AMC 10 · 2013 · #22
Grade 8 geometry-3dSix spheres of radius 1 are positioned so that their centers are at the vertices of a regular hexagon of side length 2. The six spheres are internally tangent to a larger sphere whose center is the center of the hexagon. An eighth sphere is externally tangent to the six smaller spheres and internally tangent to the larger sphere. What is the radius of this eighth sphere?
Pick an answer.
AMC 10 2013 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: Six unit spheres have their centers at the vertices of a regular hexagon with side $2$. They are all internally tangent to one big sphere centered at the hexagon's center. A new sphere is externally tangent to all six unit spheres and internally tangent to the big sphere. Find its radius.
Givens: Six spheres of radius $1$, centers at the vertices of a regular hexagon of side $2$.; A big sphere centered at the hexagon's center is internally tangent to all six.; An eighth sphere is externally tangent to the six unit spheres and internally tangent to the big sphere.; Answer choices: (A) $\sqrt2$, (B) $\frac{3}{2}$, (C) $\frac{5}{3}$, (D) $\sqrt3$, (E) $2$.
Unknowns: The radius of the eighth sphere.
Understand
Restated: Six unit spheres have their centers at the vertices of a regular hexagon with side $2$. They are all internally tangent to one big sphere centered at the hexagon's center. A new sphere is externally tangent to all six unit spheres and internally tangent to the big sphere. Find its radius.
Givens: Six spheres of radius $1$, centers at the vertices of a regular hexagon of side $2$.; A big sphere centered at the hexagon's center is internally tangent to all six.; An eighth sphere is externally tangent to the six unit spheres and internally tangent to the big sphere.; Answer choices: (A) $\sqrt2$, (B) $\frac{3}{2}$, (C) $\frac{5}{3}$, (D) $\sqrt3$, (E) $2$.
Plan
Primary tool: #17 Visualize Spatial Relationships
Secondary: #1 Draw a Diagram, #4 Introduce a Variable, #7 Identify Subproblems
This is a 3D packing picture, so Tool #17 (Visualize Spatial Relationships) is the engine: seeing that the six unit spheres form a flat ring and the eighth sphere must rest on the axis right above the ring's center turns the whole solid into one flat right triangle. Tool #1 (Draw a Diagram) fixes that triangle on paper. Tool #7 (Identify Subproblems) splits the work into two easy pieces — first the big sphere's radius, then the eighth sphere. Tool #4 (Introduce a Variable) names the unknown radius $r$ and the height $h$, so the two tangency facts become equations we can solve.
Execute — Answer: B
8.G.A.5 Step 1 Set the scale of the ring
- A regular hexagon splits into six equilateral triangles meeting at its center, so the distance from the center to each vertex equals the side length, which is $2$.
- Put the hexagon's center at the origin $O$ in a horizontal plane; then the six unit-sphere centers all sit on a circle of radius $2$ around $O$, at the same height.
💡 A regular hexagon is six equilateral triangles, so center-to-corner equals the side.
8.G.B.8 Step 2 Radius of the big sphere
- The big sphere is centered at $O$ and is internally tangent to a unit sphere whose center is $2$ away.
- Internal tangency means the center distance equals the difference of the radii, so the small sphere's center is $R-1$ from $O$.
- Setting $R-1=2$ gives the big radius $R=3$.
💡 For a sphere held inside a bigger one and touching it, the gap between centers is the radius difference.
8.G.B.8 Step 3 Place the eighth sphere on the axis
- By symmetry the eighth sphere (radius $r$) sits with its center on the vertical axis through $O$, at some height $h$ above the plane of the ring.
- Its internal tangency to the big sphere gives center distance $R-r=3-r$, and that distance is just the height, so $h=3-r$.
- Its external tangency to a unit sphere gives center distance $1+r$.
💡 Each 'just touching' condition turns straight into a fixed distance between two centers.
