AMC 10 · 2021 · #21

Grade 8 geometry-2d

paper-foldingsimilar-trianglesinteger-pythagorean-triplespythagorean-theorem physical-representationidentify-subproblemsconvert-to-algebra ↑ Prerequisites: similar-trianglespythagorean-theorem

📏 Long solution 💡 4 insights 📊 Diagram

Problem

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 $\overline{BC}$ intersects edge $\overline{AB}$ at point $E$ . Suppose that $C'D = \frac{1}{3}$ . What is the perimeter of triangle $\bigtriangleup AEC' ?$

Pick an answer.

(A)

(B)

$~1+\frac{2}{3}\sqrt{3}$

(C)

$~\frac{13}{6}$

(D)

$~1 + \frac{3}{4}\sqrt{3}$

(E)

$~\frac{7}{3}$

View mode:

Toolkit + CCSS Solution

Understand

Restated: A unit square has corners $A$ (top-left), $B$ (bottom-left), $C$ (bottom-right), $D$ (top-right). The paper is folded so that $C$ lands on a point $C'$ on edge $\overline{AD}$ with $C'D = \tfrac{1}{3}$ (so $AC' = \tfrac{2}{3}$). The folded edge $\overline{BC}$ crosses edge $\overline{AB}$ at point $E$. Find the perimeter of right triangle $\triangle AEC'$.

Givens: Square side length $1$, corners $A, B, C, D$ in order; Fold sends $C$ to $C'$ on $\overline{AD}$ with $C'D = \tfrac{1}{3}$, so $AC' = \tfrac{2}{3}$; $E$ is on $\overline{AB}$, formed by the folded image of edge $\overline{BC}$; Answer choices: (A) $2$, (B) $1 + \tfrac{2}{3}\sqrt{3}$, (C) $\tfrac{13}{6}$, (D) $1 + \tfrac{3}{4}\sqrt{3}$, (E) $\tfrac{7}{3}$

Unknowns: Perimeter $AE + AC' + EC'$ of $\triangle AEC'$

Understand

Plan

Primary tool: #1 Draw a Diagram

Secondary: #10 Create a Physical Representation, #7 Identify Subproblems, #13 Convert to Algebra, #3 Eliminate Possibilities

Tool #10 (Physical) — fold an actual square of paper to feel the symmetry. Tool #1 (Diagram) — place the square on a coordinate grid so each named segment has clear length. Tool #7 (Subproblems) — split into (a) find $AE$, then (b) use the Pythagorean theorem in right $\triangle AEC'$. Tool #13 (Algebra) — call $AE = x$ and write the fold-symmetry equation $EC = EC'$. Tool #3 (Eliminate) — the perimeter must be one of five clean values, so after we compute it, just match.

Execute — Answer: A

#1 Draw a Diagram 6.NS.C.8 Step 1

Set up coordinates so $A = (0, 1)$, $B = (0, 0)$, $C = (1, 0)$, $D = (1, 1)$.
Then $C' = (\tfrac{2}{3}, 1)$ because $C'$ is on $\overline{AD}$ with $C'D = \tfrac{1}{3}$.
Place $E$ on $\overline{AB}$ as $E = (0, h)$ with $0 \le h \le 1$.
Then $AE = 1 - h$.

$$A=(0,1),\; B=(0,0),\; C=(1,0),\; D=(1,1),\; C'=(\tfrac{2}{3},1),\; E=(0,h)$$

💡 Coordinates turn the picture into a calculation — every length becomes arithmetic.

#10 Create a Physical Representation 8.G.A.1 Step 2

Key fact about folding (reflection): the crease passes through points that don't move, and reflection preserves distances.
The fold sends $C$ to $C'$.
The folded edge $\overline{BC}$ is the image of the original edge $\overline{BC}$, and $E$ lies on this image.
So $E$ corresponds (under the fold) to some pre-image point $P$ on the original $\overline{BC}$.
Reflection preserves distance, so $|EC'| = |PC|$ where $|PC|$ is the distance from $P$ to $C$.
Since $P$ is on segment $BC$ (the bottom edge) and the segment $EP$ is bisected perpendicularly by the crease, we get the cleaner symmetry $|EC'| = |EC|$ (a standard fold identity used for problems like this).

$$\text{fold} : C \mapsto C',\; \text{distances preserved} \Rightarrow EC' = EC$$

💡 Folding paper preserves distance — the new edge is just a copy of the old.

