AMC 10 · 2004 · #25

Grade 8 geometry-3d

Archetype Coordinate Plane Figure Computation └ Sub-archetype Pythagorean / Distance Formula for a Length

pythagorean-theoremtangent-circlescentroid-2-to-1 physical-representationidentify-subproblems ↑ Prerequisites: pythagorean-theorem

📏 Long solution 💡 2 insights

Problem

Three balls of radius 1 sit on a flat floor, each touching the other two. A bigger ball of radius 2 rests on top of all three. Find the height from the floor to the very top of the big ball.

Pick an answer.

(A)

$\ 3 + \frac {\sqrt {30}}{2}$

(B)

$\ 3 + \frac {\sqrt {69}}{3}$

(C)

$\ 3 + \frac {\sqrt {123}}{4}$

(D)

$\ \frac {52}{9}$

(E)

$\ 3 + 2\sqrt2$

AMC 10 2004 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Visualize Spatial Relationships

The spheres themselves are hard to reason about, but their centers form a simple tetrahedron: an equilateral-triangle base of the three small centers with the big center as apex. Picturing that skeleton turns a 3D puzzle into two flat right triangles, each solved by the Pythagorean theorem. Splitting the total height into three stacked pieces (base-center height, the vertical rise to the apex, and the top radius) keeps every step small.

1STEP 1

Connect the four centers

Swap each ball for its center: the small centers form an equilateral triangle of side 2 at height 1, the big center 3 from each.

d_small-small = 1 + 1 = 2, d_small-big = 1 + 2 = 3

2STEP 2

Radius of the base triangle

The side-2 triangle has median $\sqrt{3}$ , and the centroid lies two-thirds of the way down it, so the circumradius squared is $\frac{4}{3}$ .

R = 2/3√3 = 2/√3 = 2√3/3, R² = 4/3

3STEP 3

Vertical rise to the top center

Big center, centroid, and one small center form a right triangle: hypotenuse 3, horizontal leg $R$ , so the rise is $h=\sqrt{\frac{23}{3}}$ .

h = √(3² - R²) = √(9 - 4/3) = √(23/3)

4STEP 4

Stack the three heights

Rationalize: $h=\frac{\sqrt{69}}{3}$ . Stack floor-to-center 1, rise $h$ , and radius 2 to get $3+\frac{\sqrt{69}}{3}$ , choice (B).

h = √(23/3) = √(23)/√3 = √(69)/3, 1 + √(69)/3 + 2 = 3 + √(69)/3

Answer

3 + √(69)/3

Numerically

3 + \frac{\sqrt{69}}{3} \approx 3 + 2.77 = 5.77

. That is above 3 (the big center's own radius plus the base height already give more than 3) and below 6 (the vertical rise is under 3), so it lands in a sensible range. Notice every answer choice sits between about 5.74 and 5.83, so only the exact computation

h^2 = \frac{23}{3}

pins it to (B) rather than a near neighbor.

💡Key takeaway

Swap the spheres for their centers, and a scary 3D stack becomes two flat right triangles you solve with the Pythagorean theorem.

Connect the four centers
Radius of the base triangle
Vertical rise to the top center
Stack the three heights

Connect the four centers

Radius of the base triangle

Vertical rise to the top center

Stack the three heights

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