AMC 10 · 2004 · #21
Grade 8 geometry-2dTwo distinct lines pass through the center of three concentric circles of radii 3, 2, and 1. The area of the shaded region in the diagram is 138 of the area of the unshaded region. What is the radian measure of the acute angle formed by the two lines? (Note: π radian is 180 degree.)
Pick an answer.
AMC 10 2004 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: Three circles share the same center and have radii $3$, $2$, and $1$. Two different straight lines both pass through that center, cutting the whole picture into wedges. Following the diagram's shading, the shaded part turns out to be $\frac{8}{13}$ of the unshaded part. Find the size, in radians, of the acute angle between the two lines.
Givens: Three concentric circles with radii $3$, $2$, and $1$, so the circles split the big disk into three rings: the inner disk (radius up to $1$), the middle ring (between radius $1$ and $2$), and the outer ring (between radius $2$ and $3$).; Two distinct lines pass through the common center; the acute angle between them is the unknown.; The two lines cut every ring into four wedges: two equal 'narrow' wedges opposite each other, and two equal 'wide' wedges opposite each other.; The shaded region equals $\frac{8}{13}$ of the unshaded region.; Reminder given in the problem: $\pi$ radians $= 180$ degrees.; Answer choices: (A) $\frac{\pi}{8}$, (B) $\frac{\pi}{7}$, (C) $\frac{\pi}{6}$, (D) $\frac{\pi}{5}$, (E) $\frac{\pi}{4}$
Unknowns: The radian measure of the acute angle formed by the two lines.
Understand
Restated: Three circles share the same center and have radii $3$, $2$, and $1$. Two different straight lines both pass through that center, cutting the whole picture into wedges. Following the diagram's shading, the shaded part turns out to be $\frac{8}{13}$ of the unshaded part. Find the size, in radians, of the acute angle between the two lines.
Givens: Three concentric circles with radii $3$, $2$, and $1$, so the circles split the big disk into three rings: the inner disk (radius up to $1$), the middle ring (between radius $1$ and $2$), and the outer ring (between radius $2$ and $3$).; Two distinct lines pass through the common center; the acute angle between them is the unknown.; The two lines cut every ring into four wedges: two equal 'narrow' wedges opposite each other, and two equal 'wide' wedges opposite each other.; The shaded region equals $\frac{8}{13}$ of the unshaded region.; Reminder given in the problem: $\pi$ radians $= 180$ degrees.; Answer choices: (A) $\frac{\pi}{8}$, (B) $\frac{\pi}{7}$, (C) $\frac{\pi}{6}$, (D) $\frac{\pi}{5}$, (E) $\frac{\pi}{4}$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #7 Identify Subproblems, #1 Draw a Diagram, #13 Convert to Algebra
The whole shaded area depends on just one thing: the acute angle. So name it $\theta$ (Tool #4) and everything else can be written in terms of it. The picture looks busy, so break it into the three rings the circles create (Tool #7); each ring's area is easy, and inside each ring the shaded part is just a fraction set by $\theta$. Reading the diagram carefully (Tool #1) shows the shading flips between rings — narrow wedges in the outer and inner rings, wide wedges in the middle ring — which is exactly what makes the totals interesting. Once the shaded area is a formula in $\theta$, the condition 'shaded is $\frac{8}{13}$ of unshaded' becomes one equation to solve (Tool #13).
Execute — Answer: B
7.G.B.4 Step 1 Split the picture into three rings
- The three circles have radii $3$, $2$, and $1$, so the full picture is one big disk of radius $3$ with total area $\pi(3)^2 = 9\pi$.
- The circles carve it into three rings.
- The inner disk (radius $1$) has area $\pi(1)^2 = \pi$.
- The middle ring (between radii $1$ and $2$) has area $\pi(2)^2 - \pi(1)^2 = 4\pi - \pi = 3\pi$.
- The outer ring (between radii $2$ and $3$) has area $\pi(3)^2 - \pi(2)^2 = 9\pi - 4\pi = 5\pi$.
- As a check, $\pi + 3\pi + 5\pi = 9\pi$, the whole disk.
💡 A ring's area is the big circle minus the smaller circle inside it, so three subtractions give all three pieces.
7.RP.A.2 Step 2 Name the angle and measure each wedge
- Let $\theta$ be the acute angle between the two lines.
- The two lines cross at the center and split each ring into four wedges: two 'narrow' wedges (opposite each other) each of angle $\theta$, and two 'wide' wedges (opposite each other) each of angle $\pi-\theta$, since going once around is $2\pi$ and $\theta+(\pi-\theta)+\theta+(\pi-\theta)=2\pi$.
- A wedge takes up the same fraction of its ring as its angle takes up of a full turn.
- So the two narrow wedges together are the fraction $\frac{2\theta}{2\pi}=\frac{\theta}{\pi}$ of a ring, and the two wide wedges together are $\frac{2(\pi-\theta)}{2\pi}=\frac{\pi-\theta}{\pi}$.
