AMC 10 · 2013 · #22

Grade 8 geometry-3d
Archetype Coordinate Plane Figure Computation └ Sub-archetype Pythagorean / Distance Formula for a Length
pythagorean-theoremtangent-circlesspatial-visualization convert-to-algebra ↑ Prerequisites: pythagorean-theorem
📏 Long solution 💡 3 insights
Problem
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.

Pick an answer.

(A)
$\sqrt2$
(B)
$\frac{3}{2}$
(C)
$\frac{5}{3}$
(D)
$\sqrt3$
(E)
2

AMC 10 2013 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Visualize Spatial Relationships

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.

1STEP 1

Set the scale of the ring

A regular hexagon is six equilateral triangles, so its center O sits a distance of 2 from every unit sphere's center.

OC_small = side = 2
2STEP 2

Radius of the big sphere

Internal tangency puts a unit center R-1 from O, so R-1=2 and the big sphere has radius 3.

R - 1 = 2 → R = 3
3STEP 3

Place the eighth sphere on the axis

By symmetry the eighth sphere (radius r) rides the axis above O at height 3-r, and 1+r from each small center.

h = R - r = 3 - r, distance to a small center = 1 + r
4STEP 4

Build the right triangle

Leg 2 (O out to a small center) and leg 3-r (O up to the new center) are perpendicular, hypotenuse 1+r: 2²+(3-r)²=(1+r)².

2² + h² = (1+r)² → 4 + (3-r)² = (1+r)²
5STEP 5

Solve for the radius

Expanding cancels the r² terms, leaving 13-6r=1+2r, so 8r=12 and r=3/2 — choice (B).

13-6r=1+2r → 8r=12 → r=3/2 → (B)
Answer
3/2
The eighth sphere must fit inside the big sphere of radius 3, so its radius has to be well under 3: 3/2 passes. Its center height is h=3-r=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 3/2 give hypotenuse √(4+9/4)=√(25/4)=5/2, and indeed 1+r=1+3/2=5/2. Both tangency conditions hold exactly, so 3/2 is correct and the other choices can be ruled out.
💡Key takeaway

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=3/2, choice (B).

  • Set the scale of the ring
  • Radius of the big sphere
  • Place the eighth sphere on the axis
  • Build the right triangle
  • Solve for the radius

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