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2015 AMC 10B Problem 9

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2015 AMC 10B Problem 9AMC 10 Step by Step
Name:                             Date: October 28, 2026
Circles · difficulty 2 of 5 · about 2 min · position P1-10

The shaded region below is called a shark's fin falcata, a figure studied by Leonardo da Vinci. It is bounded by the portion of the circle of radius 33 and center (0,0)(0,0) that lies in the first quadrant, the portion of the circle with radius 32\tfrac{3}{2} and center (0,32)(0,\tfrac{3}{2}) that lies in the first quadrant, and the line segment from (0,0)(0,0) to (3,0)(3,0) . What is the area of the shark's fin falcata?

figure

  1. A)4π5\dfrac{4\pi}{5}
  2. B)9π8\dfrac{9\pi}{8}
  3. C)4π3\dfrac{4\pi}{3}
  4. D)7π5\dfrac{7\pi}{5}
  5. E)3π2\dfrac{3\pi}{2}

Answer

B

Key insight

The shaded fin is a quarter of the big circle with half of the small circle cut out: 9pi/4 - 9pi/8 = 9pi/8.

Solution

The fin is what is left of a quarter circle after a smaller round piece is cut out of it. So find the two areas and subtract.

1. The big piece. The circle of radius 33 centered at (0,0)(0,0), cut to the first quadrant, is a quarter of a circle. Its area is
14⋅π⋅32=9π4.\frac14 \cdot \pi \cdot 3^2 = \frac{9\pi}{4}.
2. The piece cut out. The small circle has center (0,32)(0, \tfrac32) and radius 32\tfrac32. Its lowest point is (0,0)(0, 0) and its highest point is (0,3)(0, 3), so it sits on the yy-axis. The yy-axis cuts it exactly in half, and the right half is the part inside the first quadrant. Its area is
12⋅π⋅(32)2=12⋅9π4=9π8.\frac12 \cdot \pi \cdot \left(\frac32\right)^2 = \frac12 \cdot \frac{9\pi}{4} = \frac{9\pi}{8}.
3. Subtract.
9π4−9π8=18π8−9π8=9π8.\frac{9\pi}{4} - \frac{9\pi}{8} = \frac{18\pi}{8} - \frac{9\pi}{8} = \frac{9\pi}{8}.

The answer is (B) 9π8\boxed{\textbf{(B)}\ \frac{9\pi}{8}}.

Why this works

Shapes bounded by circle arcs are usually "a bigger round piece minus a smaller round piece". For each arc, find its circle, then decide what fraction of that circle you need by looking at where the center is. Here the small circle's center is on the yy-axis, so the yy-axis splits it into two equal halves.

The trap

Cutting out a quarter of the small circle instead of half. Its center is on the y-axis, so exactly half of it lies in the first quadrant.

Concise solution

The large circle, centered at the origin with radius 33, contributes the quarter disk in the first quadrant:
14⋅π⋅32=9π4.\frac14 \cdot \pi \cdot 3^2 = \frac{9\pi}{4}.

The small circle has center (0,32)(0, \tfrac32) and radius 32\tfrac32, so it passes through (0,0)(0,0) and (0,3)(0,3) and its center sits on the yy-axis. Exactly half of it lies in the first quadrant (the right half), and that half disk is cut out of the quarter disk:
12⋅π(32)2=9π8.\frac12 \cdot \pi \left(\frac32\right)^2 = \frac{9\pi}{8}.

Falcata area:
9π4−9π8=9π8.\frac{9\pi}{4} - \frac{9\pi}{8} = \frac{9\pi}{8}.

The answer is (B) 9π8\boxed{\textbf{(B)}\ \frac{9\pi}{8}}.

Review in the AoPS books

Techniques

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