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2008 AMC 10A Problem 17

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2008 AMC 10A Problem 17AMC 10 Step by Step
Name:                             Date: October 28, 2026
Circles · difficulty 3 of 5 · about 4 min · position P16-20

An equilateral triangle has side length 6. What is the area of the region containing all points that are outside the triangle but not more than 3 units from a point on the triangle?

  1. A)36+24336+24\sqrt{3}
  2. B)54+9π54+9\pi
  3. C)54+183+6π54+18\sqrt{3}+6\pi
  4. D)(23+3)2π\left(2\sqrt{3}+3\right)^2\pi
  5. E)9(3+1)2π9\left(\sqrt{3}+1\right)^2\pi

Answer

B

Key insight

The band around the triangle is three 6 by 3 rectangles, one along each side, plus three rounded corner pieces that fit together into one full circle of radius 3.

Solution

Picture a band, 33 units wide, wrapped around the outside of the triangle. Split it into straight parts and corner parts.

1. Straight parts. Next to each side, the points within 33 units of that side form a rectangle 66 long and 33 wide. Three sides give
3×(6×3)=54.3 \times (6 \times 3) = 54.
2. Corner parts. At each corner of the triangle, the two rectangles do not meet; there is a gap. Points in the gap are within 33 units of the corner point itself, so the gap is a slice of a circle of radius 33.
3. How big is each slice? Around the corner point there are 360∘360^\circ in total. The triangle takes 60∘60^\circ, and the two rectangles each take 90∘90^\circ. The slice gets what is left:
360∘−60∘−90∘−90∘=120∘.360^\circ - 60^\circ - 90^\circ - 90^\circ = 120^\circ.
4. Three slices of 120∘120^\circ make 360∘360^\circ, a full circle of radius 33. Its area is
π×32=9π.\pi \times 3^2 = 9\pi.
5. The whole band has area 54+9π54 + 9\pi.

The answer is (B) 54+9π\boxed{\textbf{(B)}\ 54+9\pi}.

Why this works

A band of fixed width around any shape with straight sides and no dents is always rectangles on the sides plus circle slices at the corners. The slices always add up to one full circle, because walking once around the shape you turn a full 360∘360^\circ. So the band's area is (perimeter) × (width) + π\pi × (width)2^2; here 18×3+9π18 \times 3 + 9\pi.

The trap

Using 60 degree pieces at the corners (the triangle's own angle) instead of 120 degree pieces, or forgetting the corner pieces entirely.

Concise solution

Picture the band of width 33 wrapped around the outside of the triangle. It splits into two kinds of pieces.

Along each side, the points within 33 units of that side (measured perpendicularly, outside the triangle) form a 6×36 \times 3 rectangle. Three sides give
3⋅(6⋅3)=54.3 \cdot (6 \cdot 3) = 54.

At each vertex the two rectangles leave a wedge-shaped gap. Points there are within 33 of the vertex itself, so the gap is a sector of radius 33. Its angle is 360∘360^\circ minus the triangle's 60∘60^\circ interior angle minus the two 90∘90^\circ rectangle corners, which is 120∘120^\circ. Three such sectors total 360∘360^\circ, a full disk of radius 33:
π⋅32=9π.\pi \cdot 3^2 = 9\pi.

The region has area 54+9π54 + 9\pi, so the answer is (B) 54+9π\boxed{\textbf{(B)}\ 54+9\pi}.

Review in the AoPS books

Techniques

Cut the figure into known shapes (triangles, rectangles, sectors)