AMC 10 · 2010 · #16
Grade 8 geometry-2dA square of side length 1 and a circle of radius 33 share the same center. What is the area inside the circle, but outside the square?
Pick an answer.
AMC 10 2010 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 with side length $1$ and a circle with radius $\dfrac{\sqrt{3}}{3}$ are drawn so they share the same center. Find the area of the region that is inside the circle but outside the square.
Givens: Square side length $= 1$, so its half-side is $\dfrac{1}{2}$; Circle radius $r = \dfrac{\sqrt{3}}{3} = \dfrac{1}{\sqrt{3}} \approx 0.577$; The square and circle share the same center
Unknowns: The area lying inside the circle but outside the square
Understand
Restated: A square with side length $1$ and a circle with radius $\dfrac{\sqrt{3}}{3}$ are drawn so they share the same center. Find the area of the region that is inside the circle but outside the square.
Givens: Square side length $= 1$, so its half-side is $\dfrac{1}{2}$; Circle radius $r = \dfrac{\sqrt{3}}{3} = \dfrac{1}{\sqrt{3}} \approx 0.577$; The square and circle share the same center
Plan
Primary tool: #7 Identify Subproblems
Secondary: #1 Draw a Diagram, #4 Introduce a Variable, #16 Change Focus / Count the Complement, #17 Visualize Spatial Relationships
The wanted region has a curvy, awkward shape, so the first job is to see it (Tool #1, Draw a Diagram): compare the radius with the square's half-side and half-diagonal to learn exactly how the circle sits over the square. That comparison shows the circle bulges out past the middle of each of the four sides while the four corners of the square stick out of the circle. By symmetry the leftover region is four identical little pieces, so the problem breaks into 'find one piece, then multiply by four' (Tool #7, Identify Subproblems). Each piece is a circular segment, and the clean way to measure a segment is to take a pie-slice (sector) and cut away the triangle inside it (Tool #16, Change Focus / Count the Complement). The one measurement we need — how wide the circle's opening is at each side — comes from a right triangle and the Pythagorean theorem. Seeing that the resulting triangle is equilateral (Tool #17) hands us the $60^\circ$ angle for free.
Execute — Answer: B
8.NS.A.2 Step 1 See how the circle sits
- Place the shared center at the origin.
- The square reaches out $\dfrac{1}{2} = 0.5$ to each side and $\dfrac{\sqrt{2}}{2} \approx 0.707$ to each corner.
- The circle's radius is $r = \dfrac{\sqrt{3}}{3} \approx 0.577$.
- Since $0.5 < 0.577 < 0.707$, the circle reaches past the middle of every side but not as far as the corners.
- So the circle pokes out through the four sides, while the four corners of the square poke out of the circle.
💡 Comparing the radius to the half-side and the half-diagonal tells you exactly where the circle spills over the square and where it tucks inside.
7.G.B.6 Step 2 Break into four equal caps
- The picture has four-fold symmetry: what happens at the top side matches the bottom, left, and right.
- The region inside the circle but outside the square is the four bulges that stick out past the four sides.
- Each bulge is a circular segment (the sliver of a circle cut off by a straight chord).
- So the total area is $4$ times the area of one segment, and we only need to measure one.
💡 Because the four pieces are identical copies, solving one and multiplying by four beats measuring the whole ragged region at once.
8.G.B.7 Step 3 Find the circle's opening at a side
- Look at the right side of the square, the vertical line $x = \dfrac{1}{2}$.
- Drop a radius from the center to a point where the circle crosses this line.
- That makes a right triangle: the horizontal leg runs from the center to the side, length $\dfrac{1}{2}$; the hypotenuse is the radius $r = \dfrac{1}{\sqrt{3}}$; the vertical leg is the half-height of the opening, call it $h$.
- By the Pythagorean theorem, $h = \sqrt{r^2 - \left(\tfrac{1}{2}\right)^2}$.
💡 The center, the side, and the crossing point form a right triangle, so the Pythagorean theorem measures how far the crossing point sits above the middle of the side.
8.G.A.5 Step 4 Spot the equilateral triangle
- The circle crosses the side at two points, one at height $+h$ and one at $-h$, so the full chord between them has length $2h = \dfrac{2}{2\sqrt{3}} = \dfrac{1}{\sqrt{3}} = r$.
- Now connect the center to those two crossing points with two radii.
- The triangle formed by the two radii and the chord has all three sides equal to $r$, so it is equilateral.
- An equilateral triangle has three $60^\circ$ angles, so the central angle of our segment is $60^\circ$, which is $\dfrac{1}{6}$ of a full turn.
💡 When all three sides of a triangle are the same length, all three angles must be $60^\circ$, which pins the sector's angle without any measuring.
7.G.B.4 Step 5 Area of the pie slice
- The sector (pie slice) reaching from the center out to the two crossing points spans $60^\circ$, which is $\dfrac{60}{360} = \dfrac{1}{6}$ of the whole circle.
