# Tetrahedron

# Tetrahedron bounding box size:

# Figure 1: Side View
#
#       *_        -
#      /  \       ^
#     / |  \_
#    /  |    \    depth
#   /   |     \_  v
#  *----0-------* -
#
# Figure 2: Top View
#
#  *_B
#  | \_
#  |\  \_
#  | \   \_
#  |  \    \_
#  |   \A    \__
#  |<a> *-<b>-__*D
#  |   /    _/
#  |  /   _/
#  | /  _/
#  |/ _/
#  |_/
#  *C

# From these diagrams, you can tell that the width is the height of an
# equilateral triangle. If we assume the edge of the tetrahedron is s, then:

define @ns s - 4
define @half_height s / 2

# For the depth, a few calculations are necessary:
#
# a + b = s * sin(⅓π)
# a² + (s/2)² = b²
define @a (s/2)*tan(⅙π)
define @na (@ns/2)*tan(⅙π)
define @b s * (sin(⅓π) - tan(⅙π)/2)
# a² + d² = (s*sin(⅓π))²
define @depth s * sqrt((sin(⅓π))^2 - ((1/2)*tan(⅙π))^2)
define @ndepth @ns * sqrt((sin(⅓π))^2 - ((1/2)*tan(⅙π))^2)

# Considering that the origin is in the center of the base, the min x is -a,
# the min y is -s/2, and the min z is 0
-@a < x < @b
-@half_height < y < @half_height
0 < z < @depth

{{ρ ≤ @a * sec(φ - ⅓π) * (1 - z/@depth) ∧ φ ≥ 0 ∧ φ ≤ 2 * ⅓π} ∨ \
{ρ ≤ @a * sec(φ - π) * (1 - z/@depth) ∧ {φ ≥ 2 * ⅓π ∨ φ ≤ -2 * ⅓π}} ∨ \
{ρ ≤ @a * sec(φ + ⅓π) * (1 - z/@depth) ∧ φ ≤ 0 ∧ φ ≥ -2 * ⅓π} ∨ ρ = 0} \
⊻ \
{z ≥ 1 ∧ z ≤ @ndepth ∧ \
{{ρ ≤ @na * sec(φ - ⅓π) * (1 - z/@ndepth) ∧ φ ≥ 0 ∧ φ ≤ 2 * ⅓π} ∨ \
{ρ ≤ @na * sec(φ - π) * (1 - z/@ndepth) ∧ {φ ≥ 2 * ⅓π ∨ φ ≤ -2 * ⅓π}} ∨ \
{ρ ≤ @na * sec(φ + ⅓π) * (1 - z/@ndepth) ∧ φ ≤ 0 ∧ φ ≥ -2 * ⅓π} ∨ ρ = 0}}