AMC 10 · 2007 · #18
Grade 8 geometry-2dA circle of radius 1 is surrounded by 4 circles of radius r as shown. What is r?
Pick an answer.
AMC 10 2007 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 unit circle (radius $1$) is ringed by four equal circles of radius $r$. Each outer circle touches the central circle, and each touches its two neighbors. Find $r$.
Givens: A central circle has radius $1$; Four surrounding circles each have radius $r$, arranged symmetrically around the center; Each outer circle is tangent to the central circle and to its two neighboring outer circles; Answer choices: (A) $\sqrt{2}$, (B) $1+\sqrt{2}$, (C) $\sqrt{6}$, (D) $3$, (E) $2+\sqrt{2}$
Unknowns: The radius $r$ of each of the four outer circles
Understand
Restated: A unit circle (radius $1$) is ringed by four equal circles of radius $r$. Each outer circle touches the central circle, and each touches its two neighbors. Find $r$.
Givens: A central circle has radius $1$; Four surrounding circles each have radius $r$, arranged symmetrically around the center; Each outer circle is tangent to the central circle and to its two neighboring outer circles; Answer choices: (A) $\sqrt{2}$, (B) $1+\sqrt{2}$, (C) $\sqrt{6}$, (D) $3$, (E) $2+\sqrt{2}$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #17 Visualize Spatial Relationships, #13 Convert to Algebra, #4 Introduce a Variable, #3 Eliminate Possibilities
The picture is the whole problem, so first draw the centers (Tool #1) and turn each "touch" into a distance between centers. The symmetry places the four outer centers at the corners of a square (Tool #17), which hands us a right angle at the middle. That right triangle converts the geometry into a Pythagorean equation (Tool #13) with the single unknown $r$ (Tool #4). Solving and matching the exact value to a listed choice finishes it (Tool #3).
Execute — Answer: B
7.G.B.4 Step 1 Turn each touch into a distance
- Put the central circle's center at $O$.
- An outer circle touches the central circle on the outside, so the distance from $O$ to that outer center is the sum of the radii, $1+r$.
- Two neighboring outer circles also touch, so the distance between their centers is $r+r=2r$.
💡 When two circles touch on the outside, the gap between their centers is just their two radii added.
4.MD.C.5 Step 2 Read the right angle from symmetry
- The four outer circles are equally spaced around $O$, so their centers form a square with $O$ at its center.
- Stepping from one outer center to the next around $O$ splits the full turn into four equal parts, so the angle between neighbors at $O$ is $360^\circ\div4=90^\circ$.
- Take two neighboring centers $P$ and $Q$: triangle $OPQ$ has $OP=OQ=1+r$, side $PQ=2r$, and a right angle at $O$.
💡 Four evenly placed circles split the full turn into four equal right angles.
8.G.B.7 Step 3 Apply the Pythagorean theorem
- Triangle $OPQ$ has its right angle at $O$, so the legs $OP$ and $OQ$ and the hypotenuse $PQ$ satisfy $OP^2+OQ^2=PQ^2$.
- Substitute $OP=OQ=1+r$ and $PQ=2r$.
💡 In a right triangle the two legs squared add up to the hypotenuse squared.
7.EE.B.4 Step 4 Simplify to a clean equation
- Divide both sides by $2$ to get $(1+r)^2=2r^2$.
- Both sides are positive, so take the positive square root: $1+r=r\sqrt{2}$.
- Gather the $r$ terms on one side: $1=r\sqrt{2}-r=r(\sqrt{2}-1)$.
💡 Collecting the $r$ terms on one side leaves a single equation to solve for $r$.
8.EE.A.2 Step 5 Solve and rationalize
- Divide by $\sqrt{2}-1$: $r=\dfrac{1}{\sqrt{2}-1}$.
- Multiply top and bottom by the conjugate $\sqrt{2}+1$ to clear the radical from the denominator, using $(\sqrt{2}-1)(\sqrt{2}+1)=2-1=1$.
- This gives $r=1+\sqrt{2}$, which is choice $\textbf{(B)}$.
💡 Multiplying by the conjugate turns the messy denominator into a difference of squares that collapses to $1$.
7.G.B.4 Put the central circle's center at $O$. An outer circle touches the central circ 4.MD.C.5 The four outer circles are equally spaced around $O$, so their centers form a sq 8.G.B.7 Triangle $OPQ$ has its right angle at $O$, so the legs $OP$ and $OQ$ and the hyp 7.EE.B.4 Divide both sides by $2$ to get $(1+r)^2=2r^2$. Both sides are positive, so take 8.EE.A.2 Divide by $\sqrt{2}-1$: $r=\dfrac{1}{\sqrt{2}-1}$. Multiply top and bottom by th Review
Reasonableness: Check $r=1+\sqrt{2}\approx2.414$. Then $O$ to an outer center is $1+r\approx3.414$ and neighboring centers are $2r\approx4.828$ apart. Test the right triangle: $2\cdot(3.414)^2\approx23.3$ and $(4.828)^2\approx23.3$ — they match. The outer circles must be bigger than the unit circle, and $2.414>1$ fits. The nearby decoy $\sqrt{6}\approx2.449$ is close but fails the exact equation, so the geometry pins the value to $1+\sqrt{2}$, choice (B).
Alternative: Use the diagonal directly instead of a right triangle. Two opposite outer centers and $O$ are collinear, so the distance across is $(1+r)+(1+r)=2+2r$; this is the diagonal of the square whose side is $2r$. A square's diagonal equals side $\times\sqrt{2}$, so $2+2r=2r\sqrt{2}$, giving $1+r=r\sqrt{2}$ and again $r=1+\sqrt{2}$.
CCSS standards used (min grade 8)
7.G.B.4Know the formulas for area and circumference of a circle (Using tangency of circles to read the distance between two centers as the sum of their radii ($1+r$ and $2r$).)4.MD.C.5Recognize angles as geometric shapes formed when two rays share an endpoint (Splitting the full turn around $O$ into four equal $90^\circ$ angles to get the right angle at $O$.)8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Relating the legs $OP=OQ=1+r$ to the hypotenuse $PQ=2r$ in right triangle $OPQ$.)7.EE.B.4Use variables to represent quantities and construct simple equations and inequalities (Simplifying $2(1+r)^2=4r^2$ and collecting terms into $1=r(\sqrt{2}-1)$.)8.EE.A.2Use square root and cube root symbols to represent solutions (Taking the square root of $(1+r)^2=2r^2$ and rationalizing $\tfrac{1}{\sqrt{2}-1}$ to $1+\sqrt{2}$.)
⭐ When circles are packed so they just touch, the distance between two centers is the sum of their radii — turn every touch into that length and the picture becomes an equation.
⭐ When circles are packed so they just touch, the distance between two centers is the sum of their radii — turn every touch into that length and the picture becomes an equation.
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