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2014 AMC 10A Problem 12

2014 AMC 10A Problem 12AMC 10 Step by Step

Name: Date: October 28, 2026

Circles · difficulty 2 of 5 · about 3 min · position P11-15

A regular hexagon has side length 6. Congruent arcs with radius 3 are drawn with the center at each of the vertices, creating circular sectors as shown. The region inside the hexagon but outside the sectors is shaded as shown. What is the area of the shaded region?

A) $27\sqrt{3}-9\pi$
B) $27\sqrt{3}-6\pi$
C) $54\sqrt{3}-18\pi$
D) $54\sqrt{3}-12\pi$
E) $108\sqrt{3}-9\pi$

Answer

Key insight

Each sector opens as wide as the hexagon's corner, 120 degrees, which is one third of a circle. Six thirds make two whole circles of radius 3.

Solution

The shaded part is the hexagon with the six sectors taken out, so find the two areas and subtract.

The hexagon.

A regular hexagon with side $6$ splits into six equilateral triangles with side $6$ by drawing lines from the center to each corner.
An equilateral triangle with side $s$ has area $\frac{\sqrt3}{4}s^2$ . With $s = 6$ that is $\frac{\sqrt3}{4} \cdot 36 = 9\sqrt3$ .
Six of them: $6 \cdot 9\sqrt3 = 54\sqrt3$ .

The sectors.

Each sector sits in a corner of the hexagon, between the two sides that meet there. So it opens exactly as wide as the corner angle.
The corner angles of a hexagon add up to $(6 - 2) \cdot 180^\circ = 720^\circ$ , so each corner is $720^\circ \div 6 = 120^\circ$ . That is $\frac{120}{360} = \frac13$ of a full circle.
A full circle of radius $3$ has area $9\pi$ , so each sector has area $\frac13 \cdot 9\pi = 3\pi$ . Six sectors: $18\pi$ .
The sectors do not overlap: the radius $3$ is half of a side, so two neighbouring sectors just meet at the middle of the side between them.

Shaded area. $54\sqrt3 - 18\pi$ .

The answer is $\boxed{\textbf{(C)}\ 54\sqrt{3}-18\pi}$ .

Why this works

"Inside the shape but outside the sectors" means subtract. The only thing to be careful about is the angle of each sector, which is the polygon's corner angle, not the angle at the center. A quick check: the six corner angles add to $720^\circ$ , which is exactly two full circles, so the sectors together make two circles of radius $3$ , area $2 \cdot 9\pi = 18\pi$ .

The trap

Using 60 degrees for each sector (the angle at the hexagon's center) instead of the 120-degree corner angle, which gives 54 root 3 minus 9 pi.

Concise solution

Hexagon. A regular hexagon of side $6$ splits into six equilateral triangles of side $6$ , each of area $\frac{\sqrt3}{4}\cdot 36 = 9\sqrt3$ . Total: $54\sqrt3$ .

Sectors. At each vertex the sector is bounded by the two sides meeting there, so its angle is the interior angle of the hexagon, $120^\circ$ , one third of a circle. Each sector has area $\frac13 \cdot \pi \cdot 3^2 = 3\pi$ . Six of them give $18\pi$ . (Since the radius $3$ is half the side, neighboring sectors just touch at the midpoints of the sides and do not overlap.)

Shaded region. Subtract:
$54\sqrt3 - 18\pi.$

The answer is $\boxed{\textbf{(C)}\ 54\sqrt{3}-18\pi}$ .

Review in the AoPS books

Introduction to Geometry, Chapter 11: Circles
Introduction to Geometry, Chapter 8: Quadrilaterals
Introduction to Geometry, Chapter 2: Angles
Introduction to Geometry, Chapter 4: Perimeter and Area (§4.2, §4.3)

Techniques

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