AMC 10 · 2004 · #16
Grade 8 geometry-2dThree circles of radius 1 are externally tangent to each other and internally tangent to a larger circle. What is the radius of the large circle?
Pick an answer.
AMC 10 2004 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: Three circles, each of radius $1$, are placed so that every pair touches on the outside (externally tangent), and all three sit inside one larger circle, each touching it from within (internally tangent). Find the radius of the larger circle.
Givens: There are three small circles, each of radius $1$.; Each pair of small circles is externally tangent (they touch on the outside).; Each small circle is internally tangent to one big circle (touches it from the inside).; Answer choices: (A) $\frac{2 + \sqrt{6}}{3}$, (B) $2$, (C) $\frac{2 + 3\sqrt{2}}{2}$, (D) $\frac{3 + 2\sqrt{3}}{3}$, (E) $\frac{3 + \sqrt{3}}{2}$.
Unknowns: The radius $R$ of the large circle.
Understand
Restated: Three circles, each of radius $1$, are placed so that every pair touches on the outside (externally tangent), and all three sit inside one larger circle, each touching it from within (internally tangent). Find the radius of the larger circle.
Givens: There are three small circles, each of radius $1$.; Each pair of small circles is externally tangent (they touch on the outside).; Each small circle is internally tangent to one big circle (touches it from the inside).; Answer choices: (A) $\frac{2 + \sqrt{6}}{3}$, (B) $2$, (C) $\frac{2 + 3\sqrt{2}}{2}$, (D) $\frac{3 + 2\sqrt{3}}{3}$, (E) $\frac{3 + \sqrt{3}}{2}$.
Plan
Primary tool: #1 Draw a Diagram
Secondary: #4 Introduce a Variable, #7 Identify Subproblems
The whole problem is about how circles touch, so the first move is Tool #1 (Draw a Diagram) and mark every center. Once the picture is drawn, tangency turns into clean distance facts: two touching unit circles have centers exactly $2$ apart, so the three small centers form an equilateral triangle of side $2$. Tool #4 (Introduce a Variable) lets us drop coordinates onto that triangle and name the big center's height, and Tool #7 (Identify Subproblems) splits the job into two bite-sized pieces: first find the distance from the big center to a small center, then add one small radius to reach the big circle's edge.
Execute — Answer: D
7.G.A.2 Step 1 Turn tangency into a triangle
- Draw the three unit circles and mark their centers.
- When two circles of radius $1$ touch on the outside, the point where they meet lies on the line joining the centers, one radius from each, so the centers are $1+1=2$ apart.
- This is true for all three pairs, so the three centers are equally spaced $2$ apart — they are the corners of an equilateral triangle with side length $2$.
💡 Two touching circles are stuck exactly one radius apart on each side, so equal circles give equal spacing.
8.G.B.7 Step 2 Place coordinates on the triangle
- Put two of the small centers at $B=(-1,0)$ and $C=(1,0)$, so they are $2$ apart.
- The third center $A$ sits directly above the midpoint.
- Its height is the altitude of the equilateral triangle: drop a perpendicular from $A$ to $BC$, splitting the side-$2$ triangle into a right triangle with legs $1$ and $h$ and hypotenuse $2$.
- The Pythagorean theorem gives $h=\sqrt{2^2-1^2}=\sqrt{3}$, so $A=(0,\sqrt{3})$.
- By symmetry the big circle's center $O$ lies on the vertical line through the middle, at $O=(0,y)$ for some $y$.
💡 Cutting the equilateral triangle down its middle makes a right triangle, and the Pythagorean theorem hands over the height.
8.G.B.8 Step 3 Find the big center by equal distances
- The big center $O$ must be the same distance from all three small centers.
- Setting its distance to $A$ equal to its distance to $B$ pins down $y$.
- The distance from $O=(0,y)$ to $A=(0,\sqrt{3})$ is $\sqrt{3}-y$, and to $B=(-1,0)$ is $\sqrt{1+y^2}$.
