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2008 AMC 10A Problem 16

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2008 AMC 10A Problem 16AMC 10 Step by Step
Name:                             Date: September 13, 2026
Circles · difficulty 3 of 5 · about 3 min · position P16-20

Points AA and BB lie on a circle centered at OO , and ∠AOB=60∘\angle AOB = 60^\circ . A second circle is internally tangent to the first and tangent to both OA‾\overline{OA} and OB‾\overline{OB} . What is the ratio of the area of the smaller circle to that of the larger circle?

  1. A)

    116\frac{1}{16}

  2. B)

    19\frac{1}{9}

  3. C)

    18\frac{1}{8}

  4. D)

    16\frac{1}{6}

  5. E)

    14\frac{1}{4}

Answer

B

Key insight

The small center lies on the 30-degree bisector, so its distance from O is twice its radius r; internal tangency gives 2r + r = R, hence r = R/3.

Solution

Let the large circle have radius RR and the small one radius rr with center PP.

Since the small circle is tangent to both OAOA and OBOB, its center is equidistant from the two rays, so PP lies on the bisector of ∠AOB\angle AOB, making ∠POA=30∘\angle POA = 30^\circ. Drop the perpendicular from PP to OAOA; its foot is the tangency point, so the perpendicular has length rr. In the resulting 3030-6060-9090 triangle the hypotenuse OPOP is twice the short leg:
OP=2r.OP = 2r.

Internal tangency means the two centers are R−rR - r apart, i.e. OP=R−rOP = R - r. Therefore 2r=R−r2r = R - r, so r=R3r = \frac{R}{3}.

The area ratio is (rR)2=19\left(\frac{r}{R}\right)^2 = \frac{1}{9}.

The answer is (B) 19\boxed{\textbf{(B)}\ \frac{1}{9}}.

Why this works

Two standard facts do all the work: a circle tangent to both sides of an angle has its center on the bisector (giving a right triangle with a known angle), and internally tangent circles have centers separated by the difference of the radii. Draw the center, the tangency point and the line of centers; the rest is a 3030-6060-9090 triangle. Area ratios are the square of length ratios.

The trap

Forgetting to add the small radius when using internal tangency (writing OP = R instead of R - r), which gives r = R/2 and a ratio of 1/4.

Common mistakes

  • Forgetting to add the small radius when using internal tangency (writing OP = R instead of R - r), which gives r = R/2 and a ratio of 1/4.
  • Reporting the radius ratio 13\frac13 rather than the area ratio 19\frac19.

Techniques

Add construction lines/points (drop altitudes, extend segments, connect centers)