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
\[7 \cdot \frac{\sqrt{3}}{4 \sqrt{7}} = \boxed{\frac{\sqrt{3}}{4}}.\]
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 $P$?
1 answer