AMC 10 · 2007 · #15
Grade 8 geometry-2dFour circles of radius 1 are each tangent to two sides of a square and externally tangent to a circle of radius 2, as shown. What is the area of the square?
Pick an answer.
AMC 10 2007 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 square holds four circles of radius $1$, each tucked into a corner so it touches two sides of the square. A bigger circle of radius $2$ sits in the middle and just touches each of the four small circles from the outside. Find the area of the square.
Givens: Each of the four small circles has radius $1$; Each small circle is tangent to two sides of the square (so it sits in a corner); The central circle has radius $2$; The central circle is externally tangent to each small circle; By symmetry the center of the square and the center of the big circle are the same point; Answer choices: (A) $32$, (B) $22 + 12\sqrt{2}$, (C) $16 + 16\sqrt{3}$, (D) $48$, (E) $36 + 16\sqrt{2}$
Unknowns: The area of the square
Understand
Restated: A square holds four circles of radius $1$, each tucked into a corner so it touches two sides of the square. A bigger circle of radius $2$ sits in the middle and just touches each of the four small circles from the outside. Find the area of the square.
Givens: Each of the four small circles has radius $1$; Each small circle is tangent to two sides of the square (so it sits in a corner); The central circle has radius $2$; The central circle is externally tangent to each small circle; By symmetry the center of the square and the center of the big circle are the same point; Answer choices: (A) $32$, (B) $22 + 12\sqrt{2}$, (C) $16 + 16\sqrt{3}$, (D) $48$, (E) $36 + 16\sqrt{2}$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems, #4 Introduce a Variable
This is a placement puzzle, so Tool #1 (Draw a Diagram) with coordinates pins down where every center sits. Put the shared center of the square and the big circle at the origin; then a corner circle's center is forced by the fact that it must stay one radius from each side it touches. Tool #7 (Identify Subproblems) isolates the one number that unlocks everything: the distance from the origin to a corner circle's center, which the external-tangency condition fixes at $2+1=3$. Tool #4 (Introduce a Variable) names the side length $s$, turns that distance into an equation, and solves it; squaring gives the area.
Execute — Answer: B
6.G.A.3 Step 1 Set up coordinates
- Put the center of the square (which is also the center of the big circle) at the origin, and let $s$ be the side length.
- Then the four sides lie along $x = \pm\tfrac{s}{2}$ and $y = \pm\tfrac{s}{2}$.
- Look at the small circle in the top-right corner.
- It touches the top side and the right side, and a circle touching a straight line sits one radius away from it, so its center is $1$ unit left of the right side and $1$ unit below the top side.
- That center is at $\left(\tfrac{s}{2}-1,\ \tfrac{s}{2}-1\right)$.
💡 A circle hugging two sides must sit one radius in from each, which pins its center in terms of the square's size.
7.G.A.2 Step 2 Read the distance from tangency
- The big circle (radius $2$) and the corner circle (radius $1$) touch on the outside.
- When two circles are externally tangent, they meet at a single point on the segment joining their centers, so the distance between centers is exactly the sum of the radii.
- That distance is $2 + 1 = 3$.
- So the origin is exactly $3$ units from the corner circle's center.
💡 Two circles touching outside are as far apart as their radii stacked end to end.
8.G.B.8 Step 3 Build and solve the equation
- The corner center $\left(\tfrac{s}{2}-1,\ \tfrac{s}{2}-1\right)$ has both coordinates equal; call that common value $a = \tfrac{s}{2}-1$.
- Its distance from the origin, by the Pythagorean theorem, is $\sqrt{a^2 + a^2} = a\sqrt{2}$.
- Set this equal to $3$: $a\sqrt{2} = 3$, so $a = \tfrac{3}{\sqrt{2}} = \tfrac{3\sqrt{2}}{2}$.
- Then $\tfrac{s}{2} = 1 + \tfrac{3\sqrt{2}}{2}$, which gives $s = 2 + 3\sqrt{2}$.
💡 A point $(a,a)$ sits $a\sqrt{2}$ from the origin, so a known distance hands back the coordinate.
8.EE.A.2 Step 4 Square the side to get the area
- The area of the square is $s^2 = (2 + 3\sqrt{2})^2$.
- Expand it: $2^2 + 2\cdot 2 \cdot 3\sqrt{2} + (3\sqrt{2})^2 = 4 + 12\sqrt{2} + 9\cdot 2 = 4 + 12\sqrt{2} + 18 = 22 + 12\sqrt{2}$.
- That is choice (B).
💡 Squaring $2+3\sqrt2$ keeps a plain part and a $\sqrt2$ part because the cross term carries the radical.
6.G.A.3 Put the center of the square (which is also the center of the big circle) at the 7.G.A.2 The big circle (radius $2$) and the corner circle (radius $1$) touch on the outs 8.G.B.8 The corner center $\left(\tfrac{s}{2}-1,\ \tfrac{s}{2}-1\right)$ has both coordi 8.EE.A.2 The area of the square is $s^2 = (2 + 3\sqrt{2})^2$. Expand it: $2^2 + 2\cdot 2 Review
Reasonableness: Estimate to make sure the answer is sane. Since $\sqrt{2} \approx 1.414$, the side is $s \approx 2 + 3(1.414) = 2 + 4.24 = 6.24$, so the area is about $6.24^2 \approx 39$. Choice (B) gives $22 + 12(1.414) \approx 22 + 17 = 39$, which matches. It also passes a sanity floor: the big circle alone has diameter $4$ across the middle, and the small circles add margin on each side, so a side near $6$ and an area near $39$ is exactly what the picture demands. Choices (A) $32$ and (D) $48$ have no radical and are too clean for this slanted setup; (C) uses $\sqrt{3}$, which never appears here; (E) $\approx 59$ is far too big.
Alternative: Skip coordinates and walk the diagonal of the square. Start at the top-right corner circle's center and go straight through the origin to the bottom-left corner circle's center. That run is one small radius, then the big circle's diameter, then another small radius: $1 + 4 + 1 = 6$, and it is the diagonal of the small square that joins opposite corner-circle centers. A square with diagonal $6$ has side $6/\sqrt{2} = 3\sqrt{2}$, and that small square's side is shorter than the big square's side by $1$ on each end (from center to the touched side), so the big side is $3\sqrt{2} + 2$. Squaring again gives $22 + 12\sqrt{2}$.
CCSS standards used (min grade 8)
6.G.A.3Draw polygons in the coordinate plane given coordinates for the vertices (Placing the square and circle centers on a coordinate grid and writing the corner circle's center as $\left(\tfrac{s}{2}-1,\ \tfrac{s}{2}-1\right)$.)7.G.A.2Draw geometric shapes with given conditions including triangles (Using the external-tangency condition to fix the center-to-center distance at $2+1=3$.)8.G.B.8Apply the Pythagorean theorem to find distance between two points in a coordinate system (Turning the distance from the origin to $(a,a)$ into $a\sqrt{2}=3$ and solving for the side length $s = 2 + 3\sqrt{2}$.)8.EE.A.2Use square root and cube root symbols to represent solutions (Squaring the radical side length $(2+3\sqrt{2})^2$ to get the area $22 + 12\sqrt{2}$.)
⭐ Pin every center on a grid, let each tangency hand you a distance, and one Pythagorean step turns those distances into the square's side.
⭐ Pin every center on a grid, let each tangency hand you a distance, and one Pythagorean step turns those distances into the square's side.
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