[/quote]
,
be complex numbers such that [center]
for any 
with 
[/center]
and 
for any 
,
,
we have ![\[ 3(a+b+c) \ge 8\sqrt[3]{abc} + \sqrt[3]{\frac{a^3+b^3+c^3}{3}}. \]](http://latex.artofproblemsolving.com/a/9/8/a98b6c29802355c491016bcce58228a020423526.png)
and two lines 
and 
are given in the space. For each nonnegative integer 
we denote by 
the projection of 
on 
and by 
the projection of 
Prove that there exist two segments ![$[A'A''] \subset d_1$](http://latex.artofproblemsolving.com/8/8/6/88667ee717ddd630efd34a9c4c83ee359c4897ed.png)
and ![$[B'B''] \subset d_2$](http://latex.artofproblemsolving.com/3/7/4/374927492266c8adaed5d34e3871d250baba9c1b.png)
of length 
and a nonnegative integer 
such that ![$A_n \in [A'A'']$](http://latex.artofproblemsolving.com/4/8/5/4851110dd8f0bf0ee7e42239ee051da76f5fbd65.png)
and ![$B_n \in [B'B'']$](http://latex.artofproblemsolving.com/5/e/7/5e7d78dd940f5c81673b779a9b45e83f4dc05b4d.png)
for any 
,
,
,
be a sequence of positive integers such that the greatest common divisor of any two consecutive terms is greater than the preceding term; in symbols, 
.
for all 
.
opposite the vertex 
be tangent to the side 
at the point 
.
on 
and 
on 
analogously, using the excircles opposite 
and 
,
lies on the circumcircle of triangle 
on the side 
and 
are the feet of the perpendiculars from 
,
lies on the circumcircle of the triangle 
and 
.
be the circumcenter of triangle 
.
.
be a parallelogram such that 
Let 
and 
be the midpoints of 
respectively. Assuming that 
is a cyclic quadrilateral, find 
be a triangle, and let 
,
and 
as the orthocenters of triangles 
and 
.
passes through a fixed point which is independent of the choice of 
be a triangle inscribed in a circle, where the tangents to the circle at points 
be a point on the arc 
and 
.
intersect at point 
be the reflection of 
and 
intersect at point 
.
,
be the orthocenter of 
at 
.
be the midpoint of 
,
meet 
.
at 
.
is tangent to 
,
and 
,
is the intersection point of the diagonals 
.
intersects 
and 
.
is tangent to the circles with diameters 
and 
.
.
.
.
and 
.
Y by Adventure10, Mango247, ItsBesi, cubres
Consider decompositions of an 
chessboard into 
non-overlapping rectangles subject to the following conditions:
(i) Each rectangle has as many white squares as black squares.
(ii) If
is the number of white squares in the 
-th rectangle, then 
.
Find the maximum value of for which such a decomposition is possible. For this value of , determine all possible sequences 
.
chessboard into 
non-overlapping rectangles subject to the following conditions:(i) Each rectangle has as many white squares as black squares.
(ii) If
is the number of white squares in the 
-th rectangle, then 
.Find the maximum value of
for which such a decomposition is possible. For this value of , determine all possible sequences 
.
.
for 
to 
so 
so this forces 
.
rectangles, as we will show later. But first let us find all such sequences 
.
white squares, it will have an area of 
so it's dimensions are either 
or 
- neither of which would fit on a 
.
(which could fit as a 
rectangle) then 
.
so 
.
are 6 numbers among 1-7. If 
is the number that is not equal to any 
so 
.
.
squares horizontally and 
vertically. Then directly below use a 
and a 
rectangle to cover the 3 rows below it. It's simple from there.
or 
or 
.
![\[2+4+6+8+10+12+14+16=72>64. \]](http://latex.artofproblemsolving.com/5/1/3/513b3ee869bdbd2dc44837c9951460f1d12afed3.png)
There are five ways to write 
as sum of ![\[ \begin{aligned} 2+4+6+8+10+12+22 & ;\\ 2+4+6+8+10+14+20 & ;\\ 2+4+6+8+10+16+18 & ;\\ 2+4+6+8+12+14+18 & ;\\ 2+4+6+10+12+14+16 \end{aligned} \]](http://latex.artofproblemsolving.com/0/5/2/052096c1dd1ff49d7c1c4e6ecd62459f6cc2954a.png)
with the first one being eliminated, since a rectangle with 
.![\[
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline
2 & 2 & 2 & 2 & 2 & 2 & 2 & 4 \\ \hline
2 & 2 & 2 & 2 & 2 & 2 & 2 & 4 \\ \hline
1 & 1 & 1 & 1 & 1 & 5 & 5 & 4 \\ \hline
1 & 1 & 1 & 1 & 1 & 5 & 5 & 4 \\ \hline
1 & 1 & 1 & 1 & 1 & 5 & 5 & 4 \\ \hline
1 & 1 & 1 & 1 & 1 & 6 & 6 & 4 \\ \hline
3 & 3 & 3 & 3 & 3 & 6 & 6 & 4 \\ \hline
3 & 3 & 3 & 3 & 3 & 7 & 7 & 4 \\ \hline
\end{tabular}\]](http://latex.artofproblemsolving.com/6/8/8/6880f7cd7f0fad91dfb07daa2a188b93e96c5342.png)
achieved with 
since otherwise 
which is a clear contradiction. For 
the only possible ordered tuples 
such that 
and 
are 
The last tuple does not have a construction since that would require us to have a rectangle of area 
which is clearly impossible. The other four tuples are possible, as shown by the pictures below, proving our claim.
.
squares altogether. Let 
and 
but the total number of cells in the table is 8.8 = 64, but since 72 is larger than 64, 
.
.
,
.
.
.
s.t. 
After a check we can just find four viable pairs becuase one has 
which is clearly not possible. The others are possible but I don't know how to asy.
,
.![\[ 1 + 2 + \dots + p = \dfrac{p(p + 1)}{2} \le a_1 + a_2 + \dots + a_p = 32 \implies p \le 7. \]](http://latex.artofproblemsolving.com/a/b/9/ab9e962481298c734d74e46381709694dac60d2d.png)
Now we show attainability at ![\[ (1, 2, 3, 5, 6, 7, 8), (1, 2, 3, 4, 5, 6, 11), (1, 2, 3, 4, 5, 7, 10), (1, 2, 3, 4, 5, 8, 9), (1, 2, 3, 4, 6, 7, 9). \]](http://latex.artofproblemsolving.com/5/4/9/549df24b35039d6618724083ebb4da89b595de28.png)
Clearly 
does not work as one of the rectangles would have a side length that is a multiple of 
grid. It is well-known that a rectangle with even area must have an equal number of white and black squares when checkerboard-colored, below are constructions:
(clearly 
)
,
Sequence 
fails because we cannot fit a 
could result in a decomposition. The sequence with smallest sum for 
for 
.
,
tiles in total that the rectangles must cover. Since we have an 
tiles to be covered by the rectangles as a minimum. We now want to figure out all the possible ways to add 
,
,
,
.
