AMC 8 · 2025 · #12
Grade 8 geometry-2dProblem
The region shown below consists of 24 squares, each with side length 1 centimeter. What is the area, in square centimeters, of the largest circle that can fit inside the region, possibly touching the boundaries?
Pick an answer.
AMC 8 2025 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.
Toolkit + CCSS Solution
Understand
Restated: A plus-shaped region is made of $24$ unit squares (each side $1$ cm), symmetric horizontally and vertically. We want the area, in square centimeters, of the largest circle that fits entirely inside this region (it may touch the boundary).
Givens: The region consists of $24$ unit squares, each with side length $1$ cm; The region is symmetric both horizontally and vertically (a plus/diamond shape); Reading the grid: the region spans $x \in [0,6]$ and $y \in [1,7]$ on a coordinate grid; Answer choices: (A) $3\pi$, (B) $4\pi$, (C) $5\pi$, (D) $6\pi$, (E) $8\pi$ (square cm)
Unknowns: The area (in square cm) of the largest circle that fits inside the region
Understand
Restated: A plus-shaped region is made of $24$ unit squares (each side $1$ cm), symmetric horizontally and vertically. We want the area, in square centimeters, of the largest circle that fits entirely inside this region (it may touch the boundary).
Givens: The region consists of $24$ unit squares, each with side length $1$ cm; The region is symmetric both horizontally and vertically (a plus/diamond shape); Reading the grid: the region spans $x \in [0,6]$ and $y \in [1,7]$ on a coordinate grid; Answer choices: (A) $3\pi$, (B) $4\pi$, (C) $5\pi$, (D) $6\pi$, (E) $8\pi$ (square cm)
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems, #3 Eliminate Possibilities
The figure is geometric and symmetric, so Tool #1 (Draw a Diagram) — overlaying a coordinate grid on the plus shape — instantly reveals the symmetry center and the boundary vertices to watch. Tool #7 (Identify Subproblems) splits the question into three clean pieces: (a) Where is the circle's center? (b) What is the radius? (c) What is the area? Each piece is easy by itself. Tool #3 (Eliminate Possibilities) is the safety net: once we find $r^2 = 5$, the answer must be $5\pi$, knocking out the other four choices immediately.
Execute — Answer: C
5.G.A.1 Step 1 - Place the figure on a coordinate plane and locate the center using symmetry.
- The plus shape is symmetric left-right and top-bottom, so the largest inscribed circle must share that center.
- The horizontal span is $x = 0$ to $x = 6$ and the vertical span is $y = 1$ to $y = 7$, so the center of symmetry is at $C = \left(\tfrac{0+6}{2}, \tfrac{1+7}{2}\right) = (3, 4)$.
💡 Setting up coordinate axes on a grid and reading off a midpoint is a Grade 5 coordinate-plane skill.
5.G.A.1 Step 2 - Identify the boundary points closest to the center.
- By symmetry, the four "inner notch" corners closest to $C$ are at $(5,5)$, $(4,6)$, $(2,6)$, $(1,5)$ (and their reflections below).
- These concave corners pinch the circle tighter than the flat outer edges, so one of them sets the radius.
💡 Reading lattice-point coordinates straight off a coordinate grid is Grade 5 graphing.
8.G.B.8 Step 3 - Find the radius as the distance from $C = (3,4)$ to one of those nearest corners — say $P = (5,5)$.
- Each such corner sits $2$ units across and $1$ unit up from $C$, forming a right triangle with legs $2$ and $1$.
- By the Pythagorean theorem, the hypotenuse (which is the distance $CP$) satisfies $r^2 = 2^2 + 1^2 = 5$, so $r = \sqrt{5}$ cm.
- Checking $Q = (4,6)$ gives $r^2 = 1^2 + 2^2 = 5$ — the same value, as symmetry predicts.
💡 Finding the distance between two coordinate points via $\sqrt{(\Delta x)^2 + (\Delta y)^2}$ is the Grade 8 Pythagorean-distance standard.
