Let $P$ be a regular nonagon with side length $2,$ and let $S$ be the set of points that are within a distance of $1$ from $P.$ (That is, a point $X$ is in $S$ if there exists a point $Y$ in $P$ such that $XY \le 1.$) What is the area of $P$?
We can view $P$ as inscribed in a unit circle centered at the origin. Then the units will conveniently match.
[asy]
unitsize(3 cm);
pair A, B, C, D, E, F, G, H, I, O, P, X, Y;
A = dir(20);
B = dir(60);
C = dir(100);
D = dir(140);
E = dir(180);
F = dir(220);
G = dir(260);
H = dir(300);
I = dir(340);
O = origin;
P = (0.5)*(A + B + C + D + E + F + G + H + I);
X = (sqrt(3)/2,0);
Y = 2*P - X;
draw(Circle(O,1));
draw(X--Y);
draw(A--B--C--D--E--F--G--H--I--cycle);
label("$P$", P, NW);
label("$X$", X, E);
label("$Y$", Y, NW);
[/asy]
Thus, we want to find the area that $X = \overline{XY}$ sweeps out, as $X$ moves around the circle counterclockwise from $X = B$ to $X = K.$
Note that as $X$ ranges from $X = B$ to $X = C,$ $M(X)$ takes on four values, and as $X$ ranges from $X = C$ to $X = D,$ $M(X)$ takes on two values, as illustrated below.
[asy]
unitsize(3 cm);
pair A, B, C, D, E, F, G, H, I, O, P, S, T, X, Y;
A = dir(20);
B = dir(60);
C = dir(100);
D = dir(140);
E = dir(180);
F = dir(220);
G = dir(260);
H = dir(300);
I = dir(340);
O = origin;
P = (0.5)*(A + B + C + D + E + F + G + H + I);
X = (sqrt(3)/2,0);
Y = 2*P - X;
path arc;
arc = arc(O, 1, A, B);
draw(arc);
arc = arc(O, 1, B, C);
draw(arc);
arc = arc(O, 1, C, D);
draw(arc);
label("$A$", A, dir(A));
label("$B$", B, N);
label("$C$", C, N);
label("$D$", D, N);
label("$X$", X, E);
S = intersectionpoint(arc(O, 1, A, B), arc(O, 1, D, E));
T = intersectionpoint(arc(O, 1, F, G), arc(O, 1, I, A));
draw((1,0.25)--(-1,0.25),red,dashed);
draw((1,0)--(-1,0),red,dashed);
draw((1,0) + 0.5*dir(30)--(-1,0) + 0.25*dir(30),red,dashed);
draw((1,0) + 0.5*dir(330)--(-1,0) + 0.25*dir(330),red,dashed);
label("$M(X)$", (1,0.25), NE, red);
label("$M(X)$", (1,0), NE, red);
label("$M(X)$", (1,0) + 0.5*dir(30), NNE, red);
label("$M(X)$", (1,0) + 0.5*dir(330), SSE, red);
[/asy]
Note that $M(X) = 1$ when $0^\circ < \angle MOX \le 60^\circ,$ $M(X) = 2$ when $60^\circ < \angle MOX \le 120^\circ,$ $M(X) = 1$ when $120^\circ < \angle MOX \le 180^\circ,$ $M(X) = 0$ when $180^\circ < \angle MOX \le 240^\circ,$ $M(X) = -1$ when $240^\circ < \angle MOX \le 300^\circ,$ and $M(X) = 0$ when $300^\circ < \angle MOX \le 360^\circ.$
These values correspond $\angle MOX$ to values of $\angle XOY$ as shown. Let $OY = d.$
[asy]
unitsize(3 cm);
pair A, B, C, D, E, F, G, H, I, O, P, S, T, X, Y;
A = dir(20);
B = dir(60);
C = dir(100);
D = dir(140);
E = dir(180);
F = dir(220);
G = dir(260);
H = dir(300);
I = dir(340);
O = origin;
P = (0.5)*(A + B + C + D + E + F + G + H + I);
X = (sqrt(3)/2,0);
Y = 2*P - X;
fill(arc(O,1,A,B)--arc(O,1,B,C)--cycle,gray(0.7));
