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2006 AMC 10B Problem 6

2006 AMC 10B Problem 6AMC 10 Step by Step

Name: Date: September 13, 2026

Circles · difficulty 1 of 5 · about 1 min · position P1-10

A region is bounded by semicircular arcs constructed on the side of a square whose sides measure $\frac{2}{\pi}$ , as shown. What is the perimeter of this region?

A)
$\frac{4}{\pi}$
B)
$2$
C)
$\frac{8}{\pi}$
D)
$4$
E)
$\frac{16}{\pi}$

Answer

Key insight

Each side of the square is a diameter, so each semicircular arc has length (pi/2) times the side, which is exactly 1; four arcs give 4.

Solution

The boundary of the region consists only of the four semicircular arcs; the square's sides are interior to the region.

Each arc is built on a side of the square, so that side is the arc's diameter, $d = \frac{2}{\pi}$ . A full circle of diameter $d$ has circumference $\pi d$ , so a semicircle has arc length
$\frac{\pi d}{2} = \frac{\pi}{2} \cdot \frac{2}{\pi} = 1.$

Four arcs give a perimeter of $4 \cdot 1 = 4$ .

The answer is $\boxed{\textbf{(D)}\ 4}$ .

Why this works

The side length $\frac{2}{\pi}$ is chosen so that $\pi$ cancels, a strong hint that circumference (which carries a factor of $\pi$ ) is the quantity in play. Perimeter means the outer boundary only, so segments that lie inside the region contribute nothing.

The trap

Including the sides of the square in the perimeter, or using the side length as the radius instead of the diameter.

Common mistakes

Including the sides of the square in the perimeter, or using the side length as the radius instead of the diameter.
Computing a full circumference per arc ( $2$ each) and answering $8/\pi$ or $8$ .

Techniques

Set up the equation/formula and compute; no special trick needed