2025 AMC 8 Problem 18

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To help you master 2025 AMC 8 Problem 18, 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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2025 AMC8 Problem 18

Solution

Solution 1

Direct Area Calculation

We will calculate the shaded area for both the left and right figures using the area formulas for a circle and a square. Then, we will set these two expressions equal to each other to solve for the unknown radius $R$ .

The area of a circle is $A = π r^{2}$ . A square inscribed in a circle can be divided into four identical right triangles by drawing its diagonals, allowing its area to be calculated exactly using only the circle’s radius.

Step 1: Find the shaded area in the left circle

The left circle has a radius of $1$ . Its total area is found using the circle area formula. The inscribed square’s diagonals meet at the center, dividing the square into four identical right triangles. The base and height of each triangle are both equal to the radius ( $1$ ). We subtract the square’s area from the circle’s area to find the shaded region.

A_{left circle} = π (1)^{2} = π

A_{left square} = 4 \times (\frac{1}{2} \times 1 \times 1) = 2

Shaded_{left} = π - 2

Step 2: Find the shaded area in the right circle

The right circle has an unknown radius of $R$ . Using the exact same logic, we find the area of the right circle and its inscribed square in terms of $R$ . The total gap between the circle and the square is their difference. The problem states that only one-quarter of this total gap is shaded.

A_{right circle} = π R^{2}

A_{right square} = 4 \times (\frac{1}{2} \times R \times R) = 2 R^{2}

Total gap = π R^{2} - 2 R^{2} = R^{2} (π - 2)

Shaded_{right} = \frac{1}{4} R^{2} (π - 2)

Step 3: Equate the shaded areas and solve for $R$

Set the two shaded area expressions equal based on the problem’s core condition. Since the expression inside the parentheses, $π - 2$ , represents a positive area, we can safely divide both sides of the equation by it. This leaves a simple equation for $R^{2}$ , which we solve to find the positive radius.

π - 2 = \frac{1}{4} R^{2} (π - 2)

1 = \frac{1}{4} R^{2}

R^{2} = 4

R = 2

Correct answer: B. $2$
The radius of the right circle is

2

, which corresponds to answer choice B. Final check: If

R = 2

, the right circle has an area of

4 π

and the inscribed square has an area of

8

. The total gap is

4 π - 8 = 4 (π - 2)

. One quarter of this gap is

π - 2

, which perfectly matches the fully shaded gap of the left figure.

Solution 2

Similarity and Area Scaling

Because any circle with an inscribed square has the exact same proportions, the two figures are geometrically similar. We can compare their areas using scale factors without ever calculating $π$ .

For any two similar figures, if their lengths scale by a factor of $k$ , their areas scale by a factor of $k^{2}$ . This means $\frac{Area _{2}}{Area _{1}} = (\frac{Length _{2}}{Length _{1}})^{2}$ .

Step 1: Compare the total gap areas of the two figures

Let’s define the ‘gap’ as the entire region between the circle and the inscribed square. In the left figure, the shaded area is this entire gap. In the right figure, the shaded area is exactly one-quarter of its gap. Because these two shaded pieces are equal in size, the total gap in the right figure must be four times as large as the total gap in the left figure.

Gap_{left} = Shaded_{left}

\frac{1}{4} Gap_{right} = Shaded_{right}

Since Shaded_{left} = Shaded_{right} :

Gap_{left} = \frac{1}{4} Gap_{right}

Gap_{right} = 4 \times Gap_{left}

Step 2: Set up a relationship using the area scaling principle

A circle paired with an inscribed square forms a geometric shape whose proportions never change, regardless of its overall size. Therefore, the left and right configurations are mathematically similar. For similar figures, the ratio of any corresponding areas equals the square of the ratio of their corresponding lengths (like the radius).

\frac{Area _{right}}{Area _{left}} = (\frac{Radius _{right}}{Radius _{left}})^{2}

4 = (\frac{R}{1})^{2}

Step 3: Calculate the new radius

Substitute the known values into the scaling equation. Taking the square root of the area scale factor gives the length scale factor, which provides the final radius directly.

R^{2} = 4

R = 2

Correct answer: B. $2$
The radius is

2

(Answer B). Final check: We can verify this scaling logic conceptually. If scaling a shape’s radius by

2

multiplies its total area by

4

, then its gap area is also multiplied by

4

. Taking one-quarter of that newly expanded gap area brings us exactly back to the size of the original gap area, perfectly satisfying the problem.

Adapted Problems to Practice

1 / 10

What Formulas Do You Need for This Problem?

Area of a Square Inscribed in a Circle:

A = 2 r^{2}

$A$ is the area of the square, and $r$ is the radius of the circumscribed circle. Drawing the two diagonals of an inscribed square splits it into four identical right isosceles triangles. The base and height of each triangle are both exactly the radius of the circle, making the total area $4 \times (\frac{1}{2} \times r \times r) = 2 r^{2}$ .

Scaling Principle for Similar 2D Figures:

\frac{A _{2}}{A _{1}} = k^{2}

$A_{1}$ and $A_{2}$ are the areas of two similar figures, and $k = \frac{L _{2}}{L _{1}}$ is the ratio of their corresponding lengths (the length scale factor). A circle with an inscribed square is a geometric configuration that always has the exact same proportions. The right figure is just a scaled-up version of the left figure, meaning every corresponding area scales proportionally by the square of the ratio of their radii.

Key Takeaway: Remember This Strategy for Next Time

A square inscribed in a circle with a known radius.

Try: Find the area of the square by splitting it into four right triangles using the circle's center, rather than trying to find the square's side length first.

Why: The diagonals of the square act as diameters of the circle. This naturally divides the square into triangles where both the base and height are exactly the circle's radius, making calculation very straightforward.

A problem comparing the areas of figures that have the exact same shape but different sizes.

Try: Use the area scaling principle rather than calculating the messy individual areas for each figure.

Why: For any completely similar figures, if the length is multiplied by a factor of $k$ , the area is always multiplied by $k^{2}$ . This lets you find missing dimensions quickly by comparing proportional parts and bypassing irrational constants like $π$ entirely.

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2025 AMC 8 Problem 18

In This Page:

2025 AMC8 Problem 18