- Squaring both and setting them equal: $(\sqrt{3}-y)^2=1+y^2$, which expands to $3-2\sqrt{3}\,y+y^2=1+y^2$.
- The $y^2$ cancels, leaving $3-2\sqrt{3}\,y=1$, so $y=\frac{1}{\sqrt{3}}$.
- The common distance is then $d=\sqrt{1+y^2}=\sqrt{1+\frac13}=\sqrt{\frac43}=\frac{2}{\sqrt{3}}=\frac{2\sqrt{3}}{3}$.
💡 The center of the big circle is the one spot balanced equally from all three little centers, and equal distances give an equation for it.
8.EE.A.2 Step 4 Step out one radius to the rim
- The big circle touches each small circle from the inside, so along the line from $O$ through a small center, the big circle's edge is exactly one small radius past that center.
- That means the big radius is the center-to-center distance plus $1$: $R=d+1=\frac{2\sqrt{3}}{3}+1$.
- Writing $1=\frac{3}{3}$ to combine the fractions gives $R=\frac{3+2\sqrt{3}}{3}$.
- The answer is (D).
💡 From the big center, you reach a small center, then keep going one more radius to hit the big circle's edge.
7.G.A.2 Draw the three unit circles and mark their centers. When two circles of radius $ 8.G.B.7 Put two of the small centers at $B=(-1,0)$ and $C=(1,0)$, so they are $2$ apart. 8.G.B.8 The big center $O$ must be the same distance from all three small centers. Setti 8.EE.A.2 The big circle touches each small circle from the inside, so along the line from Review
Reasonableness: As a decimal, $\frac{3+2\sqrt{3}}{3}\approx\frac{3+3.46}{3}\approx 2.15$. That is comfortably bigger than $1$ (it must contain three radius-$1$ circles) yet not huge, which matches the picture. Checking the other choices numerically, (A) $\approx1.48$ is too small to hold the arrangement, (B) $=2$ is a touch too small, (C) $\approx3.12$ and (E) $\approx2.37$ are too big; only (D) $\approx2.15$ fits, so the answer is stable.
Alternative: Skip coordinates and use the known fact that the center of an equilateral triangle lies $\tfrac{2}{3}$ of the way down each median from a vertex. The median (altitude) here is $\sqrt{3}$, so the distance from the center $O$ to a small center is $\tfrac{2}{3}\cdot\sqrt{3}=\frac{2\sqrt{3}}{3}$ — the same $d$ as before. Adding the small radius $1$ again gives $R=\frac{2\sqrt{3}}{3}+1=\frac{3+2\sqrt{3}}{3}$.
CCSS standards used (min grade 8)
7.G.A.2Draw geometric shapes with given conditions (Turning the three externally tangent unit circles into an equilateral triangle of side $2$ formed by their centers.)8.G.B.7Apply the Pythagorean Theorem to determine unknown side lengths in right triangles (Finding the altitude $\sqrt{3}$ of the side-$2$ equilateral triangle to locate the third center.)8.G.B.8Apply the Pythagorean Theorem to find the distance between two points in a coordinate system (Writing the distances from the big center $O$ to the small centers and solving for $O$ and the distance $d=\frac{2\sqrt{3}}{3}$.)8.EE.A.2Use square root symbols and evaluate square roots (Simplifying $\frac{2}{\sqrt{3}}$ to $\frac{2\sqrt{3}}{3}$ and combining it with $1$ into $\frac{3+2\sqrt{3}}{3}$.)
⭐ Touching circles fix the distances between their centers, so the three unit circles make an equilateral triangle; measure from that triangle's center out to a small center and add one radius to reach the big circle's edge.
⭐ Touching circles fix the distances between their centers, so the three unit circles make an equilateral triangle; measure from that triangle's center out to a small center and add one radius to reach the big circle's edge.
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