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

A regular pentagon with area $\sqrt{5}+1$ is printed on paper and cut out. The five vertices of the pentagon are folded into the center of the pentagon, creating a smaller pentagon. What is the area of the new 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.

Try it yourself first — the explanation is most useful after you’ve attempted it.

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Toolkit + CCSS Solution

Understand

Restated: Start with a regular pentagon of area $\sqrt{5} + 1$. 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.

Givens: Original shape: regular pentagon, area $\sqrt{5} + 1$; Each vertex folds to the center; the crease for vertex $V$ is the perpendicular bisector of segment $OV$ (where $O$ is the center); The five creases bound a smaller pentagon, which is also regular and centered at $O$ (by the $5$-fold symmetry of the original); Answer choices: (A) $4 - \sqrt{5}$, (B) $\sqrt{5} - 1$, (C) $8 - 3\sqrt{5}$, (D) $\tfrac{\sqrt{5}+1}{2}$, (E) $\tfrac{2+\sqrt{5}}{3}$

Unknowns: Area of the inner (folded) pentagon

Understand

Plan

Primary tool: #10 Create a Physical Representation

Secondary: #1 Draw a Diagram, #7 Identify Subproblems, #16 Change Focus / Count the Complement

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.

Execute — Answer: B

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

Fold a paper pentagon.
With $O$ at the center, folding vertex $V$ onto $O$ produces a crease that is the perpendicular bisector of segment $OV$.
The five creases (one per vertex) bound the inner pentagon.
By the $72^{\circ}$ rotational symmetry of the original, the inner pentagon is also regular and shares center $O$.

$$\text{Crease for }V = \{\text{perpendicular bisector of }OV\}$$

💡 Grade 8 'congruence via rigid motions' — folding $V$ onto $O$ is a reflection across the crease line; that line must be the perpendicular bisector of $OV$.

#1 Draw a Diagram 7.G.B.5 Step 2

Read off the small pentagon's apothem.
Each crease is the perpendicular bisector of $OV$, so the foot of perpendicular from $O$ to the crease is the midpoint of $OV$, which sits at distance $\tfrac{|OV|}{2} = \tfrac{R}{2}$ from $O$.
That foot is exactly the apothem of the small pentagon.
So $r_{s} = \tfrac{R}{2}$, where $R$ is the circumradius of the big pentagon.

$$r_{s} = \dfrac{R}{2}$$

💡 Grade 7 'use facts about perpendicular and adjacent angles' — the apothem is just the foot of the perpendicular from $O$, which is the midpoint of $OV$.

#7 Identify Subproblems 8.G.B.7 Step 3

Find the big pentagon's apothem.
The pentagon splits into $5$ isoceles triangles from the center, each with apex angle $\tfrac{360^{\circ}}{5} = 72^{\circ}$.
Drop the apothem $r$ from $O$ onto a side; it bisects the apex angle into $36^{\circ}$ and lies along the leg of a right triangle with hypotenuse $R$ (the circumradius).
So $\cos 36^{\circ} = \tfrac{r}{R}$, i.e.
$r = R \cos 36^{\circ}$.

$$r = R \cos 36^{\circ}$$

💡 Grade 8 'Pythagorean / right-triangle ratios' — split the slice in half to expose a right triangle.

#7 Identify Subproblems 7.G.A.1 Step 4

Build the linear ratio of the two pentagons.
Both pentagons are regular and centered at $O$, so they are similar.
Their similarity ratio is the ratio of any matching linear dimension — take apothems.
$\dfrac{r_{s}}{r} = \dfrac{R/2}{R \cos 36^{\circ}} = \dfrac{1}{2 \cos 36^{\circ}}$.

$$\dfrac{r_{s}}{r} = \dfrac{1}{2 \cos 36^{\circ}}$$

💡 Grade 7 'scale drawings of geometric figures' — similar polygons share one scaling factor across every dimension.

#7 Identify Subproblems 7.G.A.1 Step 5

Square the ratio for areas.
Areas of similar figures scale by the square of linear ratio.
So $\dfrac{\text{Area}_{s}}{\text{Area}_{\text{big}}} = \dfrac{1}{(2 \cos 36^{\circ})^{2}} = \dfrac{1}{4 \cos^{2} 36^{\circ}}$.

$$\dfrac{\text{Area}_{s}}{\text{Area}_{\text{big}}} = \dfrac{1}{4 \cos^{2} 36^{\circ}}$$

💡 Grade 7 'similarity' — area ratio is the square of the linear ratio.

#16 Change Focus / Count the Complement 8.NS.A.2 Step 6

Plug in the known value $\cos 36^{\circ} = \tfrac{1 + \sqrt{5}}{4}$ (a famous golden-ratio identity).
Squaring: $\cos^{2} 36^{\circ} = \tfrac{(1 + \sqrt{5})^{2}}{16} = \tfrac{1 + 2\sqrt{5} + 5}{16} = \tfrac{6 + 2\sqrt{5}}{16} = \tfrac{3 + \sqrt{5}}{8}$.
So $4 \cos^{2} 36^{\circ} = \tfrac{3 + \sqrt{5}}{2}$, and the area ratio is $\dfrac{2}{3 + \sqrt{5}}$.
Rationalize by multiplying by $\tfrac{3 - \sqrt{5}}{3 - \sqrt{5}}$: $\dfrac{2(3 - \sqrt{5})}{9 - 5} = \dfrac{2(3 - \sqrt{5})}{4} = \dfrac{3 - \sqrt{5}}{2}$.

$$\dfrac{\text{Area}_{s}}{\text{Area}_{\text{big}}} = \dfrac{3 - \sqrt{5}}{2}$$

💡 Grade 8 'rational approximations of irrationals' — $\cos 36^{\circ}$ has a clean radical form thanks to the golden-ratio link.

