AMC 8 · 2013 · #23

Grade 8 geometry-2d

Archetype Arithmetic Expression Decomposition

pythagorean-theoremarea-circlesperimeter identify-subproblems ↑ Prerequisites: pythagorean-theoremarea-circlesperimeter

📏 Medium solution 💡 4 insights 📊 Diagram

Problem

Angle $ABC$ of $\triangle ABC$ is a right angle. The sides of $\triangle ABC$ are the diameters of semicircles as shown. The area of the semicircle on $\overline{AB}$ equals $8\pi$ , and the arc of the semicircle on $\overline{AC}$ has length $8.5\pi$ . What is the radius of the semicircle on $\overline{BC}$ ?

Pick an answer.

(A)

(B)

7.5

(C)

(D)

8.5

(E)

View mode:

Toolkit + CCSS Solution

Understand

Restated: Right triangle $\triangle ABC$ has its right angle at $B$. Each side is the diameter of a semicircle drawn outward. The semicircle on $\overline{AB}$ has area $8\pi$, and the semicircle on $\overline{AC}$ has arc length $8.5\pi$. Find the radius of the semicircle on $\overline{BC}$.

Givens: $\angle ABC = 90^\circ$, so $\overline{AC}$ is the hypotenuse; Each side of $\triangle ABC$ is the diameter of a semicircle on that side; Area of the semicircle on $\overline{AB}$ $= 8\pi$; Arc length of the semicircle on $\overline{AC}$ $= 8.5\pi$; Answer choices: (A) $7$, (B) $7.5$, (C) $8$, (D) $8.5$, (E) $9$

Unknowns: The radius of the semicircle on $\overline{BC}$

Understand

Plan

Primary tool: #1 Draw a Diagram

Secondary: #7 Identify Subproblems

The problem describes a figure, so Tool #1 (Draw a Diagram) — sketch the right triangle and label each side with its semicircle — makes the chain $\text{semicircle measurement} \to \text{radius} \to \text{diameter} \to \text{side of triangle}$ visible. Tool #7 (Identify Subproblems) then splits the work into three clean pieces: (1) recover $AB$ from the area, (2) recover $AC$ from the arc length, (3) use the Pythagorean theorem to get $BC$ and halve it for the radius. No algebra heavier than $r^2 = 16$ is needed.

Execute — Answer: B

#1 Draw a Diagram 7.G.B.4 Step 1

Sketch the figure: right triangle $\triangle ABC$ with the right angle at $B$, plus a semicircle on each side.
Label the radius of each semicircle $r_{AB}$, $r_{AC}$, $r_{BC}$.
Since each side IS the diameter, the side length equals $2r$.

$$AB = 2r_{AB}, \quad AC = 2r_{AC}, \quad BC = 2r_{BC}$$

💡 Drawing the diagram and writing each side as $2r$ makes it obvious what we need from each semicircle: just its radius.

#7 Identify Subproblems 7.G.B.4 Step 2

Use the semicircle area formula on $\overline{AB}$.
A full circle has area $\pi r^2$, so a semicircle has area $\tfrac{1}{2}\pi r^2$.
Setting this equal to $8\pi$ gives $r_{AB}^2 = 16$, so $r_{AB} = 4$ and $AB = 2 \cdot 4 = 8$.

$$\tfrac{1}{2}\pi r_{AB}^2 = 8\pi \;\Rightarrow\; r_{AB}^2 = 16 \;\Rightarrow\; r_{AB} = 4 \;\Rightarrow\; AB = 8$$

💡 Inverting the Grade 7 circle-area formula recovers the radius, and doubling gives the side length — a one-step subproblem.

#7 Identify Subproblems 7.G.B.4 Step 3

Use the semicircle arc-length formula on $\overline{AC}$.
A full circle's circumference is $2\pi r$, so a semicircle's arc is half of that: $\pi r$.
Setting $\pi r_{AC} = 8.5\pi$ gives $r_{AC} = 8.5$ and $AC = 2 \cdot 8.5 = 17$.

$$\pi r_{AC} = 8.5\pi \;\Rightarrow\; r_{AC} = 8.5 \;\Rightarrow\; AC = 17$$

💡 Same Grade 7 formula family — arc length instead of area — gives the hypotenuse directly.

