📦 3D Geometry Light

3D geometry appears occasionally in AMC 12 problems. Focus on projections, basic volumes, and cross-sections for contest success.

🎯 Key Concepts

Basic 3D Shapes

  • Cube: All faces are squares
  • Rectangular Prism: All faces are rectangles
  • Cylinder: Circular base and top
  • Cone: Circular base, point top
  • Sphere: All points equidistant from center

Volume Formulas

  • Cube: $V = s^3$ (side length $s$)
  • Rectangular Prism: $V = lwh$ (length, width, height)
  • Cylinder: $V = \pi r^2 h$ (radius $r$, height $h$)
  • Cone: $V = \frac{1}{3}\pi r^2 h$ (radius $r$, height $h$)
  • Sphere: $V = \frac{4}{3}\pi r^3$ (radius $r$)

Surface Area Formulas

  • Cube: $A = 6s^2$ (side length $s$)
  • Rectangular Prism: $A = 2(lw + lh + wh)$
  • Cylinder: $A = 2\pi r^2 + 2\pi rh$ (lateral + bases)
  • Cone: $A = \pi r^2 + \pi rl$ (base + lateral, $l$ = slant height)
  • Sphere: $A = 4\pi r^2$ (radius $r$)

Projections

Orthogonal Projection: Project 3D figure onto 2D plane Key Insight: Many 3D problems can be solved by considering 2D cross-sections

🔍 Micro-Examples

Cube Volume

Cube with side length 4:

  • Volume = $4^3 = 64$
  • Surface area = $6 \cdot 4^2 = 96$

Cylinder Volume

Cylinder with radius 3 and height 5:

  • Volume = $\pi \cdot 3^2 \cdot 5 = 45\pi$
  • Surface area = $2\pi \cdot 3^2 + 2\pi \cdot 3 \cdot 5 = 18\pi + 30\pi = 48\pi$

Sphere Volume

Sphere with radius 2:

  • Volume = $\frac{4}{3}\pi \cdot 2^3 = \frac{32\pi}{3}$
  • Surface area = $4\pi \cdot 2^2 = 16\pi$

⚠️ Common Traps

Pitfall: Wrong volume formula

  • Fix: Remember that cone has $\frac{1}{3}$ factor, sphere has $\frac{4}{3}$ factor

Pitfall: Forgetting units

  • Fix: Volume is cubic units, surface area is square units

Pitfall: Wrong projection setup

  • Fix: Project perpendicular to the plane

Pitfall: Confusing radius and diameter

  • Fix: Volume formulas use radius, not diameter

🎯 AMC-Style Worked Example

Problem: A cube has side length 6. What is the volume of the largest sphere that can fit inside the cube?

Solution: For a sphere to fit inside a cube, the sphere’s diameter must equal the cube’s side length.

Given:

  • Cube side length = 6
  • Sphere diameter = 6
  • Sphere radius = 3

Using the sphere volume formula: $V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi \cdot 3^3 = \frac{4}{3}\pi \cdot 27 = 36\pi$

Answer: $36\pi$

🔗 Related Topics

💡 Quick Reference

Volume Formulas

  • Cube: $V = s^3$
  • Rectangular Prism: $V = lwh$
  • Cylinder: $V = \pi r^2 h$
  • Cone: $V = \frac{1}{3}\pi r^2 h$
  • Sphere: $V = \frac{4}{3}\pi r^3$

Surface Area Formulas

  • Cube: $A = 6s^2$
  • Rectangular Prism: $A = 2(lw + lh + wh)$
  • Cylinder: $A = 2\pi r^2 + 2\pi rh$
  • Cone: $A = \pi r^2 + \pi rl$
  • Sphere: $A = 4\pi r^2$

Common Applications

  • Projections: 3D to 2D conversion
  • Cross-sections: 2D slices of 3D figures
  • Volumes: Finding volumes of composite figures
  • Surface areas: Finding surface areas of composite figures

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