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AMC 8 · 2011 · #25

Grade 8 geometry-2d
Archetype Compound Area by Decomposition / Difference
area-circlesarea-rectanglesestimation identify-subproblemsarea-difference ↑ Prerequisites: area-circlesarea-rectangles
📏 Medium solution 💡 3 insights 📊 Diagram

Problem

A circle with radius 11 is inscribed in a square and circumscribed about another square as shown. Which fraction is closest to the ratio of the circle's shaded area to the area between the two squares?

Pick an answer.

(A)
$\frac{1}2$
(B)
1
(C)
$\frac{3}2$
(D)
2
(E)
$\frac{5}2$
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Toolkit + CCSS Solution

Understand

Restated: A circle of radius $1$ sits inside a larger square (touching all four sides) and outside a smaller square (whose four corners are on the circle). The picture has two shaded pieces: the part of the circle outside the small square, and the part of the large square outside the circle. We need the ratio (circle-shaded area) : (area between the two squares), and pick the answer choice closest to that ratio.

Givens: Circle radius $r = 1$; Large square is circumscribed about the circle (each side touches the circle); Small square is inscribed in the circle (each vertex lies on the circle); Answer choices: (A) $\tfrac{1}{2}$, (B) $1$, (C) $\tfrac{3}{2}$, (D) $2$, (E) $\tfrac{5}{2}$

Unknowns: The fraction closest to $\dfrac{\text{circle's shaded area}}{\text{area between the two squares}}$

Understand

Restated: A circle of radius $1$ sits inside a larger square (touching all four sides) and outside a smaller square (whose four corners are on the circle). The picture has two shaded pieces: the part of the circle outside the small square, and the part of the large square outside the circle. We need the ratio (circle-shaded area) : (area between the two squares), and pick the answer choice closest to that ratio.

Givens: Circle radius $r = 1$; Large square is circumscribed about the circle (each side touches the circle); Small square is inscribed in the circle (each vertex lies on the circle); Answer choices: (A) $\tfrac{1}{2}$, (B) $1$, (C) $\tfrac{3}{2}$, (D) $2$, (E) $\tfrac{5}{2}$

Plan

Primary tool: #7 Identify Subproblems

Secondary: #1 Draw a Diagram, #3 Eliminate Possibilities

The picture has three nested shapes — large square, circle, small square — so Tool #7 (Identify Subproblems) splits the work into three clean area calculations and a final ratio. Tool #1 (Draw a Diagram) is built into the problem's figure: reading off that the circle's diameter equals the large square's side and the small square's diagonal is what unlocks every dimension. With a clean numerical ratio in hand, Tool #3 (Eliminate Possibilities) finishes the job by comparing one decimal to the five answer choices.

Execute — Answer: A

#1 Draw a Diagram 7.G.B.4 Step 1
  • Read the two key length facts from the figure.
  • The circle is inscribed in the large square, so the large square's side equals the circle's diameter: $\text{side} = 2r = 2$.
  • The small square is inscribed in the circle, so its diagonal equals the diameter: $\text{diagonal} = 2r = 2$.
$$\text{side}_{\text{large}} = 2, \quad \text{diagonal}_{\text{small}} = 2$$

💡 Tool #1: the figure is doing work for us — touching points and corners are how the circle's radius transfers to the squares.

#7 Identify Subproblems 3.MD.C.7 Step 2
  • Subproblem 1: area of the large square.
  • A square's area is side squared.
$$A_{\text{large}} = 2^2 = 4$$

💡 Grade 3 "area = side $\times$ side" — the easiest of the three pieces.

#7 Identify Subproblems 8.G.B.7 Step 3
  • Subproblem 2: area of the small square.
  • Let its side be $s$.
  • The diagonal of a square splits it into two right triangles with legs $s, s$ and hypotenuse $s\sqrt{2}$.
  • So $s\sqrt{2} = 2$, meaning $s^2 = 2$ — which is exactly the area.
$$s\sqrt{2} = 2 \;\Rightarrow\; s^2 = \dfrac{2^2}{2} = 2, \quad A_{\text{small}} = s^2 = 2$$

💡 Pythagorean theorem on the half-square turns the diagonal into the side-squared, which is the area itself.

#7 Identify Subproblems 7.G.B.4 Step 4

Subproblem 3: area of the circle.

$$A_{\text{circle}} = \pi r^2 = \pi (1)^2 = \pi$$

💡 Grade 7 circle-area formula. The $r = 1$ trick makes the answer just $\pi$.