8.G.B.7 Step 4 Build the right triangle
- Look at the triangle formed by $O$, one small center $C$, and the eighth sphere's center $P$.
- Going from $O$ to $C$ is horizontal with length $2$; going from $O$ up to $P$ is vertical with length $h$.
- These legs are perpendicular, so $CP$ is the hypotenuse, and $CP=1+r$ from the external tangency.
- By the Pythagorean theorem, $2^2+h^2=(1+r)^2$.
- Substituting $h=3-r$ gives $4+(3-r)^2=(1+r)^2$.
💡 Horizontal reach and vertical rise are perpendicular legs, so the center-to-center line is the hypotenuse.
8.EE.C.7 Step 5 Solve for the radius
- Expand both sides: $4+9-6r+r^2=1+2r+r^2$.
- The $r^2$ terms cancel, leaving the linear equation $13-6r=1+2r$.
- Then $12=8r$, so $r=\frac{12}{8}=\frac{3}{2}$.
- This matches choice $\textbf{(B)}$.
💡 The squared terms cancel, turning a scary-looking equation into a one-step linear one.
8.G.A.5 A regular hexagon splits into six equilateral triangles meeting at its center, s 8.G.B.8 The big sphere is centered at $O$ and is internally tangent to a unit sphere who 8.G.B.8 By symmetry the eighth sphere (radius $r$) sits with its center on the vertical 8.G.B.7 Look at the triangle formed by $O$, one small center $C$, and the eighth sphere' 8.EE.C.7 Expand both sides: $4+9-6r+r^2=1+2r+r^2$. The $r^2$ terms cancel, leaving the li Review
Reasonableness: The eighth sphere must fit inside the big sphere of radius $3$, so its radius has to be well under $3$: $\frac{3}{2}$ passes. Its center height is $h=3-r=\frac{3}{2}$, which is positive, so the sphere really does sit above the ring rather than through it — a good sign. A quick check of the triangle: legs $2$ and $\frac{3}{2}$ give hypotenuse $\sqrt{4+\frac{9}{4}}=\sqrt{\frac{25}{4}}=\frac{5}{2}$, and indeed $1+r=1+\frac{3}{2}=\frac{5}{2}$. Both tangency conditions hold exactly, so $\frac{3}{2}$ is correct and the other choices can be ruled out.
Alternative: Skip naming $h$ separately. Put coordinates with $O$ at the origin, a small center at $(2,0)$, and the eighth center at $(0,y)$ with $y\ge 0$. External tangency: $\sqrt{2^2+y^2}=1+r$. Internal tangency to the big sphere: $y=3-r$. Substituting the second into the first and squaring gives the same equation $4+(3-r)^2=(1+r)^2$, hence $r=\frac{3}{2}$.
CCSS standards used (min grade 8)
8.G.A.5Use informal arguments to establish facts about angle sum and exterior angles (Splitting the regular hexagon into six equilateral triangles to show center-to-vertex distance equals the side length $2$.)8.G.B.8Apply the Pythagorean theorem to find the distance between two points in a coordinate system (Turning internal and external tangency into exact center-to-center distances ($R-1=2$, $3-r$, and $1+r$).)8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Relating the horizontal leg $2$, vertical leg $h$, and hypotenuse $1+r$ in the right triangle $O C P$.)8.EE.C.7Solve linear equations in one variable (Canceling the $r^2$ terms and solving $13-6r=1+2r$ to get $r=\frac{3}{2}$.)
⭐ Flatten the 3D picture into one right triangle: the eighth sphere sits on the axis, and the legs $2$ and $3-r$ with hypotenuse $1+r$ give $r=\frac{3}{2}$, choice $\textbf{(B)}$.
⭐ Flatten the 3D picture into one right triangle: the eighth sphere sits on the axis, and the legs $2$ and $3-r$ with hypotenuse $1+r$ give $r=\frac{3}{2}$, choice $\textbf{(B)}$.
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