2023 AMC 12B Problem 25

Let the original pentagon have circumradius $R.$ Folding a vertex to the center creases along the perpendicular bisector of the segment from the center to that vertex, a line at distance $\tfrac{R}{2}$ from the center. The five creases bound a regular pentagon with apothem $\tfrac{R}{2},$ whereas the original has apothem $R\cos 36^\circ.$ Areas scale as the square of the apothem, so the ratio is $$\frac{(\frac{R}{2})^2}{(R\cos 36^\circ)^2}=\frac{1}{4\cos^2 36^\circ}.$$ Since $\cos 36^\circ=\tfrac{1+\sqrt5}{4},$ this ratio is $\tfrac{4}{(1+\sqrt5)^2}$ $=\tfrac{4}{6+2\sqrt5}$ $=\tfrac{2}{3+\sqrt5}$ $=\tfrac{3-\sqrt5}{2}.$ Multiplying by the original area $\sqrt5+1$ gives $\tfrac{(3-\sqrt5)(\sqrt5+1)}{2}$ $=\tfrac{2\sqrt5-2}{2}$ $=\sqrt5-1.$ Thus, the correct answer is B.