2021 AMC 10 B Problem 21

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To help you master 2021 AMC 10 B Problem 21, this page includes:

• The original AMC problem to try, plus the correct answer

• Clear, step-by-step solutions using different methods

• Key ideas, useful formulas, and takeaways you can apply to similar problems

• Related problems from past AMC exams, plus new practice problems that build the same skills

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2021 AMC10 B Problem 21

Solution

Solution 1

Geometric Similarity Method

The fold preserves the $90^{\circ }$ angle of corner $C$. This right angle at $C^{\prime }$ forces the two right triangles on edge $AD$ ($△AE{C}^{\prime }$ and $△C^{\prime }DP$) to be similar.

Step 1: Define the fold and use the Pythagorean theorem

Let the fold line intersect edge $CD$ at point $P$. Because folding reflects the paper across the fold line, the segment $PC$ folds exactly onto $P{C}^{\prime }$. Therefore, $PC=P{C}^{\prime }$. Let $PD=y$. Since the side of the square is $1$, we have $PC=1-y$, which means $P{C}^{\prime }=1-y$. Apply the Pythagorean theorem to $△C^{\prime }DP$.

$$PC=P{C}^{\prime }$$

$$P{D}^2+C^{\prime }{D}^2=P{C}^{\prime 2}$$

$$y^2+{\left(\frac{1}{3}\right)}^2=(1-y{)}^2$$

Step 2: Solve for the sides of $△C^{\prime }DP$

Expand the equation to find $y$.

$$y^2+\frac{1}{9}=1-2y+y^2$$

$$2y=1-\frac{1}{9}=\frac{8}{9}$$

$$y=\frac{4}{9}$$

Step 3: Calculate the perimeter of $△C^{\prime }DP$

Now we know $PD=\frac{4}{9}$ and $P{C}^{\prime }=1-\frac{4}{9}=\frac{5}{9}$. We can easily find the perimeter of $△C^{\prime }DP$.

$$\text{Perimeter}=C^{\prime }D+PD+P{C}^{\prime }=\frac{1}{3}+\frac{4}{9}+\frac{5}{9}=\frac{12}{9}=\frac{4}{3}$$

Step 4: Prove $△AE{C}^{\prime }\sim △C^{\prime }DP$

The points $A$, $C^{\prime }$, and $D$ lie on a straight line, so the angles around $C^{\prime }$ sum to $180^{\circ }$. The fold preserves the $90^{\circ }$ angle from corner $C$, meaning $\angle E{C}^{\prime }P=90^{\circ }$. This leaves $90^{\circ }$ for the sum of the other two angles on the line. In the right triangle $△PD{C}^{\prime }$, the acute angles are also complementary. Therefore, $\angle A{C}^{\prime }E=\angle DP{C}^{\prime }$. Since both $△AE{C}^{\prime }$ and $△C^{\prime }DP$ are right triangles, they are similar by AA similarity.

$$\angle A{C}^{\prime }E+\angle E{C}^{\prime }P+\angle P{C}^{\prime }D=180^{\circ }$$

$$\angle A{C}^{\prime }E+90^{\circ }+\angle P{C}^{\prime }D=180^{\circ }$$

$$\angle A{C}^{\prime }E=90^{\circ }-\angle P{C}^{\prime }D$$

$$\angle DP{C}^{\prime }=90^{\circ }-\angle P{C}^{\prime }D$$

Step 5: Use the similarity ratio to find the perimeter

The ratio of similarity between $△AE{C}^{\prime }$ and $△C^{\prime }DP$ can be found by comparing the corresponding legs $A{C}^{\prime }$ and $PD$. The length of $A{C}^{\prime }$ is $1-C^{\prime }D=1-\frac{1}{3}=\frac{2}{3}$. Since the triangles are similar, the ratio of their perimeters is the same as the ratio of their sides.

$$\text{Ratio}=\frac{A{C}^{\prime }}{PD}=\frac{2/3}{4/9}=\frac{6}{4}=\frac{3}{2}$$

$$\text{Perimeter of}△AE{C}^{\prime }=\frac{3}{2}\times \left(\text{Perimeter of}△C^{\prime }DP\right)$$

$$\text{Perimeter}=\frac{3}{2}\times \frac{4}{3}=2$$

Correct answer: A. $2$
The exact answer is $2$, which corresponds to choice A. A meaningful check is to calculate the individual sides of $△AE{C}^{\prime }$. Using the ratio $\frac{3}{2}$, we have $AE=\frac{3}{2}\times C^{\prime }D=\frac{3}{2}\times \frac{1}{3}=\frac{1}{2}$ and $E{C}^{\prime }=\frac{3}{2}\times P{C}^{\prime }=\frac{3}{2}\times \frac{5}{9}=\frac{5}{6}$. Checking the Pythagorean theorem: ${\left(\frac{1}{2}\right)}^2+{\left(\frac{2}{3}\right)}^2=\frac{1}{4}+\frac{4}{9}=\frac{25}{36}={\left(\frac{5}{6}\right)}^2$. The lengths are consistent.

