2023 AMC 10 B Problem 25

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To help you master 2023 AMC 10 B Problem 25, this page includes:

The original AMC problem to try, plus the correct answer

Clear, step-by-step solutions using different methods

Key ideas, useful formulas, and takeaways you can apply to similar problems

Related problems from past AMC exams, plus new practice problems that build the same skills

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2023 AMC10 B Problem 25

Solution

Solution 1

Apothem Ratio Method

Folding a vertex to the center creates a crease that acts as a side of the new pentagon. By comparing the apothems of the two pentagons, we can find the ratio of their areas.

The ratio of the areas of two similar polygons is the square of the ratio of their corresponding linear dimensions (such as their apothems).

Step 1: Model the fold lines

Let the original regular pentagon be $P_{1}$ with center $O$ and circumradius $R$ . When a vertex $V$ is folded to the center $O$ , the crease formed is the perpendicular bisector of the segment $O V$ . Therefore, the crease is a line segment at a perpendicular distance of $R /2$ from $O$ .

Step 2: Identify the new pentagon and its apothem

Because $P_{1}$ is regular, the 5 creases form a smaller regular pentagon $P_{2}$ centered at $O$ . The distance from the center $O$ to each side of $P_{2}$ is exactly the distance to the crease. This distance is the apothem of $P_{2}$ .

r_{2} = \frac{R}{2}

Step 3: Determine the apothem of the original pentagon

The apothem $r_{1}$ of the original pentagon $P_{1}$ is the distance from $O$ to its sides. Using right triangle trigonometry in $P_{1}$ , we can express $r_{1}$ in terms of the circumradius $R$ .

r_{1} = R cos 3 6^{\circ}

Step 4: Set up the area ratio

The ratio of the areas of two similar 2D figures is the square of the ratio of their corresponding linear dimensions. We use the ratio of their apothems.

\frac{Area ( P _{2} )}{Area ( P _{1} )} = (\frac{r _{2}}{r _{1}})^{2}

(\frac{R /2}{R cos 3 6 ^{\circ}})^{2} = \frac{1}{4 cos ^{2} 3 6 ^{\circ}}

Step 5: Calculate the exact area ratio

We use the known trigonometric value for $cos 3 6^{\circ}$ to evaluate the ratio. After squaring the value, we substitute it into our area ratio and rationalize the denominator.

cos 3 6^{\circ} = \frac{1 + 5}{4}

cos^{2} 3 6^{\circ} = (\frac{1 + 5}{4})^{2} = \frac{1 + 2 5 + 5}{16} = \frac{3 + 5}{8}

\frac{Area ( P _{2} )}{Area ( P _{1} )} = \frac{1}{4 ( \frac{3 + 5}{8} )} = \frac{2}{3 + 5}

\frac{2 ( 3 - 5 )}{( 3 + 5 ) ( 3 - 5 )} = \frac{2 ( 3 - 5 )}{9 - 5} = \frac{3 - 5}{2}

Step 6: Calculate the final area

Multiply the area of the original pentagon by the area ratio to find the area of the new pentagon.

Area (P_{2}) = (5 + 1) \times \frac{3 - 5}{2}

Area (P_{2}) = \frac{3 5 - 5 + 3 - 5}{2} = \frac{2 5 - 2}{2}

Area (P_{2}) = 5 - 1

Correct answer: B. $5 - 1$
The area of the new pentagon is

5

-1, which corresponds to answer choice B. We can check the magnitude: the original area is roughly 2.236 + 1 = 3.236, and the new area is roughly 2.236 – 1 = 1.236. The ratio is about 0.382, which matches the theoretical ratio of 1/\phi^2 \approx 1/2.618 \approx 0.382.

Solution 2

Scale Factor and the Golden Ratio

The ratio of the apothems simplifies beautifully when expressed in terms of the golden ratio $ϕ$ . The original area is also closely related to $ϕ$ , allowing us to bypass messy radical arithmetic.

The golden ratio $ϕ = \frac{1 + 5}{2}$ satisfies the identity $ϕ^{2} - ϕ - 1 = 0$ , which rearranges to $\frac{1}{ϕ} = ϕ - 1$ .

