AMC 10 · 2023 · #25

Grade 8 geometry-2d

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

regular-pentagongolden-ratiopaper-foldingsimilar-figuresreflection-symmetry physical-representationidentify-subproblemsreflection-unfolding ↑ Prerequisites: paper-foldingsimilar-figures

📏 Long solution 💡 4 insights

Problem

Start with a regular pentagon of given area. Fold each of its five vertices onto the centre, creating five creases that bound a smaller regular pentagon. Find the area of the inner pentagon.

Pick an answer.

(A)

$4-\sqrt{5}$

(B)

$\sqrt{5}-1$

(C)

$8-3\sqrt{5}$

(D)

$\frac{\sqrt{5}+1}{2}$

(E)

$\frac{2+\sqrt{5}}{3}$

How to solve

Strategy Create a Physical Representation

Tool #10 (Physical) is the unlock — actually fold a paper regular pentagon and see what shape the creases make. The result is a smaller regular pentagon centered at the same point. Tool #1 (Diagram) keeps a labeled sketch with O at the center, vertex V, the segment OV, and the crease as its perpendicular bisector. Tool #7 (Subproblems) splits 'find the new area' into (i) find the linear ratio between the two regular pentagons, then (ii) square that ratio and multiply by the given area. Tool #16 (Change Focus) takes the cleanest linear dimensions — small pentagon's apothem and big pentagon's apothem — instead of trying to compute either area directly.

1STEP 1

What each crease is

Each crease is a perpendicular bisector.

Crease for V = {perpendicular bisector of OV}

2STEP 2

The inner pentagon's apothem

The apothem becomes half the circumradius.

r_s = R/2

3STEP 3

The outer pentagon's apothem

Compute the outer apothem too.

r = R cos 36^°

4STEP 4

The ratio of the two

Take the ratio of the two.

r_s/r = 1/(2 cos 36^°)

5STEP 5

Square it for the area ratio

Squaring gives the area ratio.

Area_s/Area_big = 1/(4 cos² 36^°)

6STEP 6

Simplify the trigonometric value

Simplify the cosine value.

Area_s/Area_big = (3 - √(5))/2

7STEP 7

Compute the area

Multiplying gives root five minus one.

Area_s = (√(5) + 1) · (3 - √(5))/2 = √(5) - 1 → (B)

Answer

√(5)-1

Three checks. (1) The conjugate factorization (√(5) + 1)(√(5) - 1) = 5 - 1 = 4 matches the related identity: Area_big · Area_s = (√(5) + 1)(√(5) - 1) = 4, a clean integer — a strong sign the answer is right. (2) Numerically, Area_big = √(5) + 1 ≈ 3.236, Area_s = √(5) - 1 ≈ 1.236, and ratio ≈ 0.382 = (3 - √(5))/2 ✓. (3) The new pentagon is smaller than the original (1.236 ＜ 3.236), as expected from folding inward. (B) √(5) - 1 is the answer.

💡Key takeaway

This AMC 12 problem only needs Grade 8 square-root algebra you already know — fold every vertex of the pentagon to its center; each crease bisects OV perpendicularly, so the small pentagon's apothem is exactly R/2. The big pentagon's apothem is R cos 36^°. The area ratio 1/(4cos² 36^°) simplifies (via cos 36^° = (1+√5)/4) to (3 - √(5))/2, and (√(5)+1) · (3 - √(5))/2 = √(5) - 1, choice (B).