Solution 2

Coordinate Geometry and Angle Double Formula

Place the square on a coordinate plane. The folded edge is the reflection of the horizontal edge $BC$ across the fold line. We can find its slope using the double-angle tangent formula, completely bypassing the need to find the coordinates of the reflected vertex $B^{\prime }$.

Step 1: Assign coordinates to the square

Place the square on a coordinate plane with $B(0,0)$, $C(1,0)$, $D(1,1)$, and $A(0,1)$. The folded point $C^{\prime }$ lies on $AD$ (the line $y=1$). Since $C^{\prime }D=1/3$ and $D$ is at $x=1$, $C^{\prime }$ is at $x=1-1/3=2/3$.

$$C^{\prime }=\left(\frac{2}{3},1\right)$$

Step 2: Find the slope of the fold line

The fold line is the perpendicular bisector of the segment connecting $C(1,0)$ and $C^{\prime }(2/3,1)$. First, find the slope of $C{C}^{\prime }$. The fold line is perpendicular to $C{C}^{\prime }$, so its slope is the negative reciprocal.

$$m_{C{C}^{\prime }}=\frac{1-0}{2/3-1}=\frac{1}{-1/3}=-3$$

$$m_{\text{fold}}=\frac{1}{3}$$

Step 3: Find the slope of the folded edge using the double-angle formula

Let $\alpha$ be the angle the fold line makes with the positive x-axis, so $\tan \alpha =1/3$. The original edge $BC$ lies exactly on the x-axis (angle $0^{\circ }$). When $BC$ is reflected across the fold line, the angle of reflection doubles the distance from the horizontal. The new edge $B^{\prime }{C}^{\prime }$ has an angle of $2\alpha$. We find its slope $\tan (2\alpha )$ using the double-angle tangent formula.

$$\tan (2\alpha )=\frac{2\tan \alpha }{1-{\tan }^2\alpha }$$

$$\tan (2\alpha )=\frac{2(1/3)}{1-(1/3{)}^2}=\frac{2/3}{8/9}=\frac{3}{4}$$

Step 4: Find the equation of the folded edge and point $E$

The folded edge lies on a line passing through $C^{\prime }(2/3,1)$ with a slope of $3/4$. Point $E$ is the intersection of this edge with $AB$, which lies on the y-axis ($x=0$). Substitute $x=0$ to find the y-coordinate of $E$.

$$y-1=\frac{3}{4}\left(x-\frac{2}{3}\right)$$

$$y-1=\frac{3}{4}\left(0-\frac{2}{3}\right)=-\frac{1}{2}$$

$$y=\frac{1}{2}$$

Step 5: Calculate the perimeter of $△AE{C}^{\prime }$

The vertices of $△AE{C}^{\prime }$ are $A(0,1)$, $E(0,1/2)$, and $C^{\prime }(2/3,1)$. This is a right triangle at $A$. Find the lengths of the legs and the hypotenuse, and sum them to find the perimeter.

$$AE=1-\frac{1}{2}=\frac{1}{2}$$

$$A{C}^{\prime }=\frac{2}{3}$$

$$\begin{array}{rl}& E{C}^{\prime }=\sqrt{{\left(\frac{1}{2}\right)}^2+{\left(\frac{2}{3}\right)}^2}\\ & =\sqrt{\frac{1}{4}+\frac{4}{9}}=\sqrt{\frac{25}{36}}=\frac{5}{6}\end{array}$$

$$\text{Perimeter}=\frac{1}{2}+\frac{2}{3}+\frac{5}{6}=\frac{3}{6}+\frac{4}{6}+\frac{5}{6}=\frac{12}{6}=2$$

Correct answer: A. $2$
The exact answer is $2$, matching choice A. A meaningful check is verifying the position of $E$. Since $y=1/2$ is strictly between $0$ and $1$, point $E$ correctly falls on the segment $AB$, satisfying the physical folding conditions of the problem.

Similar Problems in Real Exams

The following five AMC 10 B problems are closely related to AMC 10 B 2021 Question 21. These real AMC 10 B problems test the same key concept—Geometry. Click any problem to learn and practice, then continue to the adapted practice section later in this article for more problems to try.

2023 AMC 10 B Problem 25

2025 AMC 10 B Problem 5

2025 AMC 10 B Problem 6

2025 AMC 10 B Problem 7

2025 AMC 10 B Problem 12

Adapted Problems to Practice

What Formulas Do You Need for This Problem?

• Complementary Angles on a Line:

Applies when three adjacent angles form a straight line ($180^{\circ }$) and the middle angle is a right angle. The $90^{\circ }$ angle at $C^{\prime }$ rests on the straight line $AD$, forcing $\angle A{C}^{\prime }E$ and $\angle P{C}^{\prime }D$ to be complementary, which proves the adjacent right triangles are similar.

• Double-Angle Tangent Formula:

$\alpha$ is the angle of a line. Used to efficiently find the slope of a line after it is reflected across a mirror line with angle $\alpha$, assuming the original line was horizontal.

• Paper Folding Angle Invariance:

Applies to any rigid fold mapping a 2D shape onto itself. Because the corner $C$ of the square was $90^{\circ }$, the angle at the folded corner $C^{\prime }$ is exactly $90^{\circ }$.

Key Takeaway: Remember This Strategy for Next Time

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