AMC 8 · 2012 · #24
Grade 7 geometry-2dProblem
A circle of radius 2 is cut into four congruent arcs. The four arcs are joined to form the star figure shown. What is the ratio of the area of the star figure to the area of the original circle?
Pick an answer.
AMC 8 2012 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 circle of radius $2$ is cut into four congruent arcs (each a quarter-circle). The four arcs are rejoined into a four-pointed "star" whose sides curve inward. Find the ratio of the area of this star to the area of the original circle.
Givens: Original circle has radius $r = 2$, so its area is $\pi r^2 = 4\pi$; The boundary is cut into $4$ congruent arcs, each a quarter-circle of radius $2$; The four arcs are reassembled into the star shown, with the arcs curving inward (concave); Answer choices: (A) $\tfrac{4-\pi}{\pi}$, (B) $\tfrac{1}{\pi}$, (C) $\tfrac{\sqrt{2}}{\pi}$, (D) $\tfrac{\pi-1}{\pi}$, (E) $\tfrac{3}{\pi}$
Unknowns: The ratio $\dfrac{\text{area of star}}{\text{area of original circle}}$
Understand
Restated: A circle of radius $2$ is cut into four congruent arcs (each a quarter-circle). The four arcs are rejoined into a four-pointed "star" whose sides curve inward. Find the ratio of the area of this star to the area of the original circle.
Givens: Original circle has radius $r = 2$, so its area is $\pi r^2 = 4\pi$; The boundary is cut into $4$ congruent arcs, each a quarter-circle of radius $2$; The four arcs are reassembled into the star shown, with the arcs curving inward (concave); Answer choices: (A) $\tfrac{4-\pi}{\pi}$, (B) $\tfrac{1}{\pi}$, (C) $\tfrac{\sqrt{2}}{\pi}$, (D) $\tfrac{\pi-1}{\pi}$, (E) $\tfrac{3}{\pi}$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems
The star is an odd, curved shape — we cannot apply any single area formula to it directly. Tool #1 (Draw a Diagram) is the right first move: by sketching the star inside a square of side $4$ whose corners coincide with the star's points, we can SEE that the four "bites" taken out of the square are exactly four quarter-circles of radius $2$ — which together form one full circle of radius $2$. Tool #7 (Identify Subproblems) then turns the problem into two easy pieces we already know: (a) area of a $4 \times 4$ square, and (b) area of a circle of radius $2$. The star is (a) minus (b), and the ratio falls out by simple division.
Execute — Answer: A
3.MD.C.7 Step 1 - Draw the star and place its four points at $(2,0)$, $(0,2)$, $(-2,0)$, $(0,-2)$.
- Now draw the axis-aligned square that just contains the star: its sides are vertical and horizontal lines at $x = \pm 2$, $y = \pm 2$.
- The square has side $4$, and the four points of the star touch the midpoints of the square's sides.
💡 Drawing the square around the star is the key insight — Grade 3 area of a rectangle is all we need for the square itself.
7.G.B.4 Step 2 - Look at the region INSIDE the square but OUTSIDE the star.
- There are four congruent pieces, one near each corner of the square.
- Each piece is bounded by two straight sides of the square (meeting at a corner) and one of the star's inward-curving arcs.
- Because the arc is a quarter-circle of radius $2$ and the corner of the square is exactly $2$ units from each of the two star points it connects, each corner piece is exactly a quarter-disk of radius $2$.
💡 The picture shows that the arc, plus the two sides of the square meeting at that corner, form a perfect quarter-circle region. Grade 7 circle-area formula handles it.
7.G.B.4 Step 3 - Break the area of the square into two subproblems: (a) area of the star, and (b) the four corner quarter-disks.
- The four quarter-disks fit together to make exactly one full circle of radius $2$ — the original circle!
💡 Four quarter-circles of the same radius reassemble into one whole circle. This makes the subtraction extra clean.
3.MD.C.7 Step 4 Now apply Tool #7: area of star = area of square minus the four corner pieces.
💡 Whole minus parts — the standard area-decomposition move.
6.RP.A.3 Step 5 Form the requested ratio and simplify by factoring $4$ from the numerator.
💡 A ratio of two areas is just one number divided by another — Grade 6 ratio reasoning, with a common factor of $4$ to clean up.
3.MD.C.7 Draw the star and place its four points at $(2,0)$, $(0,2)$, $(-2,0)$, $(0,-2)$. 7.G.B.4 Look at the region INSIDE the square but OUTSIDE the star. There are four congru 7.G.B.4 Break the area of the square into two subproblems: (a) area of the star, and (b) 3.MD.C.7 Now apply Tool #7: area of star = area of square minus the four corner pieces. 6.RP.A.3 Form the requested ratio and simplify by factoring $4$ from the numerator. Review
Reasonableness: Numerically, $\pi \approx 3.14$, so $\dfrac{4 - \pi}{\pi} \approx \dfrac{0.86}{3.14} \approx 0.27$. That says the star takes up about $27\%$ of the original circle's area — which matches the picture: the star is clearly much smaller than the original circle (more than half of the circle's area sits outside the star in those four "crescent" gaps). Also, all five answer choices have $\pi$ in the denominator, which is consistent with our setup (we divided by $4\pi$).
Alternative: Tool #3 (Eliminate Possibilities) plus a quick estimate: the star is clearly smaller than the original circle, so the ratio must be less than $1$. Plugging $\pi \approx 3.14$ into each choice gives (A) $\approx 0.27$, (B) $\approx 0.32$, (C) $\approx 0.45$, (D) $\approx 0.68$, (E) $\approx 0.95$. The star looks like roughly a quarter of the circle, ruling out (D) and (E) immediately, and a careful sketch supports the smallest credible value — choice (A).
CCSS standards used (min grade 7)
3.MD.C.7Relate area to the operations of multiplication and addition (Computing the bounding square's area as $4 \times 4 = 16$ and using whole-minus-parts decomposition ($A_{\text{star}} = A_{\text{square}} - A_{\text{four corners}}$).)7.G.B.4Know the formulas for the area and circumference of a circle and use them to solve problems (Computing each corner quarter-disk as $\tfrac{1}{4}\pi(2)^2 = \pi$ and recognizing the four quarter-disks reassemble into one circle of area $4\pi$.)6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Forming and simplifying the final ratio $\dfrac{16 - 4\pi}{4\pi} = \dfrac{4 - \pi}{\pi}$.)
⭐ Draw a square around the star: the four "bites" out of the square fit together to make the original circle, so the star is just square minus circle — Grade 7 circle-area reasoning does the rest.
⭐ Draw a square around the star: the four "bites" out of the square fit together to make the original circle, so the star is just square minus circle — Grade 7 circle-area reasoning does the rest.
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