Competition · AMC preparation · step 4 of 4

AMC 8 · 2013 · #23

Grade 8 geometry-2d
Archetype Arithmetic Expression Decomposition └ Sub-archetype Multi-Segment Path or Count Sum
pythagorean-theoremarea-circlesperimeter identify-subproblems ↑ Prerequisites: pythagorean-theoremarea-circlesperimeter
📏 Medium solution 💡 4 insights 📊 Diagram
Problem
Right triangle △ ABC has its right angle at B. Each side is the diameter of a semicircle drawn outward. The semicircle on AB has area 8π, and the semicircle on AC has arc length 8.5π. Find the radius of the semicircle on BC.

Pick an answer.

(A)
7
(B)
7.5
(C)
8
(D)
8.5
(E)
9

AMC 8 2013 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Draw a Diagram

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 semicircle measurement → radius → diameter → 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 r² = 16 is needed.

1STEP 1

Sketch the triangle and semicircles

Sketch the right triangle with a semicircle on each side; since each side IS the diameter, every side length equals 2r.

AB = 2r_AB, AC = 2r_AC, BC = 2r_BC
2STEP 2

Get AB from its area

Semicircle area 12\frac{1}{2}π r² = 8π gives r_AB = 4, so AB = 8.

1/2π r_AB² = 8π → r_AB² = 16 → r_AB = 4 → AB = 8
3STEP 3

Get AC from its arc

Semicircle arc length π r = 8.5π gives r_AC = 8.5, so AC = 17.

π r_AC = 8.5π → r_AC = 8.5 → AC = 17
4STEP 4

Use the Pythagorean theorem

AC is the hypotenuse, so 8² + BC² = 17² — the 8-15-17 triple gives BC = 15.

8² + BC² = 17² → BC² = 289 - 64 = 225 → BC = 15
5STEP 5

Halve BC for the radius

BC is the diameter, so halve it: r_BC = 152\frac{15}{2} = 7.5.

r_BC = BC/2 = 15/2 = 7.5 → (B)
Answer
7.5
Check the triangle sides are consistent: 8² + 15² = 64 + 225 = 289 = 17² ✓ — the famous 8-15-17 right triangle. The semicircle on 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.
💡Key takeaway

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.

  • Sketch the triangle and semicircles
  • Get AB from its area
  • Get AC from its arc
  • Use the Pythagorean theorem
  • Halve BC for the radius

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