#7 Identify Subproblems 8.G.B.7 Step 4

Apply the Pythagorean theorem with $AB$ and $AC$ in hand.
$\overline{AC}$ is the hypotenuse (it sits opposite the right angle at $B$), so $AB^2 + BC^2 = AC^2$.
With $AB = 8$ and $AC = 17$ this is the famous $8\text{-}15\text{-}17$ Pythagorean triple, giving $BC = 15$.

$$8^2 + BC^2 = 17^2 \;\Rightarrow\; BC^2 = 289 - 64 = 225 \;\Rightarrow\; BC = 15$$

💡 Recognizing $8$-$15$-$17$ as a Pythagorean triple skips the square-root step — pattern recognition at Grade 8 level.

#7 Identify Subproblems 7.G.B.4 Step 5

Finally, halve $BC$ to get the radius of its semicircle.
Since $BC$ is the diameter, $r_{BC} = BC/2 = 15/2 = 7.5$.

$$r_{BC} = \dfrac{BC}{2} = \dfrac{15}{2} = 7.5 \;\Rightarrow\; \textbf{(B)}$$

💡 The very last subproblem — diameter divided by two — closes the loop we set up in Step 1.

[1] #1 7.G.B.4 Sketch the figure: right triangle $\triangle ABC$ with the right angle at $B$, p

[2] #7 7.G.B.4 Use the semicircle area formula on $\overline{AB}$. A full circle has area $\pi

[3] #7 7.G.B.4 Use the semicircle arc-length formula on $\overline{AC}$. A full circle's circum

[4] #7 8.G.B.7 Apply the Pythagorean theorem with $AB$ and $AC$ in hand. $\overline{AC}$ is the

[5] #7 7.G.B.4 Finally, halve $BC$ to get the radius of its semicircle. Since $BC$ is the diame

Review

Reasonableness: Check the triangle sides are consistent: $8^2 + 15^2 = 64 + 225 = 289 = 17^2$ ✓ — the famous $8\text{-}15\text{-}17$ right triangle. The semicircle on $\overline{BC}$ should sit between the other two in size: $r_{AB} = 4$, $r_{BC} = 7.5$, $r_{AC} = 8.5$, and indeed $4 < 7.5 < 8.5$. Choice (B) lands in the expected range and equals exactly half of $15$.

Alternative: Tool #3 (Eliminate Possibilities) on the choices: each candidate $r_{BC}$ gives $BC = 2r_{BC}$, so we test whether $AB^2 + BC^2 = AC^2$ holds with $AB = 8$, $AC = 17$. (A) $r=7$ gives $BC=14$, $8^2 + 14^2 = 260 \neq 289$. (B) $r=7.5$ gives $BC=15$, $8^2 + 15^2 = 289 = 17^2$ ✓. (C)-(E) give $BC = 16, 17, 18$, all too large. Only (B) satisfies the Pythagorean check.

CCSS standards used (min grade 8)

7.G.B.4 Know the formulas for the area and circumference of a circle and use them to solve problems (Inverting the semicircle area formula $\tfrac{1}{2}\pi r^2 = 8\pi$ to get $r_{AB} = 4$, inverting the semicircle arc length $\pi r = 8.5\pi$ to get $r_{AC} = 8.5$, and halving $BC$ to get $r_{BC}$.)
8.G.B.7 Apply the Pythagorean Theorem to determine unknown side lengths in right triangles (Solving $8^2 + BC^2 = 17^2$ for $BC = 15$, the missing leg of the right triangle (the classic $8$-$15$-$17$ Pythagorean triple).)

⭐ Each side is a diameter, so each semicircle's measurement (area or arc length) hands you a side of the triangle — then the Pythagorean theorem finishes the job. It's a tidy Grade 8 idea built from two Grade 7 circle formulas.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 8)

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