2014 AMC 10A Problem 23

The ratio does not depend on scale, so take width $1$ and length $\sqrt3$, giving $A = \sqrt3$. Place the rectangle with $A = (0,0)$, $B = (\sqrt3, 0)$, $C = (\sqrt3, 1)$, $D = (0, 1)$. The crease joins $P = \left(\frac{\sqrt3}{3}, 0\right)$ on the bottom edge to $Q = \left(\frac{2\sqrt3}{3}, 1\right)$ on the top edge.

The crease rises $1$ over a run of $\frac{\sqrt3}{3}$, so it makes a $60^\circ$ angle with the bottom edge, and
$$PQ = \sqrt{\tfrac13 + 1} = \tfrac{2\sqrt3}{3} = PB = DQ = DP.$$
In particular triangle $DPQ$ is equilateral, and $\angle QPB = 60^\circ = \angle QPD$.

Folding reflects the right flap $PBCQ$ across line $PQ$. Because $PB = PD$ and the two segments make equal angles with the crease, $B$ lands exactly on $D$. The flap's other two vertices, $Q$ (fixed) and the image of $C$, lie on or above the top edge, so the part of the folded flap that lies on the left piece is precisely triangle $DPQ$.

The overlap has area
$$[DPQ] = \frac12 \cdot DQ \cdot 1 = \frac12 \cdot \frac{2\sqrt3}{3} = \frac{\sqrt3}{3} = \frac{A}{3}.$$
Folding does not change the flap's area, so the new shape has area $B = A - \frac{A}{3} = \frac{2A}{3}$.

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

A flat fold is a reflection across the crease, and the visible area afterwards is the original area minus the region covered twice. Locating where a corner lands (here $B \mapsto D$) identifies the overlap without any messy intersection computation. The special ratio $\sqrt3$ is what makes the crease sit at $60^\circ$ and produces the equilateral triangle; expect such "nice" givens to create special triangles.

Trace the flap directly: the reflection of $C$ across $PQ$ is $C' = \left(\frac{\sqrt3}{2}, \frac32\right)$, so the folded flap is quadrilateral $PDC'Q$, which lies above the line $y = 1$ except for triangle $DPQ$. The final shape is the left piece $APQD$ (area $\frac{A}{2}$) plus triangle $DQC'$ (area $\frac{A}{6}$), total $\frac{2A}{3}$.

• Assuming the folded flap covers exactly half of the paper (ratio 1:2), or computing the overlap as the whole flap.
• Mixing up which third-marks the crease connects (first divider on one side to second on the other), which changes the crease's angle and the overlap.
• Measuring triangle $DPQ$ with base $DQ = \frac{2\sqrt3}{3}$ but the wrong height, or using side $\frac{2}{3}$ of the length instead of $\frac{2\sqrt3}{3}$.