[/quote]
,
and 
.
that satisfy![\[
f(x^2-y)+2yf(x)=f(f(x))+f(y)
\]](http://latex.artofproblemsolving.com/f/b/b/fbb693770cd90d8a144026608074122e54da5802.png)
for all 
.
![$\mathbb{R}[x]$](http://latex.artofproblemsolving.com/a/8/8/a88b9f4858016b1771635c611d44b56161a08009.png)
be the set of all polynomials with real coefficients. Find all functions ![$f: \mathbb{R}[x] \rightarrow \mathbb{R}[x]$](http://latex.artofproblemsolving.com/0/3/3/033be3ab043bd47e537f01791c813739ad5c7da5.png)
satisfying the following conditions:
maps the zero polynomial to itself,![$P \in \mathbb{R}[x]$](http://latex.artofproblemsolving.com/f/7/c/f7c4c970908aa5570a9afcc45fc16c1d36cc3d5d.png)
,
,![$P, Q \in \mathbb{R}[x]$](http://latex.artofproblemsolving.com/c/3/0/c301578bade5b93056d77454921cba0a9812212c.png)
,
and 
have the same set of real roots.
be a square of center 
and let 
be the symmetric of the point 
with respect to point 
.
be the intersection of 
and 
,
be the intersection of 
and 
.
is the angle bisector of 
.
be reals such that 
Show that
Let 
and 
Prove that
be a polygon formed by the edges of an infinite chessboard, which does not intersect itself. Let the numbers 
represent the number of unit squares that have exactly 
edges on the boundary of 
such that the inequality 
holds for each polygon constructed with these conditions.
and 
Find the value range of 
and 
with real coefficients satisfying
Y by pifinity, AbdulelahAltaf, Adventure10, Mango247, and 1 other user
We define a chessboard polygon to be a polygon whose sides are situated along lines of the form 
or 
, where 
and 
are integers. These lines divide the interior into unit squares, which are shaded alternately grey and white so that adjacent squares have different colors. To tile a chessboard polygon by dominoes is to exactly cover the polygon by non-overlapping 
rectangles. Finally, a tasteful tiling is one which avoids the two configurations of dominoes shown on the left below. Two tilings of a 
rectangle are shown; the first one is tasteful, while the second is not, due to the vertical dominoes in the upper right corner.
![[asy]size(300); pathpen = linewidth(2.5);
void chessboard(int a, int b, pair P){
for(int i = 0; i < a; ++i) for(int j = 0; j < b; ++j)
if((i+j) % 2 == 1) fill(shift(P.x+i,P.y+j)*unitsquare,rgb(0.6,0.6,0.6));
D(P--P+(a,0)--P+(a,b)--P+(0,b)--cycle);
}
chessboard(2,2,(2.5,0));fill(unitsquare,rgb(0.6,0.6,0.6));fill(shift(1,1)*unitsquare,rgb(0.6,0.6,0.6)); chessboard(4,3,(6,0)); chessboard(4,3,(11,0)); MP("\mathrm{Distasteful\ tilings}",(2.25,3),fontsize(12));
/* draw lines */
D((0,0)--(2,0)--(2,2)--(0,2)--cycle); D((1,0)--(1,2)); D((2.5,1)--(4.5,1)); D((7,0)--(7,2)--(6,2)--(10,2)--(9,2)--(9,0)--(9,1)--(7,1)); D((8,2)--(8,3)); D((12,0)--(12,2)--(11,2)--(13,2)); D((13,1)--(15,1)--(14,1)--(14,3)); D((13,0)--(13,3));[/asy]](//latex.artofproblemsolving.com/e/7/1/e712b33951fe4d8c30f4664d0857d871b13fbfb6.png)
a) Prove that if a chessboard polygon can be tiled by dominoes, then it can be done so tastefully.
b) Prove that such a tasteful tiling is unique.
or 
, where 
and 
are integers. These lines divide the interior into unit squares, which are shaded alternately grey and white so that adjacent squares have different colors. To tile a chessboard polygon by dominoes is to exactly cover the polygon by non-overlapping 
rectangles. Finally, a tasteful tiling is one which avoids the two configurations of dominoes shown on the left below. Two tilings of a 
rectangle are shown; the first one is tasteful, while the second is not, due to the vertical dominoes in the upper right corner.![[asy]size(300); pathpen = linewidth(2.5);
void chessboard(int a, int b, pair P){
for(int i = 0; i < a; ++i) for(int j = 0; j < b; ++j)
if((i+j) % 2 == 1) fill(shift(P.x+i,P.y+j)*unitsquare,rgb(0.6,0.6,0.6));
D(P--P+(a,0)--P+(a,b)--P+(0,b)--cycle);
}
chessboard(2,2,(2.5,0));fill(unitsquare,rgb(0.6,0.6,0.6));fill(shift(1,1)*unitsquare,rgb(0.6,0.6,0.6)); chessboard(4,3,(6,0)); chessboard(4,3,(11,0)); MP("\mathrm{Distasteful\ tilings}",(2.25,3),fontsize(12));
/* draw lines */
D((0,0)--(2,0)--(2,2)--(0,2)--cycle); D((1,0)--(1,2)); D((2.5,1)--(4.5,1)); D((7,0)--(7,2)--(6,2)--(10,2)--(9,2)--(9,0)--(9,1)--(7,1)); D((8,2)--(8,3)); D((12,0)--(12,2)--(11,2)--(13,2)); D((13,1)--(15,1)--(14,1)--(14,3)); D((13,0)--(13,3));[/asy]](http://latex.artofproblemsolving.com/e/7/1/e712b33951fe4d8c30f4664d0857d871b13fbfb6.png)
a) Prove that if a chessboard polygon can be tiled by dominoes, then it can be done so tastefully.b) Prove that such a tasteful tiling is unique.