- The whole circle has area $\pi r^2 = \pi \cdot \dfrac{1}{3} = \dfrac{\pi}{3}$.
- So the sector has area one-sixth of that.
💡 A $60^\circ$ slice is exactly one-sixth of the pie, so its area is just one-sixth of the circle's area.
6.G.A.1 Step 6 Cut off the triangle, then multiply by four
- The segment is the sector with the equilateral triangle removed.
- An equilateral triangle of side $r$ has area $\dfrac{\sqrt{3}}{4}r^2 = \dfrac{\sqrt{3}}{4}\cdot\dfrac{1}{3} = \dfrac{\sqrt{3}}{12}$.
- So one segment is $\dfrac{\pi}{18} - \dfrac{\sqrt{3}}{12}$.
- Multiply by the four sides to get the whole region: $4\left(\dfrac{\pi}{18} - \dfrac{\sqrt{3}}{12}\right) = \dfrac{2\pi}{9} - \dfrac{\sqrt{3}}{3}$.
- That matches choice $\textbf{(B)}$.
💡 A circular segment is just a pie slice with its straight-edged triangle taken away, so subtract the triangle from the sector and scale up by four.
8.NS.A.2 Place the shared center at the origin. The square reaches out $\dfrac{1}{2} = 0. 7.G.B.6 The picture has four-fold symmetry: what happens at the top side matches the bot 8.G.B.7 Look at the right side of the square, the vertical line $x = \dfrac{1}{2}$. Drop 8.G.A.5 The circle crosses the side at two points, one at height $+h$ and one at $-h$, s 7.G.B.4 The sector (pie slice) reaching from the center out to the two crossing points s 6.G.A.1 The segment is the sector with the equilateral triangle removed. An equilateral Review
Reasonableness: A quick sanity check: the circle's area is $\dfrac{\pi}{3} \approx 1.047$ and the square's area is $1$, so the circle has only about $0.047$ more area than the square. The region inside the circle but outside the square must be smaller than that tiny surplus, because the corners of the square (which lie outside the circle) also carry area. Our answer $\dfrac{2\pi}{9} - \dfrac{\sqrt{3}}{3} \approx 0.698 - 0.577 = 0.121$ is small and positive, consistent with four thin bulges. Choice (E) $\dfrac{2\pi}{9}$ is just the four sectors without removing the triangles, and (A) $\dfrac{\pi}{3}-1 \approx 0.047$ is the full circle minus the full square, which wrongly assumes the square sits entirely inside the circle. Both traps confirm (B) is the honest count.
Alternative: Instead of subtracting a triangle from a sector, integrate or use the standard circular-segment formula $\tfrac{1}{2}r^2(\theta - \sin\theta)$ with $\theta = \dfrac{\pi}{3}$: one segment $= \tfrac{1}{2}\cdot\dfrac{1}{3}\left(\dfrac{\pi}{3} - \sin 60^\circ\right) = \dfrac{1}{6}\left(\dfrac{\pi}{3} - \dfrac{\sqrt{3}}{2}\right) = \dfrac{\pi}{18} - \dfrac{\sqrt{3}}{12}$. Multiplying by four again gives $\dfrac{2\pi}{9} - \dfrac{\sqrt{3}}{3}$, confirming the answer by a second route.
CCSS standards used (min grade 8)
8.NS.A.2Use rational approximations of irrational numbers to compare their size (Comparing the radius $\dfrac{\sqrt{3}}{3}\approx 0.577$ with the half-side $0.5$ and half-diagonal $\dfrac{\sqrt{2}}{2}\approx 0.707$ to see that the circle pokes past the sides but not the corners.)7.G.B.6Solve real-world problems involving area, surface area, and volume (Using four-fold symmetry to split the target region into four identical circular segments and reducing the problem to measuring one.)8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Finding the half-height $h = \sqrt{r^2 - \tfrac{1}{4}} = \dfrac{1}{2\sqrt{3}}$ of the circle's opening at a side of the square.)8.G.A.5Use informal arguments to establish facts about angle sum and exterior angles (Recognizing that the triangle with all three sides equal to $r$ is equilateral, so its central angle is $60^\circ$.)7.G.B.4Know the formulas for area and circumference of a circle (Computing the circle's area $\pi r^2 = \dfrac{\pi}{3}$ and taking the one-sixth sector area $\dfrac{\pi}{18}$.)6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Computing the equilateral triangle's area $\dfrac{\sqrt{3}}{12}$ and subtracting it from the sector to get one segment, then scaling by four.)
⭐ The circle bulges past the middle of each side, so add up four matching slivers: take a sixth-of-the-circle pie slice, cut away its triangle, and multiply by four.
⭐ The circle bulges past the middle of each side, so add up four matching slivers: take a sixth-of-the-circle pie slice, cut away its triangle, and multiply by four.
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