AMC 10 · 2023 · #25

Grade 8 geometry-2d

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 area √(5) + 1. Fold each of its 5 vertices onto the center, creating 5 creases that bound a smaller regular pentagon. Find the area of this 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)

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

AMC 10 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

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

Folding vertex V onto center O makes a crease — the perpendicular bisector of OV; the five creases bound a regular inner pentagon at O.

Crease for V = {perpendicular bisector of OV}

2STEP 2

Each crease is the perpendicular bisector of OV, so O's distance to it is |OV|/2 — the small pentagon's apothem is r_s = R/2.

r_s = R/2

3STEP 3

Split the pentagon into 5 isosceles slices (apex 72°); halving one apex gives a right triangle with cos 36° = r/R, so r = R cos 36°.

r = R cos 36^°

4STEP 4

Both pentagons are regular and centered at O, hence similar; their similarity ratio via apothems is r_s/r = 1/(2 cos 36°).

r_s/r = 1/(2 cos 36^°)

5STEP 5

Areas of similar figures scale by the square of the linear ratio, so Area_s/Area_big = 1/(4 cos² 36°).

Area_s/Area_big = 1/(4 cos² 36^°)

6STEP 6

Put in cos 36° = (1+√(5))/4: then 4 cos² 36° = (3+√(5))/2, and rationalizing gives Area_s/Area_big = (3 - √(5))/2.

Area_s/Area_big = (3 - √(5))/2

7STEP 7

Multiply by the given area: (√(5)+1)·(3 - √(5))/2 expands to (2√(5) - 2)/2 = √(5) - 1, choice (B).

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 10 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).