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such that for all 
the following equality holds ![\[
f(\left\lfloor x\right\rfloor y)=f(x)\left\lfloor f(y)\right\rfloor \]](http://latex.artofproblemsolving.com/f/3/a/f3ac860e90d671f9fb592a6bd55c102a3000daec.png)
where 
is greatest integer not greater than 
such that
find the number of ordered real triples 
such that![\[\begin{aligned}
\begin{cases}
x+y^2+z^2=a,\\
x^2+y+z^2=a,\\
x^2+y^2+z=a.
\end{cases}
\end{aligned}\]](http://latex.artofproblemsolving.com/9/8/6/986593653bf128bf7b3d497abed78e1459134c5f.png)
be an integer. Rowan and Colin play a game on an 
grid of squares, where each square is colored either red or blue. Rowan is allowed to permute the rows of the grid and Colin is allowed to permute the columns. A grid coloring is orderly if: [list] [*]no matter how Rowan permutes the rows of the coloring, Colin can then permute the columns to restore the original grid coloring; and [*]no matter how Colin permutes the columns of the coloring, Rowan can then permute the rows to restore the original grid coloring. [/list] In terms of 
, how many orderly colorings are there?
is selected inside acute 
so that 
and 
. Point 
is chosen on ray 
so that 
. Let 
be the midpoint of 
.
is tangent to the circumcircle of triangle 
. that satisfy![\[
f(x^2-y)+2yf(x)=f(f(x))+f(y)
\]](http://latex.artofproblemsolving.com/f/b/b/fbb693770cd90d8a144026608074122e54da5802.png)
for all . and 
be positive integers. The cells of an 
grid are colored amber and bronze such that there are at least 
amber cells and at least 
bronze cells. Prove that it is possible to choose amber cells and bronze cells such that no two of the 
chosen cells lie in the same row or column.
and 
are erected outside an acute triangle 
Suppose that ![\[\angle BC_1C+\angle CA_1A+\angle AB_1B=180^{\circ}.\]](http://latex.artofproblemsolving.com/3/3/7/3378f0c4409d14bc0afddf2616992c13f624be3b.png)
Prove that lines 
and 
are concurrent.
-by- grid. He colors 
of the unit squares silver and the remaining squares gold. Kei then folds the grid in half vertically, forming pairs of overlapping unit squares. Let 
and equal equal the least and greatest possible number of gold-on-gold pairs, respectively. What is the value of 
?
-by- board of unit squares for some odd positive integer . We say that a collection 
of identical dominoes is a maximal grid-aligned configuration on the board if consists of 
dominoes where each domino covers exactly two neighboring squares and the dominoes don't overlap: then covers all but one square on the board. We are allowed to slide (but not rotate) a domino on the board to cover the uncovered square, resulting in a new maximal grid-aligned configuration with another square uncovered. Let 
be the number of distinct maximal grid-aligned configurations obtainable from by repeatedly sliding dominoes. Find the maximum value of as a function of .
be a trapezoid with 
. A point 
is chosen inside the trapezoid, and a point 
is chosen inside the triangle 
such that 
, 
. Prove that triangles 
and 
have equal areas.
, 
, 
, 
and 
, 
(we are assuming 
is parallel to 
-axis), then we should have ![\[ \frac{m_2-y_1}{m_1-a} = \frac{y_2-n_2}{c-n_1} \quad \text{and} \quad \frac{n_2-y_1}{n_1-d} = \frac{y_2-m_2}{b-m_1} \quad \quad (1) \]](http://latex.artofproblemsolving.com/7/a/1/7a1775e52fe0926ceea74b805ee37841cd53c2fd.png)
as and .![\[ \frac{1}{2} \begin{vmatrix}
1 & 1 & 1\\
a & b & n_1 \\
y_1 & y_2 & n_2
\end{vmatrix} = \frac{1}{2} \begin{vmatrix}
1 & 1 & 1\\
c & d & m_1 \\
y_2 & y_1 & m_2
\end{vmatrix} \quad \quad (2) \]](http://latex.artofproblemsolving.com/7/6/3/763b1763b0ab067d37f099290a7f892e77f235ca.png)
as they are areas of triangles (left one) and (right one).
implies 
using some algebraic manipulations.
are outside or inside the specified polygon.
as ![$[X]$](http://latex.artofproblemsolving.com/f/6/3/f631f7a53d834de52b718ed6b496689e9658eadf.png)
be a trapezoid s.t. 
; 
be a point then![$[AECD]=[BECD]$](http://latex.artofproblemsolving.com/1/a/b/1abc88febaeab84e9dd2d39d33e02d2bf0654df5.png)
![$[BAND]=[CAND ]=[CMND ]$](http://latex.artofproblemsolving.com/e/7/1/e71b0dcf4d59f6355d2aa2b23cf1651018a7e9dd.png)
but ![$[BAND]=[BAN ]+[BND ]$](http://latex.artofproblemsolving.com/9/1/b/91b2fe0229d73e2d916d02ff4b6446ea2e5af798.png)
and ![$[CMND]=[DCM ]+[DMN ] $](http://latex.artofproblemsolving.com/b/0/d/b0dfff6d5364c7d9367d43178a81f10a1a791b7a.png)
besides![$[BND ]=[DMN ](\because DN\parallel BM)$](http://latex.artofproblemsolving.com/3/4/f/34f3412f388b0beafd7eee69afbf7c4718db2250.png)
we deduces then ![$[BAN ]=[DCM ]$](http://latex.artofproblemsolving.com/f/7/c/f7c80c95eae5455ed3e2eaed13620a8ae089b562.png)
and 
are parallelograms,easy to know that 
and 
.
at 
,
at 
,
.
,
,
.
.
and 
.
and let 
.![$$[ABN]=[AYN]+[NYB]+[BYA]=\lambda(YM\cdot YN+YB\cdot YN+YB\cdot AM)=\lambda(YM\cdot YN+YB\cdot AX)$$](http://latex.artofproblemsolving.com/0/8/5/085fa07e11f01fc630d9d6a94278b21837503509.png)
Similarly ![$[CDM]=\lambda(XM\cdot XN+XD\cdot CY)$](http://latex.artofproblemsolving.com/6/f/2/6f227eca071ac3c94c8d33bb7082fda45b956a86.png)
.
and 
(because 
)
to a unit square through an affine transformation.
.
for some 
.
.
and 
,
for some 
(because 
(because 
.
we get 
.![$$\frac{[ABM]}{[ABC]}=\begin{vmatrix}1&0&0\\0&1&0\\\frac{(y+z)((t+1)x+z)}{z}&-\frac{y((t+1)x+z)}{z}&y-tx\end{vmatrix}=y-tx$$](http://latex.artofproblemsolving.com/b/2/e/b2e9a6fcaf633cb9b20c69943830d43874f12ef8.png)
and![$$\frac{[CDM]}{[ABC]}=\begin{vmatrix}0&0&1\\1&t&-t\\x&y&z\end{vmatrix}=y-tx.$$](http://latex.artofproblemsolving.com/7/f/0/7f0020c1f0c535c4766ebb4844f9786a5d76e7f0.png)
Therefore, ![$[ABM]=[CDM]$](http://latex.artofproblemsolving.com/a/5/f/a5f31256fafabff83c5b84fc662ce8058a454211.png)
,