AMC 8 · 2025 · #12

Grade 8 geometry-2d

Archetype Coordinate Plane Figure Computation

area-circlescoordinate-geometrypythagorean-theoremline-symmetry coordinate-geometryidentify-subproblems ↑ Prerequisites: area-circlescoordinate-geometrypythagorean-theorem

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

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.

(A)

$3\pi$

(B)

$4\pi$

(C)

$5\pi$

(D)

$6\pi$

(E)

$8\pi$

View mode:

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

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

#1 Draw a Diagram 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 $C = \left(\tfrac{0+6}{2}, \tfrac{1+7}{2}\right) = (3, 4)$.

$$C = (3,\, 4)$$

💡 Setting up coordinate axes on a grid and reading off a midpoint is a Grade 5 coordinate-plane skill.

#1 Draw a Diagram 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.

Inner corners near $C$: $(5,5),\ (4,6),\ (2,6),\ (1,5),\ \ldots$

💡 Reading lattice-point coordinates straight off a coordinate grid is Grade 5 graphing.

#7 Identify Subproblems 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.

$$r = \sqrt{(5-3)^2 + (5-4)^2} = \sqrt{4 + 1} = \sqrt{5}$$

💡 Finding the distance between two coordinate points via $\sqrt{(\Delta x)^2 + (\Delta y)^2}$ is the Grade 8 Pythagorean-distance standard.

#7 Identify Subproblems 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.

$\sqrt{5} \approx 2.236 < 3$ (distance to outer edge)

💡 Comparing $\sqrt{5}$ with the whole number $3$ uses Grade 8 rational approximation of an irrational number.

#3 Eliminate Possibilities 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.

$$A = \pi r^2 = \pi (\sqrt{5})^2 = 5\pi \;\Rightarrow\; \textbf{(C)}$$

💡 Plugging the radius into the area formula $A = \pi r^2$ is the Grade 7 circle-area standard.

[1] #1 5.G.A.1 Place the figure on a coordinate plane and locate the center using symmetry. The

[2] #1 5.G.A.1 Identify the boundary points closest to the center. By symmetry, the four "inner

[3] #7 8.G.B.8 Find the radius as the distance from $C = (3,4)$ to one of those nearest corners

[4] #7 8.NS.A.2 Confirm that $r = \sqrt{5}$ really is the maximum. The distance from $C$ to the

[5] #3 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.1 Use 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.8 Apply 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.2 Use 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.4 Know 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!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 8)

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