8.NS.A.2 Step 4 - Confirm that $r = \sqrt{5}$ really is the maximum.
- The distance from $C$ to the outer flat edges (for example to $x = 6$) is $6 - 3 = 3$, and $\sqrt{5} \approx 2.24 < 3$, so those edges do not restrict the circle.
- The eight inner corners (four near the top half, four near the bottom) all sit at the same distance $\sqrt{5}$, so the largest circle is tangent to all of them simultaneously.
💡 Comparing $\sqrt{5}$ with the whole number $3$ uses Grade 8 rational approximation of an irrational number.
7.G.B.4 Step 5 - Apply the area formula for a circle, $A = \pi r^2$.
- Squaring $\sqrt{5}$ gives $5$, so the area is exactly $5\pi$ square centimeters — which is choice (C).
- Tool #3 (Eliminate Possibilities) confirms the match against the five options.
💡 Plugging the radius into the area formula $A = \pi r^2$ is the Grade 7 circle-area standard.
5.G.A.1 Place the figure on a coordinate plane and locate the center using symmetry. The 5.G.A.1 Identify the boundary points closest to the center. By symmetry, the four "inner 8.G.B.8 Find the radius as the distance from $C = (3,4)$ to one of those nearest corners 8.NS.A.2 Confirm that $r = \sqrt{5}$ really is the maximum. The distance from $C$ to the 7.G.B.4 Apply the area formula for a circle, $A = \pi r^2$. Squaring $\sqrt{5}$ gives $5 Review
Reasonableness: Sanity check: the plus shape contains $24$ unit squares, so its total area is $24$ square centimeters. Our circle's area is $5\pi \approx 15.7$ square cm — comfortably less than $24$ and clearly bigger than a circle inscribed in just the central $2\times 2$ block (which would have area $\pi \approx 3.14$). The radius $\sqrt{5} \approx 2.24$ cm also makes sense: the central part of the region extends $3$ units out from $C$ horizontally and vertically, so a circle of radius $\approx 2.24$ comfortably fits while being pinched by the concave corners — exactly what we found.
Alternative: Tool #3 (Eliminate Possibilities) can also work backwards from the choices. Each option gives a candidate $r^2$: $r^2 = 3, 4, 5, 6, 8$, i.e. $r \approx 1.73, 2.00, 2.24, 2.45, 2.83$. A circle of radius $2$ centered at $(3,4)$ leaves visible gaps near the concave corners (so $r$ can grow), while a circle of radius $\geq 2.45$ would poke outside through those same corners (since their distance to $C$ is only $\sqrt{5} \approx 2.24$). The only choice that hits the inner corners exactly is $r^2 = 5$, i.e. $5\pi$.
CCSS standards used (min grade 8)
5.G.A.1Use a pair of perpendicular number lines forming a coordinate system (Placing the plus shape on a coordinate plane, reading the spans $x \in [0,6]$ and $y \in [1,7]$, finding the symmetry center $C = (3,4)$, and listing the inner-corner coordinates such as $(5,5)$ and $(4,6)$.)8.G.B.8Apply the Pythagorean theorem to find distance between two points in a coordinate system (Computing the radius $r = \sqrt{(5-3)^2 + (5-4)^2} = \sqrt{5}$ as the distance from the center $(3,4)$ to the nearest boundary corner $(5,5)$.)8.NS.A.2Use rational approximations of irrational numbers to compare their size (Verifying $\sqrt{5} \approx 2.24 < 3$, so the inner corners — not the outer flat edges — really do limit the circle.)7.G.B.4Know the formulas for area and circumference of a circle (Applying $A = \pi r^2$ with $r = \sqrt{5}$ to get the area $5\pi$ square centimeters.)
⭐ This AMC 8 problem only needs the Grade 8 Pythagorean-theorem distance formula (plus the Grade 7 circle-area formula $A = \pi r^2$) you already know!
⭐ This AMC 8 problem only needs the Grade 8 Pythagorean-theorem distance formula (plus the Grade 7 circle-area formula $A = \pi r^2$) you already know!
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