fill(arc(O,1,D,E)--arc(O,1,F,G)--cycle,gray(0.7));
fill(arc(O,1,I,A)--arc(O,1,A,B)--cycle,gray(0.4));
path arc;
arc = arc(O, 1, A, B);
draw(arc);
arc = arc(O, 1, B, C);
draw(arc);
arc = arc(O, 1, C, D);
draw(arc);
arc = arc(O, 1, F, G);
draw(arc);
arc = arc(O, 1, H, I);
draw(arc);
label("$1$", (arc(O,1,A,B)).midpoint, dir(midpoint, (arc(O,1,A,B)).midpoint), red);
label("$1$", (arc(O,1,B,C)).midpoint, dir(midpoint, (arc(O,1,C,D)).midpoint), red);
label("$2$", (arc(O,1,C,D)).midpoint, dir(midpoint, (arc(O,1,D,E)).midpoint), red);
label("$1$", (arc(O,1,F,G)).midpoint, dir(midpoint, (arc(O,1,G,H)).midpoint), red);
label("$1$", (arc(O,1,H,I)).midpoint, dir(midpoint, (arc(O,1,I,A)).midpoint), red);
draw((1,0.875)--(-1,0.875),red,dashed);
draw((1,0.625)--(-1,0.625),red,dashed);
draw((1,0.5)--(-1,0.5),red,dashed);
draw((1,0.375)--(-1,0.375),red,dashed);
draw((1,0)--(-1,0),red,dashed);
label("$d$", (1,0.5), NE, red);
draw((-1,-1)--(-1,1));
dot("$A$", A, dir(A));
dot("$B$", B, N);
dot("$C$", C, N);
dot("$D$", D, N);
dot("$X$", X, E);
dot("$Y$", Y, NW);
[/asy]
Hence,
\[1 \cdot \frac{d^2}{2} + 1 \cdot \frac{d^2}{2} + 2 \cdot d \left( \frac{d}{2} \right) = \frac{3d^2}{2} + d^2 = 1.\]Therefore, $d = \frac{1}{\sqrt{7}}.$
Then
\begin{align*}
\angle MOY &= 30^\circ + 120^\circ + 60^\circ + 150^\circ + 120^\circ \\
&= 480^\circ = -240^\circ,
\end{align*}so the area of $\triangle MOY$ is
\[\frac{1}{2} \cdot 1 \cdot \frac{1}{\sqrt{7}} \cdot \sin 240^\circ = \frac{\sqrt{3}}{4 \sqrt{7}}.\]Hence, the area of $\triangle MOY$ which we are interested in is
\[7 \cdot \frac{\sqrt{3}}{4 \sqrt{7}} = \boxed{\frac{\sqrt{3}}{4}}.\]
The area of a regular nonagon with side length $2$ can be found by dividing it into isosceles triangles.
First, let's find the area of one isosceles triangle. We can find the height of the triangle by drawing an altitude from one of the vertices to the opposite side. This altitude divides the triangle into two congruent right triangles. Since the nonagon is regular, each angle of the triangle measures $\frac{180^\circ - 360^\circ/9}{2} = 140^\circ,$ so by the Law of Sines, $\frac{h}{\sin 140^\circ} = \frac{1}{\sin 20^\circ}.$
Then by the Law of Sines, $\frac{2}{\sin \angle A} = \frac{2}{\sin 20^\circ} = \frac{h}{\sin 120^\circ},$ so
\[h = 4 \sin 120^\circ = 4 \sin 60^\circ = 4 \cdot \frac{\sqrt{3}}{2} = 2 \sqrt{3}.\]
[asy]
unitsize (2 cm);
int i;
draw(Circle((0,0),1));
draw((-2*sqrt(3)/3,-1/3)--(2*sqrt(3)/3,-1/3),red);
draw((cos(80*pi/180),sin(80*pi/180))--(0,0)--(cos(110*pi/180),sin(110*pi/180)),red);
for (i = 0; i <= 8; ++i) {
draw(dir(40*i + 20)--dir(40*(i + 1) + 20));
dot(dir(40*i + 20));
}
dot("X", (cos(120*pi/180),sin(120*pi/180)), NE);
[/asy]
Thus, the area of one triangle is
\[\frac{1}{2} \cdot 2 \sqrt{3} \cdot 2 = 2 \sqrt{3},\]so the area of the nonagon is $9 \cdot 2 \sqrt{3} = \boxed{18 \sqrt{3}}.$