AMC 10 · 2014 · #17

Grade 8 geometry-3d

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

tangent-circlesspatial-visualizationcoordinate-geometrypythagorean-theorem spatial-visualizationconvert-to-algebra ↑ Prerequisites: pythagorean-theoremcoordinate-geometry

📏 Long solution 💡 3 insights 📊 Diagram

Problem

Small spheres wedge into every corner of a box and one big sphere touches all of them. Find the box's height.

Pick an answer.

(A)

$2+2\sqrt 7$

(B)

$3+2\sqrt 5$

(C)

$4+2\sqrt 7$

(D)

$4\sqrt 5$

(E)

$4\sqrt 7$

How to solve

Strategy Visualize Spatial Relationships

The picture shows one arrangement, but a picture is not a proof. The real work is showing the nine centers have nowhere else to go (Tool #17 and Tool #14): being inside a box of width exactly 4 pins the big sphere's center to the box's central vertical line, and "tangent to three faces" pins each small sphere to a corner. Once every center is pinned, drop in coordinates (Tool #4) so that "tangent" becomes a distance equation, and split the job into two easy pieces (Tool #7): one tangency with a bottom sphere locates the big center, one tangency with a top sphere converts that into h.

1STEP 1

The big sphere has no room to wander

The big sphere is centred with no freedom.

2 ≤ x ≤ 4-2→ x=2; 2 ≤ y ≤ 4-2→ y=2; h ≥ 4

2STEP 2

Each small sphere is jammed into a corner

Each small sphere is jammed into a corner.

three mutually perpendicular faces → center =(1,1,1) measured from that corner

3STEP 3

Name every center with coordinates

Coordinates name every centre.

bottom: (1,1,1),(3,1,1),(1,3,1),(3,3,1); top: (1,1,h-1),…; big: (2,2,z)

4STEP 4

Tangency becomes a distance equation

Tangency becomes one distance equation.

(2-1)²+(2-1)²+(z-1)²=3² → 2+(z-1)²=9

5STEP 5

Find the big sphere's height

It fixes the big sphere's height.

(z-1)²=7 → z-1=±√7 → z=1+√7

6STEP 6

A top sphere converts height into h

A top sphere converts that into 2+2√7.

2+(h-1-z)²=9 → h=2+2√7 or h=2; h ≥ 4→ h=2+2√7

7STEP 7

Check the arrangement really exists

The arrangement really exists, choice (A).

√(1²+1²+(√7)²)=√9=3 for all eight small spheres, h=2+2√7→(A)

Answer

2+2√7

Numerically h=2+2√7≈ 7.29, which clears the h ≥ 4 the radius-2 sphere demands and leaves the two layers of small spheres 2√7≈ 5.29 apart, so nothing is crushed. The shape of the answer is also right: the height is one small radius at the bottom, plus one at the top, plus two equal vertical legs, so it must look like 2+2·(something). That kills 4√5 and 4√7 at a glance, and 3+2√5 would need end gaps of 3/2 each instead of the radius 1. Re-checking the tangency directly: √(2+(3.6458-1)²)=√(2+7)=3, exactly 2+1.

💡Key takeaway

Give every center coordinates: then "these two spheres touch" is just "the distance between the centers equals the sum of the radii".

The big sphere has no room to wander
Each small sphere is jammed into a corner
Name every center with coordinates
Tangency becomes a distance equation
Find the big sphere's height
A top sphere converts height into h
Check the arrangement really exists