2023 AMC 10B Problem 25

Let $O$ be the center and $R$ the circumradius (distance from $O$ to a vertex). Folding vertex $V$ onto $O$ creases the paper along the perpendicular bisector of $OV$: a line perpendicular to $OV$ at distance $\frac{R}{2}$ from $O$. The five creases are therefore tangent to a circle of radius $\frac{R}{2}$ and are spaced $72^\circ$ apart, so they bound a regular pentagon with apothem $\frac{R}{2}$ (rotated $36^\circ$ relative to the original).

The original pentagon has apothem $r = R\cos 36^\circ$, since the segment from $O$ to a side's midpoint makes a $36^\circ$ angle with the segment to an adjacent vertex. The two pentagons are similar with linear ratio
$$\frac{R/2}{R\cos 36^\circ} = \frac{1}{2\cos 36^\circ} = \frac{1}{\varphi}, \qquad \varphi = \frac{1 + \sqrt5}{2},$$
using $\cos 36^\circ = \frac{1 + \sqrt 5}{4}$.

Areas scale by the square: $\frac{1}{\varphi^2} = \frac{1}{\varphi + 1} = \frac{2}{3 + \sqrt5} = \frac{3 - \sqrt5}{2}$. Hence the new area is
$$(1 + \sqrt5)\cdot\frac{3 - \sqrt5}{2} = \frac{3 - \sqrt5 + 3\sqrt5 - 5}{2} = \frac{2\sqrt5 - 2}{2} = \sqrt5 - 1.$$

The answer is $\boxed{\textbf{(B)}\ \sqrt{5}-1}$.

A fold that carries a point $P$ to a point $Q$ is a reflection across the perpendicular bisector of $PQ$, so "fold each vertex to the center" places every crease at half the circumradius. The rest is regular-polygon bookkeeping: compare apothems (or circumradii) of the two similar pentagons and square the ratio. The golden ratio appears because $2\cos 36^\circ = \varphi$, a fact worth memorizing for pentagon problems.

Estimate: $\cos 36^\circ \approx 0.809$, so the area ratio is about $1/(4 \cdot 0.654) \approx 0.382$ and the new area is about $3.236 \times 0.382 \approx 1.236$. Among the choices, $\sqrt5 - 1 \approx 1.236$ matches, while $8 - 3\sqrt5 \approx 1.292$ and the others are farther away.

• Halving the original apothem instead of the circumradius; the crease sits at half the distance from the center to a vertex, not to a side.
• Assuming the folded pentagon is simply the original scaled by $\frac{1}{2}$ (area ratio $\frac{1}{4}$), which is not one of the choices and signals the error.