2023 AMC 10B Problem 25

Both pentagons are regular and share the same center, so one is a shrunken copy of the other. Find the shrink factor by comparing one distance in each, then square it for the area.

Let $O$ be the center and $R$ the distance from $O$ to a corner.

1. Where a crease goes. Folding a corner $V$ onto $O$ makes a crease that is exactly halfway between $V$ and $O$, and at right angles to the line $OV$. So each crease is at distance $\frac{R}{2}$ from $O$.
2. The new pentagon. The five creases are all at distance $\frac{R}{2}$ from $O$ and evenly spread around it, so they are the sides of a regular pentagon whose center-to-side distance is $\frac{R}{2}$.
3. The old pentagon's center-to-side distance. Draw the line from $O$ to the midpoint $M$ of a side, and from $O$ to an end $V$ of that side. The angle at the center between neighboring corners is $360^\circ \div 5 = 72^\circ$, so $\angle MOV = 36^\circ$. In the right triangle $OMV$,
$$OM = R \cos 36^\circ.$$
4. The shrink factor. Compare the two center-to-side distances:
$$\frac{R/2}{R\cos 36^\circ} = \frac{1}{2\cos 36^\circ}.$$
5. A pentagon fact. $2\cos 36^\circ = \frac{1 + \sqrt5}{2}$, the golden ratio. (Reason: in a regular pentagon with side $1$, the diagonal is $\frac{1 + \sqrt5}{2}$. The triangle made by two neighboring sides and one diagonal has a $108^\circ$ angle at the top and $36^\circ$ at each end. Cutting it in half, each half of the diagonal is $1 \cdot \cos 36^\circ$, so the diagonal is $2\cos 36^\circ$.)
So the shrink factor is $\frac{2}{1 + \sqrt5}$.
6. Area factor. Square it:
$$\left(\frac{2}{1 + \sqrt5}\right)^2 = \frac{4}{6 + 2\sqrt5} = \frac{2}{3 + \sqrt5} = \frac{2(3 - \sqrt5)}{9 - 5} = \frac{3 - \sqrt5}{2}.$$
7. New area.
$$(1 + \sqrt5) \cdot \frac{3 - \sqrt5}{2} = \frac{3 - \sqrt5 + 3\sqrt5 - 5}{2} = \frac{2\sqrt5 - 2}{2} = \sqrt5 - 1.$$

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

A fold that takes one point onto another creases along the line exactly halfway between them, at right angles to the segment joining them. So "fold every corner to the center" puts each crease at half the center-to-corner distance. After that, it is a matter of comparing two similar pentagons and squaring the ratio. Pentagon problems often bring in the golden ratio $\frac{1 + \sqrt5}{2}$ through the fact that a diagonal is that many times a side.

Estimate and pick the closest choice. $\cos 36^\circ \approx 0.809$, so the shrink factor is $\frac{1}{1.618} \approx 0.618$ and the area factor is about $0.618^2 \approx 0.382$. The old area is $1 + \sqrt5 \approx 3.236$, so the new area is about $3.236 \times 0.382 \approx 1.236$. The choices are about $1.764$, $1.236$, $1.292$, $1.618$ and $1.412$, and only $\sqrt5 - 1 \approx 1.236$ matches.

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