Competition · AMC preparation · step 4 of 4
AMC 8 · 2025 · #12
Grade 8 geometry-2d
Pick an answer.
AMC 8 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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 r² = 5, the answer must be 5π, knocking out the other four choices immediately.
Find the center by symmetry
By symmetry the largest inscribed circle must share the figure's center, at C = (3, 4).
Setting up coordinate axes on a grid and reading off a midpoint is a Grade 5 coordinate-plane skill.
5.G.A.1Draw A DiagramFind the nearest corners
The concave inner corners nearest C, such as (5, 5), pinch the circle more than the outer edges — so one of them sets the radius.
Reading lattice-point coordinates straight off a coordinate grid is Grade 5 graphing.
5.G.A.1Draw A DiagramCompute the radius
Each nearest corner is 2 across and 1 up from C, so by the Pythagorean theorem r² = 2² + 1² = 5 and r = √(5) cm.
Finding the distance between two coordinate points via √((Δ x)² + (Δ y)²) is the Grade 8 Pythagorean-distance standard.
The largest circle that fits inside the plus-shaped region has radius √(5) cm, equal to the distance from the center (3,4) to the nearest inner corner (5,5).
▸ Why?
A circle centered at (3,4) can keep growing only until its edge first reaches the boundary, so the largest one's radius equals the shortest distance from the center to the boundary — here the concave inner corners, which sit closer than the flat outer edges.
▸ Why?
Every point of a circle lies exactly one radius from the center, so the moment the radius equals the closest center-to-boundary distance the circle just touches the boundary, and any larger radius would push part of the circle outside the region.
▸ Why?
The nearest corner (5,5) lies 2 units across and 1 unit up from the center (3,4), and those two gaps are the legs of a right triangle whose hypotenuse is the distance we want, so distance² = 2² + 1² = 5 and the distance is √(5).
▸ Why?
The 2-unit horizontal step and the 1-unit vertical step meet at a right angle, so the straight-line distance is the hypotenuse of a right triangle with legs 2 and 1; in a right triangle the two legs' squares add up to the hypotenuse's square, so distance² = 2² + 1² = 5.
Check the radius is largest
The distance to the outer edges is 3, and √(5) ≈ 2.24 ＜ 3, so they don't bind — the circle is tangent to all 8 inner corners at once.
Comparing √(5) with the whole number 3 uses Grade 8 rational approximation of an irrational number.
8.NS.A.2Identify SubproblemsApply the circle area formula
Apply A = π r²: squaring √(5) gives 5, so the area is exactly 5π square cm — choice (C).
Plugging the radius into the area formula A = π r² is the Grade 7 circle-area standard.
7.G.B.4Eliminate PossibilitiesThis AMC 8 problem only needs the Grade 8 Pythagorean-theorem distance formula (plus the Grade 7 circle-area formula A = π r²) you already know!
- Find the center by symmetry
- Find the nearest corners
- Compute the radius
- Check the radius is largest
- Apply the circle area formula
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