.
of the forced squares together with all of the lower squares with color 
squares with the color other than 
.
the weight 
.
even.
for those in horizontal dominos and 
for those in vertical dominos. A 
distasteful block with lower left corner at 
,
.
.
must be in the polygon. We can't have the domino at 
in the same tiling as the domino at 
,
must connect in another direction; either (a
in the same tiling as 
.
,
before 
is the only possibility: 
,
must be in at least one tiling. If it's in both, the pair of fixed dominoes we have forces another fixed domino at 
,
,
,
and reverse colors to get case (a'): that 
leads to a contradiction, as a ladder of fixed horizontal dominoes leads to failure.
to 
because of the domino at 
in the same tiling. If we connect 
,
:
is 
in the loop because it is distasteful with 
.
because that's case (a'). That leaves only 
-
?
square, it takes 
moves to go from the worst arrangement (no tasteful 
board.
would make the argument simpler, though. In that case, we can't keep going down in case (b).
![[asy]size(100); pathpen = linewidth(2.5);
void chessboard(int a, int b, pair P){
for(int i = 0; i < a; ++i) for(int j = 0; j < b; ++j)
if((i+j) % 2 == 1) fill(shift(P.x+i,P.y+j)*unitsquare,rgb(0.6,0.6,0.6));
D(P--P+(a,0)--P+(a,b)--P+(0,b)--cycle);
}
chessboard(2,2,(2.5,0));fill(unitsquare,rgb(0.6,0.6,0.6));fill(shift(1,1)*unitsquare,rgb(0.6,0.6,0.6)); chessboard(4,3,(6,0)); chessboard(4,3,(11,0)); MP("\mathrm{Distasteful\ tilings}",(2.25,3),fontsize(12));
/* draw lines */
D((0,0)--(2,0)--(2,2)--(0,2)--cycle); D((1,0)--(1,2)); D((2.5,1)--(4.5,1)); D((7,0)--(7,2)--(6,2)--(10,2)--(9,2)--(9,0)--(9,1)--(7,1)); D((8,2)--(8,3)); D((12,0)--(12,2)--(11,2)--(13,2)); D((13,1)--(15,1)--(14,1)--(14,3)); D((13,0)--(13,3));[/asy]](http://latex.artofproblemsolving.com/d/9/6/d96b48633a5257fd25edaca85aa4cd582d8aad02.png)
a) Prove that if a chessboard polygon can be tiled by dominoes, then it can be done so tastefully.
(with 
nonempty tasteful tilings).
(WLOG lower left with only upper and right neighbors and facing outside the interior of the polygon) bordered above and to the right by 
respectively, so either 
or 
is a domino; also let 
denote the square to the upper right of 
cannot be tiled (with dominoes), then every tiling of 
must have at least two tasteful tilings (by removing 
(with an even number of squares and including the boundary of the actual polygon) composed of two disjoint tilings ("1 mod 2 shifts" of each other) and an interior 
composed of fixed dominoes (i.e. the same for the two tilings).
and the corner 
faces outside the polygon, we need 
in order to reach 
in this order along the (boundary) path. We can show that 
cannot be tiled (this is of course also true if 
(in regards to the distasteful configurations) that 
(which is a vertical white-black domino, viewed from bottom to top) to the tasteful tiling of 
itself (since the only distasteful configuration 
,
are not in 
such that 
for the center 
of some square in 
until it "leaves" (it loops eventually, so it must) for the first time at 
on the left and 
on the right. Clearly 
,
and 
,
,
,
,
(i.e. for the corner at 
)
.
,
is not a fixed interior domino, or else 
are all dominoes in the same tiling (one of the two tasteful ones mentioned in (*)), whence tiling instead by 
,
yields that 
can be tiled, contradicting (**) (as before, for the corner at 
is not a fixed interior domino.
lies in 
(by symmetry about 
is a fixed interior domino for 
and 
is in the interior (for 
)
is a domino, which forces 
(if 
;
,
must belong to a domino (since it's in the interior). Yet 
cannot be a domino since 
is already a fixed (interior) domino, so 
(as long as 
;
,
(in this order) along the boundary, 
,
must appear on the boundary in this order. Thus 
and 
are dominoes in the same tasteful tiling of 
)
lie on a diagonal and therefore all have the same color.
from one vertex to the next, the color of the square on the left remains the same).
rectangle can be tiled by dominoes if and only if 
is even or 
is even.
are both even
.
.
be a white square in the top row. Then in any tasteful tiling, the domino which contains 
or 
.
denote the chessboard polygon which can be tiled by dominoes.
of the polygon, and WLOG is it black (other case similar). Then we have two cases:
square with two vertical dominoes to arrive in the first case. So this additional domino will not cause any distasteful tilings.
,
be any such cycle and let 
enclosed by 
.
,
,
,
,
is covered by a vertical domino. Then 
must be covered by a horizontal domino, to avoid a distasteful tiling. Then if 
is in 
lie inside 
denote the first point of this sequence after 
so the line segment 
joining 
has slope 
(in the black case use 
instead).
squares are all counterclockwise.
