2025 AMC 8 Problem 12

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To help you master 2025 AMC 8 Problem 12, 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 12

Solution

Solution 1

Symmetry and Pythagorean Theorem

By placing the center of the circle at the center of the symmetric shape, we only need to find the shortest distance from the center to the boundary. The boundary points that ‘dent inward’ (reflex angles) are the closest to the center and act as bottlenecks for the circle’s size.

The distance from the origin $(0, 0)$ to a point $(x, y)$ on a coordinate grid is found using the Pythagorean theorem: $d^{2} = x^{2} + y^{2}$ . The area of a circle is $A = π r^{2}$ .

Step 1: Locate the center of the region

The shape is perfectly symmetric horizontally and vertically. The center of the largest inscribed circle will naturally align with the geometric center of the shape. We can treat this center point as the origin of a coordinate grid.

Center = (0, 0)

Step 2: Identify the most restrictive boundary points

Imagine a circle growing from the origin. It will first hit the boundary at the points that point inward toward the center. Let’s look at the top-right quarter of the shape. From the center, we must go 1 unit right and 2 units up (or 2 units right and 1 unit up) to reach the first inward-pointing corners.

Inward corners in Quadrant I: (1, 2) and (2, 1)

Step 3: Calculate the distance to these corners

We use the Pythagorean theorem to find the squared distance from the origin to one of these inward corners. This distance establishes the maximum possible radius for our circle.

r^{2} = 1^{2} + 2^{2}

r^{2} = 5

r = 5

Step 4: Verify the circle stays inside the rest of the shape

We must confirm the circle does not bulge out past the straight exterior edges. The outermost flat edges are 3 units away from the center, which is safely further than our radius. We must also check the flat horizontal boundary located at a height of 2, which runs between horizontal coordinates 1 and 2. Because the horizontal coordinate of the circle is strictly greater than 1 in this section, its vertical coordinate is forced to be strictly less than 2, meaning it curves safely below the boundary.

r = 5 < 3

x^{2} + y^{2} = 5

y^{2} = 5 - x^{2}

For x > 1, y^{2} < 5 - 1 = 4 ⟹ y < 2

Step 5: Calculate the area of the largest circle

Since the circle fits perfectly and is strictly bounded by the 8 inward corners around the shape, its maximum radius is established. We use the standard area formula for a circle to find the final requested quantity.

Area = π r^{2}

Area = π (5)^{2}

Area = 5 π

Correct answer: C. $5 π$
The exact area of the largest inscribed circle is

5 π

, which matches answer choice C. We can verify this visually and numerically: the inner corners of the cross shape are at a distance of

5 \approx 2.236

from the center, while the outer flat edges are 3 units away. Since the circle reaches exactly to the inner corners and comfortably clears the outer edges,

5 π

is logically sound and mathematically confirmed.

Adapted Problems to Practice

1 / 10

What Formulas Do You Need for This Problem?

Center of Symmetry Strategy:

If a bounding region has horizontal and vertical reflectional symmetry, the largest inscribed circle will share its center with the shape.

Applies to highly symmetric figures such as crosses, stars, regular polygons, or centered grids. It allows us to confidently place the center of the circle at the origin and simply calculate distances from that single point, knowing the symmetry guarantees we won't find a larger circle shifted elsewhere.

Pythagorean Theorem:

a^{2} + b^{2} = c^{2}

$a$ and $b$ are the lengths of the legs of a right triangle, and $c$ is the length of the hypotenuse. It is used to calculate the direct straight-line distance from the center of the grid to the inner corners of the boundary.

Area of a Circle:

A = π r^{2}

$A$ is the area and $r$ is the radius. The problem asks for the area of the largest inscribed circle, so we apply this formula after using the grid to deduce the maximum squared radius ( $r^{2}$ ).

Key Takeaway: Remember This Strategy for Next Time

A highly symmetric boundary on a square grid.

Try: Place the center of the inscribed shape at the center of symmetry and treat the grid as a coordinate plane.

Why: Symmetry ensures that any maximal inscribed shape doesn't favor one side over another, so its center must align with the overall geometric center.

An inscribed circle constrained by a complex, zig-zagging polygon.

Try: Identify the 'inward' pointing corners (reflex angles) and use them to find the maximum possible radius.

Why: A growing circle will always hit inward-pointing vertices before it hits the outward-pointing corners, making the inward corners the true bottlenecks for the circle's size.

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2025 AMC 8 Problem 12

In This Page:

2025 AMC8 Problem 12