AMC 10 · 2021 · #21

Grade 8 geometry-2d

Archetype Paper Fold-and-Cut Symmetry └ Sub-archetype Fold Maps One Named Point Onto Another

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

📏 Long solution 💡 4 insights 📊 Diagram

Problem

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 AD with C'D =

\frac{1}{3}

(so AC' =

\frac{2}{3}

). The folded edge BC crosses edge AB at point E. Find the perimeter of right triangle △ AEC'.

Pick an answer.

(A)

(B)

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

(C)

$~\frac{13}{6}$

(D)

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

(E)

$~\frac{7}{3}$

AMC 10 2021 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

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 △ 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.

1STEP 1

Put A = (0, 1), B = (0, 0), C = (1, 0), D = (1, 1); then C' = ( $\frac{2}{3}$ , 1) sits on AD, and E = (0, h) on AB gives AE = 1 - h.

A=(0,1), B=(0,0), C=(1,0), D=(1,1), C'=(

\frac{2}{3}

,1), E=(0,h)

2STEP 2

Folding is a reflection across the crease, and reflection preserves distance — the visible folded edge is a congruent copy of BC.

fold : C ↦ C', distances preserved → EC' = EC

3STEP 3

Test the shortcut EC = EC': squaring gives 1 + h² = $\frac{4}{9}$ + (1 - h)², so h = $\frac{2}{9}$ — too close to B, so reflect the corner properly instead.

1 + h² =

\frac{4}{9}

+ (1-h)² ⟹ 2h =

\frac{5}{9}

\frac{4}{9}

4STEP 4

The crease is the perpendicular bisector of C C', giving y = $\frac{x}{3}$ + $\frac{2}{9}$ ; reflecting B = (0, 0) across it lands at B' = ( $-\frac{2}{15}$ , $\frac{2}{5}$ ).

crease: y =

\frac{x}{3}

\frac{2}{9}

; B' = (

-\frac{2}{15}

\frac{2}{5}

)

5STEP 5

The image B'C' meets edge AB (x = 0) at t = $\frac{1}{6}$ , giving E = (0, $\frac{1}{2}$ ), so AE = $\frac{1}{2}$ .

E = (0,

\frac{1}{2}

), AE =

\frac{1}{2}

6STEP 6

Right angle at A with legs AE = $\frac{1}{2}$ and AC' = $\frac{2}{3}$ ; Pythagoras gives hypotenuse EC' = $\frac{5}{6}$ — a 3-4-5 triangle scaled by $\frac{1}{6}$ .

AE =

\frac{3}{6}

, AC' =

\frac{4}{6}

, EC' =

\frac{5}{6}

7STEP 7

Add the sides: $\frac{3}{6}$ + $\frac{4}{6}$ + $\frac{5}{6}$ = $\frac{12}{6}$ = 2 — the perimeter, matching choice (A).

Perimeter =

\frac{3}{6}

\frac{4}{6}

\frac{5}{6}

= 2 → (A)

Answer

Quick sanity check. The whole square has perimeter 4, and △ AEC' is a small corner of it, so a perimeter of 2 (about half the square's perimeter) is reasonable: two legs are

\frac{1}{2}

and

\frac{2}{3}

(both less than 1), and the hypotenuse

\frac{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.

💡Key takeaway

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 = $\frac{1}{2}$ the triangle is the classic 3-4-5 shape (scaled by $\frac{1}{6}$ ) and the perimeter is just $\frac{3 + 4 + 5}{6}$ = 2.

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