AMC 10 · 2009 · #21
Grade 8 geometry-2dMany Gothic cathedrals have windows with portions containing a ring of congruent circles that are circumscribed by a larger circle, In the figure shown, the number of smaller circles is four. What is the ratio of the sum of the areas of the four smaller circles to the area of the larger circle?
Pick an answer.
AMC 10 2009 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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.
Givens: There are four congruent small circles arranged in a ring; Each small circle is internally tangent to the large circle; Neighboring small circles are tangent to each other; In the figure the four small-circle centers sit on the up, down, left, and right directions from the large circle's center; Answer choices: (A) $3-2\sqrt2$, (B) $2-\sqrt2$, (C) $4(3-2\sqrt2)$, (D) $\tfrac12(3-\sqrt2)$, (E) $2\sqrt2-2$
Unknowns: The ratio (sum of the four small circle areas) : (area of the large circle)
Understand
Restated: 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.
Givens: There are four congruent small circles arranged in a ring; Each small circle is internally tangent to the large circle; Neighboring small circles are tangent to each other; In the figure the four small-circle centers sit on the up, down, left, and right directions from the large circle's center; Answer choices: (A) $3-2\sqrt2$, (B) $2-\sqrt2$, (C) $4(3-2\sqrt2)$, (D) $\tfrac12(3-\sqrt2)$, (E) $2\sqrt2-2$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #1 Draw a Diagram, #7 Identify Subproblems
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.
Execute — Answer: C
6.EE.B.6 Step 1 Name the small radius
- Let each small circle have radius $r$.
- Two neighboring small circles are tangent, so the straight line joining their centers passes through the point where they touch, and its length is one radius plus the other: $r+r=2r$.
- So every pair of adjacent centers is exactly $2r$ apart.
- By the four-fold symmetry of the figure, the four centers are the corners of a square whose side is this distance $2r$.
💡 When two circles just touch, the gap between their centers is simply the two radii laid end to end.
8.G.B.7 Step 2 Find the center-to-center distance
- Draw the large circle's center $O$; by symmetry it is the middle of the square of small-circle centers.
- Take two adjacent centers, say $P$ (right) and $Q$ (top).
- Triangle $OPQ$ has the right angle at $O$, with $OP=OQ$ equal (call each $d$) and hypotenuse $PQ=2r$.
- The Pythagorean theorem gives $d^2+d^2=(2r)^2$, so $2d^2=4r^2$ and $d=r\sqrt2$.
- Thus each small center sits a distance $r\sqrt2$ from $O$.
💡 The center of a square reaches each corner along the half-diagonal, which the Pythagorean theorem turns into $\sqrt2$ times the half-side.
7.EE.A.1 Step 3 Build the large radius
- Follow the straight line from $O$ out through a small center to where that small circle touches the large circle.
- Because the two circles are internally tangent, $O$, the small center, and the touch point are collinear, so the large radius $R$ is the center-to-center distance plus one small radius: $R=r\sqrt2+r$.
- Factor out $r$ to keep it tidy: $R=r(1+\sqrt2)$.
💡 To reach the outer rim, walk from the center to a small circle's center, then one more small radius to its edge.
7.G.B.4 Step 4 Write both areas
- The four small circles together have area $4\cdot\pi r^2$.
- The large circle has area $\pi R^2=\pi\big(r(1+\sqrt2)\big)^2=\pi r^2(1+\sqrt2)^2$.
- Expand the square: $(1+\sqrt2)^2=1+2\sqrt2+2=3+2\sqrt2$.
- So the large area is $\pi r^2(3+2\sqrt2)$.
💡 Every circle's area is $\pi$ times its radius squared, so squaring the radius is the only real work.
8.EE.A.2 Step 5 Divide and clean up the answer
- Take the ratio; the $\pi r^2$ cancels, confirming the size never mattered: $\dfrac{4\pi r^2}{\pi r^2(3+2\sqrt2)}=\dfrac{4}{3+2\sqrt2}$.
- Clear the root from the bottom by multiplying top and bottom by $3-2\sqrt2$.
- The bottom becomes $(3+2\sqrt2)(3-2\sqrt2)=9-8=1$, so the ratio is just $4(3-2\sqrt2)$.
- That is choice (C).
💡 Multiplying by the conjugate turns $\sqrt2$ terms into whole numbers because $\sqrt2\cdot\sqrt2=2$.
6.EE.B.6 Let each small circle have radius $r$. Two neighboring small circles are tangent 8.G.B.7 Draw the large circle's center $O$; by symmetry it is the middle of the square o 7.EE.A.1 Follow the straight line from $O$ out through a small center to where that small 7.G.B.4 The four small circles together have area $4\cdot\pi r^2$. The large circle has 8.EE.A.2 Take the ratio; the $\pi r^2$ cancels, confirming the size never mattered: $\dfr Review
Reasonableness: Estimate with $\sqrt2\approx1.414$: the ratio $4(3-2\sqrt2)\approx4(3-2.828)=4(0.172)=0.686$. 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 $R=1+\sqrt2\approx2.414$, small total area $\approx12.57$, large area $\approx18.30$, ratio $\approx0.687$, matching (C).
Alternative: Skip the variable and read the drawn figure directly, where the small radius is $1$ and the centers are at distance $\sqrt2$ from the middle. Then the large radius is $1+\sqrt2$, the small circles total $4\pi$, the large circle is $\pi(1+\sqrt2)^2=\pi(3+2\sqrt2)$, and the ratio $\dfrac{4\pi}{\pi(3+2\sqrt2)}=\dfrac{4}{3+2\sqrt2}=4(3-2\sqrt2)$ gives the same (C).
CCSS standards used (min grade 8)
6.EE.B.6Use variables to represent numbers and write expressions to solve problems (Naming the small radius $r$ and writing the adjacent center distance as $2r$.)8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Turning the right isosceles triangle of side $2r$ into the center-to-center distance $r\sqrt2$.)7.EE.A.1Apply properties of operations to add, subtract, factor, and expand linear expressions (Combining $r\sqrt2+r$ and factoring it as $r(1+\sqrt2)$ for the large radius.)7.G.B.4Know the formulas for area and circumference of a circle (Writing the four small areas as $4\pi r^2$ and the large area as $\pi R^2$.)8.EE.A.2Use square root and cube root symbols to represent solutions (Expanding $(1+\sqrt2)^2$ and rationalizing $\dfrac{4}{3+2\sqrt2}$ into $4(3-2\sqrt2)$.)
⭐ 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.
⭐ 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.
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