💡 Area of a pie slice grows in step with its angle, so the angle fraction is the area fraction.
7.EE.A.1 Step 3 Add up the shaded area
- Now read the shading.
- In the outer ring the narrow wedges are shaded, giving $5\pi \cdot \frac{\theta}{\pi} = 5\theta$.
- In the middle ring the wide wedges are shaded, giving $3\pi \cdot \frac{\pi-\theta}{\pi} = 3(\pi-\theta)$.
- In the inner disk the narrow wedges are shaded again, giving $\pi \cdot \frac{\theta}{\pi} = \theta$.
- Add these three shaded pieces: $5\theta + 3(\pi-\theta) + \theta = 5\theta + 3\pi - 3\theta + \theta = 3\pi + 3\theta$.
- So the shaded area is $3\pi + 3\theta$, and the unshaded area is everything else, $9\pi - (3\pi + 3\theta) = 6\pi - 3\theta$.
💡 Each ring contributes its area times the shaded-angle fraction, and the $\pi$'s cancel so only clean multiples of $\theta$ and $\pi$ remain.
8.EE.C.7 Step 4 Set up the ratio and solve
- The shaded area is $\frac{8}{13}$ of the unshaded area, so $3\pi + 3\theta = \frac{8}{13}\,(6\pi - 3\theta)$.
- Multiply both sides by $13$ to clear the fraction: $13(3\pi + 3\theta) = 8(6\pi - 3\theta)$, that is $39\pi + 39\theta = 48\pi - 24\theta$.
- Gather the $\theta$ terms on one side and the $\pi$ terms on the other: $39\theta + 24\theta = 48\pi - 39\pi$, so $63\theta = 9\pi$.
- Dividing gives $\theta = \frac{9\pi}{63} = \frac{\pi}{7}$.
- The acute angle is $\frac{\pi}{7}$, which is choice (B).
💡 Treat $\pi$ as a fixed number and the ratio condition becomes an ordinary linear equation in $\theta$.
7.G.B.4 The three circles have radii $3$, $2$, and $1$, so the full picture is one big d 7.RP.A.2 Let $\theta$ be the acute angle between the two lines. The two lines cross at th 7.EE.A.1 Now read the shading. In the outer ring the narrow wedges are shaded, giving $5\ 8.EE.C.7 The shaded area is $\frac{8}{13}$ of the unshaded area, so $3\pi + 3\theta = \fr Review
Reasonableness: Plug $\theta=\frac{\pi}{7}$ back in. Shaded $=3\pi+3\cdot\frac{\pi}{7}=\frac{21\pi+3\pi}{7}=\frac{24\pi}{7}$. Unshaded $=6\pi-3\cdot\frac{\pi}{7}=\frac{42\pi-3\pi}{7}=\frac{39\pi}{7}$. Their ratio is $\frac{24\pi/7}{39\pi/7}=\frac{24}{39}=\frac{8}{13}$, exactly as required. Also shaded plus unshaded is $\frac{24\pi}{7}+\frac{39\pi}{7}=\frac{63\pi}{7}=9\pi$, the full disk, so nothing was lost or double-counted.
Alternative: Work with the whole-disk total instead of separate rings. The shaded area must equal $\frac{8}{8+13}=\frac{8}{21}$ of the total, since shaded : unshaded $=8:13$ means shaded is $8$ parts out of $21$. So shaded $=\frac{8}{21}\cdot 9\pi=\frac{24\pi}{7}$. Setting the earlier formula equal to this, $3\pi+3\theta=\frac{24\pi}{7}$, gives $3\theta=\frac{24\pi}{7}-\frac{21\pi}{7}=\frac{3\pi}{7}$, so $\theta=\frac{\pi}{7}$ again.
CCSS standards used (min grade 8)
7.G.B.4Know the formulas for area and circumference of a circle (Finding the total area $9\pi$ and the three ring areas $\pi$, $3\pi$, and $5\pi$ by subtracting circle areas.)7.RP.A.2Recognize and represent proportional relationships between quantities (Turning each wedge's central angle into an area fraction, since sector area is proportional to angle: narrow wedges are $\frac{\theta}{\pi}$ of a ring, wide wedges are $\frac{\pi-\theta}{\pi}$.)7.EE.A.1Apply properties of operations to add, subtract, factor, and expand linear expressions (Combining the three shaded pieces $5\theta+3(\pi-\theta)+\theta$ into $3\pi+3\theta$ and forming the unshaded expression $6\pi-3\theta$.)8.EE.C.7Solve linear equations in one variable (Solving $3\pi+3\theta=\frac{8}{13}(6\pi-3\theta)$ to reach $63\theta=9\pi$ and $\theta=\frac{\pi}{7}$.)
⭐ When wedges meet at a center, a slice's area is just its angle's share of a full turn, so you can trade angles for areas and let one equation finish the job.
⭐ When wedges meet at a center, a slice's area is just its angle's share of a full turn, so you can trade angles for areas and let one equation finish the job.
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