#13 Convert to Algebra 8.G.B.7 Step 3

Compute both sides with coordinates.
$EC$ is the distance from $E = (0, h)$ to $C = (1, 0)$: $EC = \sqrt{1 + h^2}$.
$EC'$ is the distance from $E = (0, h)$ to $C' = (\tfrac{2}{3}, 1)$: $EC' = \sqrt{\tfrac{4}{9} + (1 - h)^2}$.
Setting $EC^2 = (EC')^2$ gives $1 + h^2 = \tfrac{4}{9} + (1 - h)^2 = \tfrac{4}{9} + 1 - 2h + h^2$.
The $1 + h^2$ terms cancel, leaving $0 = \tfrac{4}{9} - 2h$, so $h = \tfrac{2}{9}$.
Wait — that puts $E$ very close to $B$.
Let us re-check using the alternate identity $EB' = EB$ from the fold of the corner near $B$.

$$1 + h^2 = \tfrac{4}{9} + (1-h)^2 \;\Longrightarrow\; 2h = \tfrac{5}{9} - \tfrac{4}{9}?$$

💡 Squaring both sides removes the radical and leaves a linear equation in $h$.

#1 Draw a Diagram 8.G.A.3 Step 4

More careful: the fold identity $EC' = EC$ holds only when $E$ is on the crease.
Here $E$ is on $\overline{AB}$, not on the crease.
The correct identity comes from following the fold of the *edge*: under reflection, the image of $B$ is some point $B'$, and the image of segment $BC$ is the segment $B'C'$.
$E$ lies on this image segment $\overline{B'C'}$.
Using direct reflection across the crease (perpendicular bisector of $CC'$): the crease has midpoint $M = (\tfrac{5}{6}, \tfrac{1}{2})$ and slope $\tfrac{1}{3}$ (perpendicular to $CC'$, which has slope $-3$).
Equation of crease: $y - \tfrac{1}{2} = \tfrac{1}{3}(x - \tfrac{5}{6})$, i.e., $y = \tfrac{x}{3} + \tfrac{2}{9}$.
Reflecting $B = (0,0)$ across this line gives $B' = (-\tfrac{2}{15}, \tfrac{2}{5})$.

$$\text{crease}: y = \tfrac{x}{3} + \tfrac{2}{9};\quad B' = (-\tfrac{2}{15}, \tfrac{2}{5})$$

💡 The crease is the perpendicular bisector of $C C'$ — that pins it down exactly.

#7 Identify Subproblems 8.G.A.3 Step 5

The image of $\overline{BC}$ is the segment from $B' = (-\tfrac{2}{15}, \tfrac{2}{5})$ to $C' = (\tfrac{2}{3}, 1)$.
Parametrize: $(x, y) = B' + t (C' - B') = (-\tfrac{2}{15} + \tfrac{4t}{5},\; \tfrac{2}{5} + \tfrac{3t}{5})$ for $t \in [0, 1]$.
We want the intersection with line $x = 0$: $-\tfrac{2}{15} + \tfrac{4t}{5} = 0 \Rightarrow t = \tfrac{1}{6}$.
Then $y = \tfrac{2}{5} + \tfrac{3}{5} \cdot \tfrac{1}{6} = \tfrac{2}{5} + \tfrac{1}{10} = \tfrac{1}{2}$.
So $E = (0, \tfrac{1}{2})$ and $AE = 1 - \tfrac{1}{2} = \tfrac{1}{2}$.

$$E = (0, \tfrac{1}{2}),\quad AE = \tfrac{1}{2}$$

💡 Linear walk from $B'$ to $C'$ — find where it crosses the $y$-axis.

#7 Identify Subproblems 8.G.B.7 Step 6

Apply the Pythagorean theorem in $\triangle AEC'$, which has a right angle at $A$ because $\overline{AB} \perp \overline{AD}$.
The legs are $AE = \tfrac{1}{2} = \tfrac{3}{6}$ and $AC' = \tfrac{2}{3} = \tfrac{4}{6}$.
The hypotenuse is $EC' = \sqrt{(\tfrac{1}{2})^2 + (\tfrac{2}{3})^2} = \sqrt{\tfrac{1}{4} + \tfrac{4}{9}} = \sqrt{\tfrac{9 + 16}{36}} = \sqrt{\tfrac{25}{36}} = \tfrac{5}{6}$.
So this is a classic $3$-$4$-$5$ right triangle scaled by $\tfrac{1}{6}$: legs $\tfrac{3}{6}, \tfrac{4}{6}$ and hypotenuse $\tfrac{5}{6}$.

$$AE = \tfrac{3}{6},\; AC' = \tfrac{4}{6},\; EC' = \tfrac{5}{6}$$

💡 Once the two legs are clean fractions, the hypotenuse falls out by Pythagoras.

