AMC 10 · 2021 · #21
Grade 8 geometry-2dProblem
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 AD at point C', and edge BC intersects edge AB at point E. Suppose that C'D = \frac{1}{3}. What is the perimeter of triangle \bigtriangleup AEC' ?
Pick an answer.
AMC 10 2021 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
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
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) $\tfrac{7}{3}$
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
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 $AE = 1 - h$.
💡 Coordinates turn the picture into a calculation — every length becomes arithmetic.
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).
💡 Folding paper preserves distance — the new edge is just a copy of the old.
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 $B$.
💡 Squaring both sides removes the radical and leaves a linear equation in $h$.
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 $B' = (-\tfrac{2}{15}, \tfrac{2}{5})$.
💡 The crease is the perpendicular bisector of $C C'$ — that pins it down exactly.
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 $AE = 1 - \tfrac{1}{2} = \tfrac{1}{2}$.
💡 Linear walk from $B'$ to $C'$ — find where it crosses the $y$-axis.
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 $\tfrac{5}{6}$.
💡 Once the two legs are clean fractions, the hypotenuse falls out by Pythagoras.
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).
💡 Same denominator means just add numerators: $3 + 4 + 5 = 12$, divided by $6$ is $2$.
6.NS.C.8 Set up coordinates so $A = (0, 1)$, $B = (0, 0)$, $C = (1, 0)$, $D = (1, 1)$. Th 8.G.A.1 Key fact about folding (reflection): the crease passes through points that don't 8.G.B.7 Compute both sides with coordinates. $EC$ is the distance from $E = (0, h)$ to $ 8.G.A.3 More careful: the fold identity $EC' = EC$ holds only when $E$ is on the crease. 8.G.A.3 The image of $\overline{BC}$ is the segment from $B' = (-\tfrac{2}{15}, \tfrac{2 8.G.B.7 Apply the Pythagorean theorem in $\triangle AEC'$, which has a right angle at $A 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.1Add 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.8Solve 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.1Verify experimentally the properties of rotations, reflections, and translations (Using that a fold is a reflection and reflections preserve distances.)8.G.A.3Describe 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.7Apply 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 $\tfrac{3 + 4 + 5}{6} = 2$.
⭐ 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 $\tfrac{3 + 4 + 5}{6} = 2$.
More like this
Same archetype — closest grade level first.
