🇺🇸 AMC 8 ⇄ switch contest
2013 AMC 8

Problem 23

Problem 23 · 2013 AMC 8 Hard
Geometry & Measurement semicircle-area-arcpythagorean-triple

Angle ABC of ▵ABC is a right angle. The sides of ▵ABC are the diameters of semicircles as shown. The area of the semicircle on AB equals 8π, and the arc of the semicircle on AC has length 8.5π. What is the radius of the semicircle on BC?

Figure for AMC 8 2013 Problem 23
Show answer
Answer: B — Radius 7.5.
Show hints
Hint 1 of 2
The semicircles are just disguises for the triangle's three sides. Each clue (an area, an arc length) is really telling you a side length — translate them back to AB and AC first, and a familiar right triangle appears.
Still stuck? Show hint 2 →
Hint 2 of 2
Pin down which radius-formula each clue uses: semicircle area = ½πr2, semicircle arc = πr. Recover the radii, double them for the diameters (= sides), then it's just the Pythagorean theorem.
Show solution
Approach: decode each semicircle into a side, then use Pythagoras
  1. Side AB: its semicircle has area ½πr2 = 8π, so r2 = 16, r = 4, and AB = 2·4 = 8.
  2. Side AC: its semicircle arc is πr = 8.5π, so r = 8.5 and AC = 2·8.5 = 17.
  3. Right angle at B makes AC the hypotenuse: BC = √(172 − 82) = √225 = 15. (Spotting the 8-15-17 triple skips the square root entirely.)
  4. The question wants the semicircle's radius on BC, so halve: 15 ÷ 2 = 7.5.
  5. Watch the trap: after all that work it's tempting to answer 15 (the side) — but the figure's circles ride on the sides, and the question asks for a radius, so the final halving matters.
Mark: · log in to save
↑