Pentagon 2023 12B Q25 dissmo.png
Let the original pentagon be $ABCDE$ centered at $O$. The dashed lines represent the fold lines. WLOG, let's focus on vertex $A$.
Since $A$ is folded onto $O$, $AM = MO$ where $M$ is the intersection of $AO$ and the creaseline between $A$ and $O$. Note that the inner pentagon is regular, and therefore similar to the original pentagon, due to symmetry.
Because of their similarity, the ratio of the inner pentagon's area to that of the outer pentagon can be represented by
$(\frac{OM}{ON})^{2} = (\frac{\frac{OA}{2}}{OA\sin (\angle OAE)})^{2} = \frac{1}{4\sin^{2}54}$

Option 1: Knowledge
Remember that $\sin54 = \frac{1+\sqrt5}{4}$.

Option 2: Angle Identities
$\sin54 = \cos36$
$4\cos^{3}18-3\cos18 = 2\sin18\cos18$
$4(1-\sin^{2}18)-3-2\sin18=0$
$4\sin^{2}18+2\sin18-1=0$
$\sin18 = \frac{-1+\sqrt5}{4}$
$\sin54 = \cos36 = 1-2\sin^{2}18 = \frac{1+\sqrt5}{4}$

$\sin^{2}54 =\frac{3+\sqrt5}{8}$
Let the inner pentagon be $Z$.
$[Z] = \frac{1}{4\sin^{2}54}[ABCDE]$
$= \frac{2(1+\sqrt5)}{3+\sqrt5}$
$= \sqrt5-1$
$\boxed{B}$
-Dissmo