw[green,thick,shift={(\i,\j)},rotate=\k,->](0,0)--(0.5,0.5);
\foreach \i/\j/\k in {0/1/0,0/3/-90,0/4/0,0/6/-90,1/0/0,2/2/180,2/3/90,2/5/180,3/0/90,3/2/-90,3/3/0,3/5/-90,4/6/-90,5/1/90,5/3/180,5/4/90,6/6/180}
\fill[green] (\i,\j)circle(2.5pt);
\end{tikzpicture}
\begin{tikzpicture}[scale=0.9]
\fill[red!10!white,thick](4,2)--(4,3)--(5,3)--(5,2)--cycle; \fill[red!10!white,thick,shift={(1,1)}](4,2)--(4,3)--(5,3)--(5,2)--cycle;
\foreach \i in {0,1,2,3,4,5,6} \draw[blue,thick](0,\i)--(3,\i); \foreach \i in {0,1,2,3,4,5,6} \draw[blue,thick](\i,0)--(\i,6); \draw[blue,thick](3,0)--(6,0); \draw[blue,thick](3,6)--(6,6); \draw[red,thick](4,3)--(6,3); \draw[red,thick](4,2)--(5,2); \draw[red,thick](5,4)--(6,4);
\foreach \i/\j/\k in {0/1/0,0/3/-90,0/6/-90,1/0/0,1/4/-90,2/2/180,2/6/180,3/0/90,3/3/-90,3/4/90,3/6/-90,4/4/0,4/1/90,5/6/180,6/0/90,6/2/180,6/5/180}
\draw[green,thick,shift={(\i,\j)},rotate=\k,->](0,0)--(0.5,0.5);
\foreach \i/\j/\k in {0/1/0,0/3/-90,0/6/-90,1/0/0,1/4/-90,2/2/180,2/6/180,3/0/90,3/3/-90,3/4/90,3/6/-90,4/4/0,4/1/90,5/6/180,6/0/90,6/2/180,6/5/180}
\fill[green] (\i,\j)circle(2.5pt);
\draw[orange,thick,<->](4.5,2.5)--(5.5,3.5); \fill[orange] (5,3) circle (3pt);
\end{tikzpicture}");[/asy]](http://latex.artofproblemsolving.com/5/9/d/59db42522c5d2a3155bdf0fd65a1a51caeca03ef.png)
can be found (its occurence is rare, too), I concluded that there needs to be 9 redeeming cells both on the right and left 
grid. It turns out that since a 