#7 Identify Subproblems 3.MD.C.7 Step 5
  • Assemble the two shaded regions.
  • The circle's shaded part is the circle minus the small square (the corners of the small square are inside the circle).
  • The region between the squares is the large square minus the small square.
$$\text{circle shaded} = \pi - 2, \quad \text{between squares} = 4 - 2 = 2$$

💡 Subtracting nested areas is the standard "region between shapes" move from Grade 3 area reasoning.

#7 Identify Subproblems 6.RP.A.3 Step 6

Form the ratio and estimate using $\pi \approx 3.14$.

$$\dfrac{\pi - 2}{2} \approx \dfrac{3.14 - 2}{2} = \dfrac{1.14}{2} = 0.57$$

💡 A ratio of two areas is a Grade 6 rate/ratio computation — same skill as miles per hour.

#3 Eliminate Possibilities 6.RP.A.3 Step 7
  • Compare $0.57$ to the answer choices and pick the closest.
  • The gaps are $|0.57 - 0.5| = 0.07$ and $|0.57 - 1| = 0.43$; every other choice is even farther away.
$$\text{(A) } 0.5,\ \text{(B) } 1,\ \text{(C) } 1.5,\ \text{(D) } 2,\ \text{(E) } 2.5 \;\Rightarrow\; \textbf{(A) } \tfrac{1}{2}$$

💡 Tool #3: "closest to" problems become a distance check between one number and five candidates.

[1] #1 7.G.B.4 Read the two key length facts from the figure. The circle is inscribed in the la
[2] #7 3.MD.C.7 Subproblem 1: area of the large square. A square's area is side squared.
[3] #7 8.G.B.7 Subproblem 2: area of the small square. Let its side be $s$. The diagonal of a s
[4] #7 7.G.B.4 Subproblem 3: area of the circle.
[5] #7 3.MD.C.7 Assemble the two shaded regions. The circle's shaded part is the circle minus th
[6] #7 6.RP.A.3 Form the ratio and estimate using $\pi \approx 3.14$.
[7] #3 6.RP.A.3 Compare $0.57$ to the answer choices and pick the closest. The gaps are $|0.57 -

Review

Reasonableness: Sanity check the size: the circle ($\pi \approx 3.14$) is a bit bigger than the small square ($2$) but smaller than the large square ($4$), which matches the picture. The circle's shaded crescents add up to a thin sliver — about $1.14$ — and the four corner pieces between the squares add up to $2$. So the shaded circle area is about half the between-squares area, and $\tfrac{1}{2}$ matches.

Alternative: Tool #16 (Change Focus): rewrite the ratio as $\dfrac{\pi - 2}{2} = \dfrac{\pi}{2} - 1$. Since $\pi/2 \approx 1.57$, the ratio is about $0.57$ — same conclusion, but no decimal subtraction needed. This also makes it obvious that the answer must be a little above $\tfrac{1}{2}$ and far below $1$, so (A) wins without even checking the other choices.

CCSS standards used (min grade 8)

  • 3.MD.C.7 Relate area to multiplication; find area of rectangles by tiling and multiplying side lengths (Computing the large square's area as $2 \times 2 = 4$ and forming the "between squares" area as $4 - 2 = 2$ by subtracting nested regions.)
  • 7.G.B.4 Know the formulas for the area and circumference of a circle and use them to solve problems (Reading the circle's diameter from the figure ($d = 2r = 2$) and computing the circle's area as $\pi r^2 = \pi$.)
  • 8.G.B.7 Apply the Pythagorean Theorem to determine unknown side lengths in right triangles in real-world and mathematical problems (Turning the small square's diagonal ($2$) into its area: $s\sqrt{2} = 2$ gives $s^2 = 2$, which is the area.)
  • 6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems (Forming the ratio $\dfrac{\pi - 2}{2} \approx 0.57$ and selecting the closest of the five fractional choices.)

⭐ Three nested shapes break into three Grade 3-8 area facts, and a $\pi \approx 3.14$ estimate turns the ratio into a single decimal you can match to the closest answer choice.

⭐ Three nested shapes break into three Grade 3-8 area facts, and a $\pi \approx 3.14$ estimate turns the ratio into a single decimal you can match to the closest answer choice.

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