#3 Eliminate Possibilities 5.NF.A.1 Step 7

Add the three sides for the perimeter of $\triangle AEC'$: $\tfrac{3}{6} + \tfrac{4}{6} + \tfrac{5}{6} = \tfrac{12}{6} = 2$.
The perimeter is $2$, matching choice (A).

$$\text{Perimeter} = \tfrac{3}{6} + \tfrac{4}{6} + \tfrac{5}{6} = 2 \Rightarrow \textbf{(A)}$$

💡 Same denominator means just add numerators: $3 + 4 + 5 = 12$, divided by $6$ is $2$.

[1] #1 6.NS.C.8 Set up coordinates so $A = (0, 1)$, $B = (0, 0)$, $C = (1, 0)$, $D = (1, 1)$. Th

[2] #10 8.G.A.1 Key fact about folding (reflection): the crease passes through points that don't

[3] #13 8.G.B.7 Compute both sides with coordinates. $EC$ is the distance from $E = (0, h)$ to $

[4] #1 8.G.A.3 More careful: the fold identity $EC' = EC$ holds only when $E$ is on the crease.

[5] #7 8.G.A.3 The image of $\overline{BC}$ is the segment from $B' = (-\tfrac{2}{15}, \tfrac{2

[6] #7 8.G.B.7 Apply the Pythagorean theorem in $\triangle AEC'$, which has a right angle at $A

[7] #3 5.NF.A.1 Add the three sides for the perimeter of $\triangle AEC'$: $\tfrac{3}{6} + \tfra

Review

Reasonableness: Quick sanity check. The whole square has perimeter $4$, and $\triangle AEC'$ is a small corner of it, so a perimeter of $2$ (about half the square's perimeter) is reasonable: two legs are $\tfrac{1}{2}$ and $\tfrac{2}{3}$ (both less than $1$), and the hypotenuse $\tfrac{5}{6}$ is also less than $1$. The $3$-$4$-$5$ identity is a strong correctness signal — if we made an arithmetic slip on $AE$, the legs would not have been in a $3:4$ ratio with the hypotenuse $5$.

Alternative: Tool #10 (Physical Representation): cut a paper unit square, mark $C' = \tfrac{2}{3}$ along the top edge, fold corner $C$ onto $C'$, and *measure* where the folded edge crosses $\overline{AB}$. You will find it crosses at the midpoint, giving $AE = \tfrac{1}{2}$ directly. Then the $3$-$4$-$5$ shape is obvious and the perimeter $2$ follows by adding.

CCSS standards used (min grade 8)

5.NF.A.1 Add and subtract fractions with unlike denominators (Adding the three sides $\tfrac{1}{2} + \tfrac{2}{3} + \tfrac{5}{6} = 2$ with a common denominator of $6$.)
6.NS.C.8 Solve real-world problems by graphing points in all four quadrants (Placing the square on a coordinate grid and labeling each named point with $(x, y)$ coordinates.)
8.G.A.1 Verify experimentally the properties of rotations, reflections, and translations (Using that a fold is a reflection and reflections preserve distances.)
8.G.A.3 Describe the effect of dilations, translations, rotations, and reflections on coordinates (Reflecting $B = (0,0)$ across the crease line to find the image of the folded edge.)
8.G.B.7 Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Computing $EC' = \sqrt{AE^2 + AC'^2}$ in the right triangle $\triangle AEC'$.)

⭐ This hard AMC 10 problem still leans on a Grade 8 fact you already know — folding paper is a reflection, and a reflection across the crease moves the corner $C$ exactly to $C'$ while keeping all distances; once you find $AE = \tfrac{1}{2}$ the triangle is the classic $3$-$4$-$5$ shape (scaled by $\tfrac{1}{6}$) and the perimeter is just $\tfrac{3 + 4 + 5}{6} = 2$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: A

Review

CCSS standards used (min grade 8)

More like this