2014 AMC 10A Problem 23

After folding, some paper lies in two layers. The new shape's area is the old area minus the part that is now covered twice. So the job is to find that doubled part.

The ratio does not depend on the size of the paper, so pick easy numbers: width $1$ and length $\sqrt3$. Then the area is $A = \sqrt3$.

1. Label the paper. Call the corners $K$ (bottom-left), $L$ (bottom-right), $M$ (top-right) and $N$ (top-left). Put $K$ at $(0,0)$, so $L = (\sqrt3, 0)$, $M = (\sqrt3, 1)$, $N = (0, 1)$. Each section is $\frac{\sqrt3}{3}$ wide.
2. The fold line. It goes from $P = \left(\frac{\sqrt3}{3}, 0\right)$ on the bottom (the first divider) to $Q = \left(\frac{2\sqrt3}{3}, 1\right)$ on the top (the second divider).
3. Some lengths. All of these come out the same, $\frac{2\sqrt3}{3}$:
- $PQ = \sqrt{\left(\tfrac{\sqrt3}{3}\right)^2 + 1^2} = \sqrt{\tfrac13 + 1} = \sqrt{\tfrac43} = \frac{2\sqrt3}{3}$.
- $PN = \sqrt{\left(\tfrac{\sqrt3}{3}\right)^2 + 1^2}$, the same numbers, so also $\frac{2\sqrt3}{3}$.
- $NQ = \frac{2\sqrt3}{3}$, read straight off the top edge.
- $PL = \sqrt3 - \frac{\sqrt3}{3} = \frac{2\sqrt3}{3}$.
4. Triangle $NPQ$ is equilateral, since its three sides are equal. So $\angle NPQ = 60^\circ$.
5. Where does corner $L$ land? Folding flips the right part over the line $PQ$, like a mirror. The fold line rises $1$ while moving $\frac{\sqrt3}{3}$ across, which is a $60^\circ$ slope, so $\angle QPL = 60^\circ$ too. $L$ is at distance $\frac{2\sqrt3}{3}$ from $P$, at $60^\circ$ on one side of the fold line. Its mirror image is at the same distance and the same angle on the other side. That is exactly corner $N$. So $L$ lands on $N$.
6. The doubled part. The folded flap now has corners $P$, $N$ (where $L$ went), $Q$, and the new position of $M$, which is above the top edge. So the only part of the flap lying on top of the left part of the paper is triangle $NPQ$.
7. Its area. Base $NQ = \frac{2\sqrt3}{3}$ along the top edge, height $1$ (down to the bottom edge where $P$ is):
$$\frac12 \cdot \frac{2\sqrt3}{3} \cdot 1 = \frac{\sqrt3}{3} = \frac{A}{3}.$$
8. The new area. $B = A - \frac{A}{3} = \frac{2A}{3}$, so $B : A = 2 : 3$.

The answer is $\boxed{\textbf{(C)}\ 2:3}$.

Folding paper flat is a mirror flip across the crease, and it never changes the paper's total area. What the new shape loses is exactly the region where two layers sit on top of each other. Finding where a corner lands is often the fastest way to see that region. The strange-looking length $\sqrt3$ is what makes the crease sit at $60^\circ$ and creates the equilateral triangle; special numbers like this usually point to special triangles.

Add up the visible pieces instead of subtracting. The left part of the paper, $KPQN$, is exactly half the rectangle, because the crease cuts it into two matching halves; its area is $\frac{A}{2}$. The part of the flap sticking out above the paper is triangle $N Q M'$, where $M' = \left(\frac{\sqrt3}{2}, \frac32\right)$ is where $M$ lands. It has base $NQ = \frac{2\sqrt3}{3}$ and height $\frac12$, so its area is $\frac12 \cdot \frac{2\sqrt3}{3} \cdot \frac12 = \frac{\sqrt3}{6} = \frac{A}{6}$. Total: $\frac{A}{2} + \frac{A}{6} = \frac{2A}{3}$.