AMC 10 · 2008 · #16

Grade 8 geometry-2d
Archetype Similar-Triangle Length / Area Ratio └ Sub-archetype Synthetic Similar-Triangle Length or Area
area-circlestangent-circlesthirty-sixty-ninety-triangle physical-representationconvert-to-algebra ↑ Prerequisites: area-circles
📏 Medium solution 💡 2 insights
Problem
Points A and B sit on a circle centered at O, with angle AOB equal to 60 degrees. A second, smaller circle touches the big circle from the inside and also touches both rays OA and OB. Find the ratio of the small circle's area to the big circle's area.

Pick an answer.

(A)
$\ \frac{1}{16}$
(B)
$\ \frac{1}{9}$
(C)
$\ \frac{1}{8}$
(D)
$\ \frac{1}{6}$
(E)
$\ \frac{1}{4}$

AMC 10 2008 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Draw a Diagram

The whole problem is about how one circle sits against two rays and inside another circle, so a clean picture is the key. Once it is drawn, two facts fall out: touching both rays forces the small center onto the line that bisects the 60 degree angle, and touching the big circle from inside links the two radii through the distance between centers. Name the two radii, build a right triangle to relate them, then compare their squares to get the area ratio.

1STEP 1

Place the small center on the bisector

A circle touching both rays sits the same distance from each, so its center P lies on the bisector: OP makes 30 degrees with each ray.

∠ AOP = ∠ BOP = 30°
2STEP 2

Relate the radii with a right triangle

Drop PT perpendicular to OA: PT is r and the angle at O is 30 degrees, so OTP is a 30-60-90 triangle and OP is 2r.

PT = r, ∠ POT = 30° → OP = 2r
3STEP 3

Use the inside-touch to link the radii

Touching from inside makes OP equal R minus r, and OP is also 2r, so 3r equals R and the radii are in ratio 1 to 3.

OP = R - r = 2r → 3r = R → r = R/3
4STEP 4

Compare the areas

The pi cancels, so the area ratio is the square of the radius ratio: 19\frac{1}{9}, which is choice (B).

(π r²)/(π R²) = (r/R)² = (1/3)² = 1/9
Answer
1/9
The small radius came out to one third of the big radius, which looks right on the picture: the small circle is clearly much less than half the big one but not tiny. Squaring the one-third gives one ninth, a sensible slice of the area. A quick sanity check on the tangency: with r = R/3, the center sits at OP = 2R/3, and 2R/3 + R/3 = R reaches exactly the far edge of the big circle, confirming the inside touch. So (B) 1/9 is consistent.
💡Key takeaway

Touching both sides of an angle puts a circle on the middle line, and once you know a radius is one third, the area is one ninth because area follows the radius squared.

  • Place the small center on the bisector
  • Relate the radii with a right triangle
  • Use the inside-touch to link the radii
  • Compare the areas

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