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2021 AMC 10B Problem 21

2021 AMC 10B Problem 21AMC 10 Step by Step

Name: Date: September 13, 2026

Transformations & Symmetry · difficulty 4 of 5 · about 6 min · position P21-25

A square piece of paper has side length $1$ and vertices $A,B,C,$ and $D$ in that order. As shown in the figure, the paper is folded so that vertex $C$ meets edge $\overline{AD}$ at point $C'$ , and edge $\overline{AB}$ at point $E$ . Suppose that $C'D = \frac{1}{3}$ . What is the perimeter of triangle $\bigtriangleup AEC' ?$

A)
$2$
B)
$1+\frac{2}{3}\sqrt{3}$
C)
$\frac{13}{6}$
D)
$1 + \frac{3}{4}\sqrt{3}$
E)
$\frac{7}{3}$

Answer

Key insight

Fold preserves lengths, so the small right triangle at D has perimeter DC' + DC = 4/3; triangle AEC' is similar to it with ratio AC'/DF = 3/2.

Solution

Let $F$ be the point where the crease meets side $CD$ . Folding is a reflection across the crease, so $C'F = CF$ and the right angle at $C$ is carried to $C'$ : $\angle EC'F = 90^\circ$ .

Triangle $DC'F$ . Let $DF = y$ , so $CF = C'F = 1 - y$ . Right triangle $C'DF$ gives
$\left(\tfrac13\right)^2 + y^2 = (1-y)^2 \quad\Longrightarrow\quad \tfrac19 = 1 - 2y \quad\Longrightarrow\quad y = \tfrac49,$
so $C'F = \frac59$ . Its perimeter is $\frac13 + \frac49 + \frac59 = \frac43$ (equivalently $DC' + DC$ , since $C'F + DF = CF + DF = 1$ ).

Similarity. At $C'$ the angles $\angle AC'E$ , $90^\circ$ , and $\angle DC'F$ lie along line $AD$ , so $\angle AC'E + \angle DC'F = 90^\circ$ . In right triangle $DC'F$ , $\angle DFC' + \angle DC'F = 90^\circ$ as well, hence $\angle AC'E = \angle DFC'$ . With right angles at $A$ and $D$ , triangles $AEC'$ and $DC'F$ are similar, and the ratio is
$\frac{AC'}{DF} = \frac{2/3}{4/9} = \frac32.$

Perimeters scale by the same ratio: perimeter of $AEC'$ is $\frac32 \cdot \frac43 = 2$ .

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

Why this works

Folding problems are reflection problems: matched segments are equal ( $C'F = CF$ ) and the folded corner keeps its right angle. A right angle placed on a straight edge produces two similar right triangles on either side; the small one at $D$ is fully determined by one Pythagorean equation, and similarity transfers its perimeter. In fact the perimeter of the cut-off triangle is always $2$ , half the square's perimeter, wherever $C'$ lands on $AD$ ; choice (A) being the only "clean" constant is a hint.

Alternative approach

Coordinates with $A = (0,1)$ , $B = (0,0)$ , $C = (1,0)$ , $D = (1,1)$ , $C' = (\frac23, 1)$ . The crease is the perpendicular bisector of $CC'$ : through $(\frac56, \frac12)$ with slope $\frac13$ . Reflecting $B$ across it gives $B' = (-\frac{2}{15}, \frac25)$ ; line $C'B'$ has slope $\frac34$ and meets $x = 0$ at $E = (0, \frac12)$ . Then $AE = \frac12$ , $AC' = \frac23$ , $EC' = \frac56$ , total $2$ .

The trap

Assuming AE = EC' or that the crease passes through E; E lies on the folded image of edge CB, not on the crease.

Common mistakes

Assuming AE = EC' or that the crease passes through E; E lies on the folded image of edge CB, not on the crease.
Writing $CF = C'F$ but then setting $DF = \frac13$ or forgetting that $CF + DF = 1$ , which breaks the Pythagorean equation.

Techniques

Set up the equation/formula and compute; no special trick needed · Exploit symmetry to reduce work or pair up objects