2023 AMC 10B Problem 25

Folding a vertex to the center creases along the perpendicular bisector of the segment from that vertex to the center, a line at distance $\tfrac{R}{2}$ from the center

Apothem $\tfrac{R}{2}$ replaces $R\cos36^\circ,$ so the area scales by $\left(\tfrac{1}{2\cos36^\circ}\right)^2;$ use $\cos36^\circ=\tfrac{1+\sqrt5}{4}$

Let the original pentagon have circumradius $R.$ Folding a vertex to the center creases along the perpendicular bisector of the center-to-vertex segment, a line at distance $\tfrac{R}{2}$ from the center. Those five creases bound the new regular pentagon, whose apothem is $\tfrac{R}{2}$ (the original apothem was $R\cos 36^\circ$). So the new pentagon is similar with ratio $\frac{1}{2\cos 36^\circ},$ and its area is the old area times $\left(\tfrac{1}{2\cos 36^\circ}\right)^2.$ Plug in $\cos 36^\circ = \tfrac{1+\sqrt5}{4}:$ the factor becomes $\tfrac{2}{3+\sqrt5} = \tfrac{3-\sqrt5}{2}.$ So the new area is $(\sqrt5+1)\cdot\tfrac{3-\sqrt5}{2} = \sqrt{5} - 1.$ Thus, B is the correct answer.