AMC 10 · 2009 · #21

Grade 8 geometry-2d
Archetype Similar-Triangle Length / Area Ratio └ Sub-archetype Synthetic Similar-Triangle Length or Area
area-circlestangent-circlespythagorean-theoremratio-proportion convert-to-algebraidentify-subproblems ↑ Prerequisites: area-circles
📏 Medium solution 💡 3 insights 📊 Diagram
Problem
A large circle holds a ring of four congruent smaller circles. Each small circle touches the large circle from the inside, and each small circle touches its two neighbors. Find the ratio of the combined area of the four small circles to the area of the large circle.

Pick an answer.

(A)
$\ 3-2\sqrt2$
(B)
$\ 2-\sqrt2$
(C)
$\ 4(3-2\sqrt2)$
(D)
$\ \frac12(3-\sqrt2)$
(E)
$\ 2\sqrt2-2$

AMC 10 2009 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Introduce a Variable

No actual size is given, and the answer is a pure ratio, so the picture works at any scale. Tool #4 (Introduce a Variable) names the small radius r; every other length is then forced by the tangency rules, and the r cancels at the end, so the ratio is a single number. Tool #1 (Draw a Diagram) adds the radii that expose a right triangle linking the two circle sizes, and Tool #7 (Identify Subproblems) splits the work into three clean pieces: find the large radius, write both areas, then divide.

1STEP 1

Name the small radius

Let each small circle have radius r. Tangent neighbors put their centers r+r apart, so the four centers form a square of side 2r.

distance between adjacent centers=r+r=2r
2STEP 2

Find the center-to-center distance

By symmetry the large circle's center O is the square's center, so two adjacent centers give d²+d²=(2r)², hence d=r√2.

d²+d²=(2r)² → 2d²=4r² → d=r√2
3STEP 3

Build the large radius

Internal tangency puts O, a small center, and its touch point on one line, so R=r√2+r=r(1+√2).

R=r√2+r=r(1+√2)
4STEP 4

Write both areas

Four small circles total 4π r²; the large one is π r²(1+√2)²=π r²(3+2√2).

4π r² and π r²(1+√2)²=π r²(3+2√2)
5STEP 5

Divide and clean up the answer

π r² cancels: 4/(3+2√2); multiply top and bottom by 3-2√2 and the bottom becomes 9-8=1, leaving 4(3-2√2) — choice (C).

4/(3+2√2)·(3-2√2)/(3-2√2)=(4(3-2√2))/(9-8)=4(3-2√2) → (C)
Answer
4(3-2√2)
Estimate with √2≈1.414: the ratio 4(3-2√2)≈4(3-2.828)=4(0.172)=0.686. So the four small circles cover about 69% of the large circle, leaving roughly 31% as the gaps between them and the rim. That is a believable amount of empty space for a ring of four circles inside one big circle, and the value sits between 0 and 1 as any area ratio must. A quick sanity check with radius numbers: if r=1 then R=1+√2≈2.414, small total area ≈12.57, large area ≈18.30, ratio ≈0.687, matching (C).
💡Key takeaway

Give the small circle a radius of r, let tangency force every other length, and the r cancels when you divide the areas so only the shape of the ratio survives.

  • Name the small radius
  • Find the center-to-center distance
  • Build the large radius
  • Write both areas
  • Divide and clean up the answer

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