Step 1: Introduce the golden ratio

The golden ratio $ϕ$ appears frequently in regular pentagons. Notice that the given area of the original pentagon can be expressed cleanly in terms of $ϕ$ .

ϕ = \frac{1 + 5}{2}

Area (P_{1}) = 5 + 1 = 2 (\frac{1 + 5}{2}) = 2 ϕ

Step 2: Express the scale factor using the golden ratio

From the previous method, the ratio of the apothems is the linear scale factor. A fundamental property of the regular pentagon is its connection to the golden ratio. Substituting this into the ratio shows that the new pentagon’s linear dimensions are scaled exactly by the reciprocal of the golden ratio.

\frac{r _{2}}{r _{1}} = \frac{1}{2 cos 3 6 ^{\circ}}

cos 3 6^{\circ} = \frac{ϕ}{2}

\frac{r _{2}}{r _{1}} = \frac{1}{2 ( ϕ /2 )} = \frac{1}{ϕ}

Step 3: Calculate the new area

The area scales by the square of the linear scale factor. We multiply the original area by this squared factor and simplify using the golden ratio identity $\frac{1}{ϕ} = ϕ - 1$ .

Area ratio = (\frac{1}{ϕ})^{2} = \frac{1}{ϕ ^{2}}

Area (P_{2}) = 2 ϕ \times \frac{1}{ϕ ^{2}} = \frac{2}{ϕ}

Area (P_{2}) = 2 (ϕ - 1)

Area (P_{2}) = 2 (\frac{1 + 5}{2} - 1) = 2 (\frac{5 - 1}{2}) = 5 - 1

Correct answer: B. $5 - 1$
The area is

5

-1, matching choice B. This method confirms the previous result while demonstrating how structural properties of the pentagon can drastically reduce computation time.

Adapted Problems to Practice

1 / 10

What Formulas Do You Need for This Problem?

Paper Folding to a Point:

When a point $A$ on a piece of paper is folded to meet a point $B$ , the resulting crease is exactly the perpendicular bisector of the line segment $A B$ .

Applies to any flat fold mapping one specific point to another. The problem states the vertices are folded to the center, meaning the creases must be the perpendicular bisectors of the segments connecting the vertices to the center.

Area of Similar Figures:

If two polygons are similar with a linear scale factor

k

, the ratio of their areas is

k^{2}

.

$k$ is the ratio of any corresponding linear dimensions (side lengths, circumradii, apothems, etc.). Both the original and the new shapes are regular pentagons, so they are similar. We found the ratio of their apothems to determine $k$ .

Golden Ratio Identity:

\frac{1}{ϕ} = ϕ - 1

Derived from the defining quadratic equation of the golden ratio: $ϕ^{2} - ϕ - 1 = 0$ . It allows us to convert division by $ϕ$ into simple subtraction, avoiding the need to rationalize denominators containing radicals.

Key Takeaway: Remember This Strategy for Next Time

A problem involves folding a specific point on a shape to another specific point.

Try: Immediately model the crease as the perpendicular bisector of the segment connecting the two points.

Why: Folding is a physical reflection across the crease line. For a point to reflect exactly onto another, the mirror line (crease) must be equidistant from both points and perpendicular to the line connecting them.

A problem asks for the area of a new regular polygon created inside an original regular polygon.

Try: Instead of calculating the absolute area of the new shape from scratch, find the linear scale factor between the two shapes (using side lengths, radii, or apothems). The area ratio is simply the square of this scale factor.

Why: All regular polygons with the same number of sides are similar. Similarity guarantees that area scales by the square of any linear dimension.

A problem involves a regular pentagon and expressions containing $5$ .

Try: Try expressing lengths and areas in terms of the golden ratio $ϕ = \frac{1 + 5}{2}$ . Memorize that $cos 3 6^{\circ} = \frac{ϕ}{2}$ .

Why: The geometry of the regular pentagon is intrinsically linked to $ϕ$ . Using $ϕ$ algebraically simplifies radical arithmetic by taking advantage of the identity $ϕ^{2} = ϕ + 1$ .

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2023 AMC 10 B Problem 25

In This Page:

2023 AMC10 B Problem 25