#16 Change Focus / Count the Complement 8.EE.A.2 Step 7

Multiply by the given big-pentagon area to finish.
$\text{Area}_{s} = (\sqrt{5} + 1) \cdot \dfrac{3 - \sqrt{5}}{2} = \dfrac{(\sqrt{5} + 1)(3 - \sqrt{5})}{2}$.
Expand the numerator: $(\sqrt{5})(3) - (\sqrt{5})(\sqrt{5}) + (1)(3) - (1)(\sqrt{5}) = 3\sqrt{5} - 5 + 3 - \sqrt{5} = 2\sqrt{5} - 2$.
So $\text{Area}_{s} = \tfrac{2\sqrt{5} - 2}{2} = \sqrt{5} - 1$, choice (B).

$$\text{Area}_{s} = (\sqrt{5} + 1) \cdot \tfrac{3 - \sqrt{5}}{2} = \sqrt{5} - 1 \;\Rightarrow\; \textbf{(B)}$$

💡 Grade 8 'square root symbols' — expand and collect the $\sqrt{5}$ terms; the $5$ from $\sqrt{5} \cdot \sqrt{5}$ cancels with the $+3$.

[1] #10 8.G.A.2 Fold a paper pentagon. With $O$ at the center, folding vertex $V$ onto $O$ produ

[2] #1 7.G.B.5 Read off the small pentagon's apothem. Each crease is the perpendicular bisector

[3] #7 8.G.B.7 Find the big pentagon's apothem. The pentagon splits into $5$ isoceles triangles

[4] #7 7.G.A.1 Build the linear ratio of the two pentagons. Both pentagons are regular and cent

[5] #7 7.G.A.1 Square the ratio for areas. Areas of similar figures scale by the square of line

[6] #16 8.NS.A.2 Plug in the known value $\cos 36^{\circ} = \tfrac{1 + \sqrt{5}}{4}$ (a famous go

[7] #16 8.EE.A.2 Multiply by the given big-pentagon area to finish. $\text{Area}_{s} = (\sqrt{5}

Review

Reasonableness: Three checks. (1) The conjugate factorization $(\sqrt{5} + 1)(\sqrt{5} - 1) = 5 - 1 = 4$ matches the related identity: $\text{Area}_{\text{big}} \cdot \text{Area}_{s} = (\sqrt{5} + 1)(\sqrt{5} - 1) = 4$, a clean integer — a strong sign the answer is right. (2) Numerically, $\text{Area}_{\text{big}} = \sqrt{5} + 1 \approx 3.236$, $\text{Area}_{s} = \sqrt{5} - 1 \approx 1.236$, and ratio $\approx 0.382 = \tfrac{3 - \sqrt{5}}{2}$ ✓. (3) The new pentagon is smaller than the original ($1.236 < 3.236$), as expected from folding inward. (B) $\sqrt{5} - 1$ is the answer.

Alternative: Tool #5 (Pattern) on the golden ratio: $\cos 36^{\circ} = \tfrac{\phi}{2}$ where $\phi = \tfrac{1 + \sqrt{5}}{2}$ is the golden ratio. The area ratio simplifies to $\tfrac{1}{\phi^{2}}$. Since $\phi^{2} = \phi + 1$, the ratio is $\tfrac{1}{\phi + 1}$, and multiplying by $\text{Area}_{\text{big}} = \sqrt{5} + 1 = 2\phi$ gives $\tfrac{2\phi}{\phi + 1} = \tfrac{2\phi}{\phi^{2}} = \tfrac{2}{\phi} = 2(\phi - 1) = \sqrt{5} - 1$. Same answer via the $\phi^{2} = \phi + 1$ identity.

CCSS standards used (min grade 8)

7.G.A.1 Solve problems involving scale drawings of geometric figures (Using the fact that similar polygons have all corresponding linear dimensions in one common ratio, so area ratio is its square.)
7.G.B.5 Use facts about supplementary, complementary, vertical, and adjacent angles (Identifying the foot of perpendicular from $O$ to a crease as the midpoint of $OV$, the apothem of the small pentagon.)
8.G.A.2 Understand that a two-dimensional figure is congruent to another using transformations (Recognizing the fold as a reflection that swaps $V$ and $O$ — the crease must be the perpendicular bisector of $OV$.)
8.G.B.7 Apply the Pythagorean Theorem to determine unknown side lengths in right triangles (Splitting one of the five central isoceles triangles in half to extract the right triangle with $\cos 36^{\circ} = r/R$.)
8.EE.A.2 Use square root and cube root symbols to represent solutions (Manipulating $\sqrt{5}$ throughout: squaring $(1 + \sqrt{5})$, expanding $(\sqrt{5} + 1)(3 - \sqrt{5})$, rationalizing denominators.)
8.NS.A.2 Use rational approximations of irrational numbers to compare their size (Treating $\cos 36^{\circ}$ as the exact algebraic value $\tfrac{1 + \sqrt{5}}{4}$ rather than a decimal.)

⭐ 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 $R \cos 36^{\circ}$. The area ratio $\tfrac{1}{4\cos^{2} 36^{\circ}}$ simplifies (via $\cos 36^{\circ} = \tfrac{1+\sqrt 5}{4}$) to $\tfrac{3 - \sqrt{5}}{2}$, and $(\sqrt{5}+1) \cdot \tfrac{3 - \sqrt{5}}{2} = \sqrt{5} - 1$, choice (B).