AMC 10 · 2004 · #12

Grade 8 geometry-2d

Archetype Compound Area by Decomposition / Difference └ Sub-archetype Nested Frame / Concentric Difference

tangent-circlespythagorean-theoremarea-difference identify-subproblemsconvert-to-algebra ↑ Prerequisites: pythagorean-theorem

📏 Short solution 💡 2 insights 📊 Diagram

Problem

An annulus is the region between two concentric circles. The concentric circles in the figure have radii $b$ and $c$ , with $b>c$ . Let $OX$ be a radius of the larger circle, let $XZ$ be tangent to the smaller circle at $Z$ , and let $OY$ be the radius of the larger circle that contains $Z$ . Let $a=XZ$ , $d=YZ$ , and $e=XY$ . What is the area of the annulus?

Pick an answer.

(A)

$\pi a^2$

(B)

$\pi b^2$

(C)

$\pi c^2$

(D)

$\pi d^2$

(E)

$\pi e^2$

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.

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Toolkit + CCSS Solution

Understand

Restated: Two circles share the same center. The bigger one has radius b, the smaller one radius c. A line from a point X on the big circle just grazes the small circle at Z, and a=XZ is that grazing length. Find the area of the ring between the two circles.

Givens: Two concentric circles: outer radius b, inner radius c, with b > c.; OX is a radius of the larger circle, so OX = b.; XZ is tangent to the smaller circle at Z, with a = XZ.; OY is a radius of the larger circle passing through Z; d = YZ and e = XY.

Unknowns: The area of the annulus (the ring between the two circles).

Understand

Plan

Primary tool: #1 Draw a Diagram

Secondary: #7 Identify Subproblems, #13 Convert to Algebra

The ring's area is the big circle minus the small circle, so it depends on b squared minus c squared. Reading the figure shows a right triangle hiding in the tangent line, and that right triangle is exactly what turns b squared minus c squared into a single labeled length. Draw the triangle, then let algebra finish.

Execute — Answer: A

#7 Identify Subproblems 7.G.B.4 Step 1

Ring equals big minus small

The annulus is what is left when you cut the small disk out of the big disk.
The big circle has area pi times b squared and the small circle has area pi times c squared, so the ring's area is their difference.

$$\text{Area} = \pi b^2 - \pi c^2 = \pi\left(b^2 - c^2\right)$$

💡 A ring is just a big pancake with a smaller pancake punched out.

#1 Draw a Diagram 8.G.B.7 Step 2

Find the right triangle

Because XZ only touches the small circle at Z, the radius OZ meets XZ at a right angle.
That makes triangle OZX a right triangle with the right angle at Z.
Its legs are OZ = c and XZ = a, and its hypotenuse is OX = b.
The Pythagorean theorem then ties the three lengths together.

$$OX^2 = OZ^2 + XZ^2 \;\Rightarrow\; b^2 = c^2 + a^2$$

💡 A tangent line always makes a square corner with the radius it touches, and a square corner means Pythagoras.

#13 Convert to Algebra 6.EE.A.3 Step 3

Swap in a single length

Rearranging the Pythagorean relation gives b squared minus c squared equal to a squared.
That difference is exactly the quantity inside the ring's area, so replace it.
The ring's area becomes pi times a squared, which is choice (A).

$$b^2 - c^2 = a^2 \;\Rightarrow\; \text{Area} = \pi\left(b^2 - c^2\right) = \pi a^2$$

💡 The messy difference of two squares is secretly one clean square, so the ring's area is a plain circle of radius a.

[1] #7 7.G.B.4 The annulus is what is left when you cut the small disk out of the big disk. The

[2] #1 8.G.B.7 Because XZ only touches the small circle at Z, the radius OZ meets XZ at a right

[3] #13 6.EE.A.3 Rearranging the Pythagorean relation gives b squared minus c squared equal to a

Review

Reasonableness: The answer pi a squared is the area of a circle with radius a, and it uses only labeled lengths, as required. It correctly ignores d and e, which never entered the ring's area. A quick sanity check: if the inner circle shrinks toward the center (c toward 0), the tangent length a grows toward b, and pi a squared grows toward pi b squared, the full big circle, exactly as a vanishing hole should give.

Alternative: Introduce numbers: pick c = 3 and b = 5. Then a squared = b squared minus c squared = 25 minus 9 = 16, so a = 4. The ring's area is pi(25 - 9) = 16 pi, and pi a squared = 16 pi as well. Testing choices (B) through (E) with these numbers fails, confirming (A).

CCSS standards used (min grade 8)

7.G.B.4 Know the formulas for area and circumference of a circle (Writing each circle's area as pi times its radius squared and subtracting to get the ring's area.)
8.G.B.7 Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Relating the radii b and c to the tangent length a in the right triangle OZX.)
6.EE.A.3 Apply the properties of operations to generate equivalent expressions (Rearranging b squared minus c squared into a squared and substituting it into the area expression.)

⭐ A tangent line makes a right angle with the radius, so the Pythagorean theorem turns the ring's area into a single circle of radius a.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: A

Ring equals big minus small

Find the right triangle

Swap in a single length

Review

CCSS standards used (min grade 8)

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