2021 AMC 10B Problem 21

Folding is like reflecting in a mirror: any piece of the paper keeps its lengths and angles after the fold.

Step 1: the small triangle at $D$.

1. Let $F$ be where the fold line meets side $CD$. The piece $FC$ folds over onto $FC'$, so $FC' = FC$.
2. Let $DF = y$. Side $CD$ has length $1$, so $FC = 1 - y$, and therefore $FC' = 1 - y$ too.
3. Triangle $C'DF$ has a right angle at $D$ (a corner of the square), with legs $C'D = \frac13$ and $DF = y$, and longest side $FC' = 1 - y$. By the Pythagorean theorem,
$$\left(\tfrac13\right)^2 + y^2 = (1 - y)^2 = 1 - 2y + y^2.$$
4. The $y^2$ cancels: $\frac19 = 1 - 2y$, so $2y = \frac89$ and $y = \frac49$. Then $FC' = 1 - \frac49 = \frac59$.
5. So triangle $C'DF$ has sides $\frac13 = \frac39$, $\frac49$, $\frac59$ (a $3$-$4$-$5$ triangle, shrunk). Its perimeter is $\frac{3 + 4 + 5}{9} = \frac{12}{9} = \frac43$.

Step 2: the two triangles have the same shape.

1. The corner $C$ of the square was a right angle, so after folding the angle $EC'F$ is also a right angle.
2. At the point $C'$, three angles sit side by side along the straight edge $AD$: angle $AC'E$, the right angle $EC'F$, and angle $DC'F$. They add to $180^\circ$, so angle $AC'E$ and angle $DC'F$ add to $90^\circ$.
3. In the right triangle $C'DF$, the two sharp angles also add to $90^\circ$: angle $DC'F$ plus angle $DFC'$ is $90^\circ$.
4. Comparing steps 7 and 8, angle $AC'E$ equals angle $DFC'$. Both triangles $AEC'$ and $DC'F$ also have a right angle (at $A$ and at $D$). So they have the same three angles, which means the same shape.

Step 3: scale up.

1. Match the sides. In triangle $AEC'$, the side $AC'$ sits between the right angle and the angle at $C'$. In triangle $DC'F$, the matching side $DF$ sits between the right angle and the angle at $F$.
2. $AC' = 1 - \frac13 = \frac23$ and $DF = \frac49$. So triangle $AEC'$ is $\frac{2/3}{4/9} = \frac{2}{3} \times \frac{9}{4} = \frac32$ times as big.
3. Its perimeter is $\frac32 \times \frac43 = 2$.

The answer is $\boxed{\textbf{(A)}\ 2}$.

A fold keeps lengths ($FC' = FC$) and keeps the right angle at the folded corner. A right angle resting on a straight edge always creates two same-shape right triangles on either side of it. One triangle can be solved with the Pythagorean theorem, and the other follows by scaling. (In fact the answer is always $2$, half the square's perimeter, wherever $C'$ lands on $AD$.)

Use coordinates: $A = (0, 1)$, $B = (0, 0)$, $C = (1, 0)$, $D = (1, 1)$, so $C' = (\frac23, 1)$.

1. The fold line is the perpendicular bisector of $CC'$. The midpoint of $CC'$ is $(\frac56, \frac12)$, and $CC'$ has slope $\frac{1 - 0}{2/3 - 1} = -3$, so the fold line has slope $\frac13$.
2. Reflecting $B$ over this line gives $B' = (-\frac{2}{15}, \frac25)$. The folded edge runs from $C'$ through $B'$, with slope $\frac34$.
3. That edge crosses the left side $x = 0$ at $E = (0, \frac12)$.
4. So $AE = \frac12$, $AC' = \frac23$, and $EC' = \sqrt{\frac14 + \frac49} = \frac56$. The perimeter is $\frac12 + \frac23 + \frac56 = 2$.