Trirectangular tetrahedron explained

In geometry, a trirectangular tetrahedron is a tetrahedron where all three face angles at one vertex are right angles. That vertex is called the right angle of the trirectangular tetrahedron and the face opposite it is called the base. The three edges that meet at the right angle are called the legs and the perpendicular from the right angle to the base is called the altitude of the tetrahedron.

Only the bifurcating graph of the

B _{3} Affine Coxeter group has a Trirectangular tetrahedron fundamental domain.

Metric formulas

If the legs have lengths x, y, z then the trirectangular tetrahedron has the volume

V = (xyz)/(6)

The altitude h satisfies^[1]

(1)/(= h ^{2}) (1)/(+ x ^{2}) (1)/(+ y ^{2}) (1)/(z ^{2})

The area

T _{0} of the base is given by^[2] T _{0= (xyz)/(2 h)} ----

There are 5 usable internal relationships in Tri-Rectangular Tetrahedrons.

Using the illustration above designate all 3 of the diagonals shown in the green triangle as | a, b, c | & it doesn't matter which of the 3 diagonals is a or b or c. What does matter is that each diagonal be mated with its corresponding rectangular leg. Each of the diagonals connects with only 2 of the rectangular legs but not the 3rd one. The 3rd one IS the mating rectangular leg for the diagonal you chose. Then the equations shown below work properly.

K = l [ (a ^{2+b} ^{2+c} ^{2})/(2) r ] = l [x ^{2+y} ^{2+z} ^{2r]=l[x} ^{2+c} ^{2r]=l[y} ^{2+b} ^{2r]=l[z} ^{2+a} ^{2r]} Given l [a,b,c r ] Given l [x,y,z r ] K = l [Constant r ] K = l [ (a ^{2+b} ^{2+c} ^{2})/(2) r ] K = l [x ^{2+y} ^{2+z} ^{2r]}

Use l [x ^{2+c} ^{2 r ] x = \sqrt{} K - c ^{2 } c = \sqrt{} K - x ^{2 }}

Use l [y ^{2+b} ^{2 r ] y = \sqrt{} K - b ^{2 } b = \sqrt{} K - y ^{2 }}

Use l [z ^{2+a} ^{2 r ] z = \sqrt{} K - a ^{2 } a = \sqrt{} K - z ^{2 }}

Volume of a Tri-Rectangular Tetrahedron & the box it'll fit in.

V _{tet} = (xyz)/(6) V _{box} = xyz

Internal height of a Tri-Rectangular Tetrahedron from its point of origin | x,y,z | to its base bounded by | a, b, c |

h _{tet} = (xyz)/(\sqrt{ x ^{2y} ^{2+z} ^{2l[x} ^{2+y} ^{2r]}) }

Area of the to its base bounded by | a, b, c | [2 formulas ] - Herons Theorem does the same thing differently.

Area of Base triangle bounded by | a, b, c |

A _{abc} = (xyz)/(2 h _{tet}) A _{abc} = (\sqrt{x ^{2y} ^{2+z} ^{2l[x} ^{2+y} ^{2r]})/(}{ 2 })

Area of all 4 surfaces of a Tri-Rectangular Tetrahedron

A _{abc} = (xyz)/(2 h _{tet}) A _{xy} = (xy)/(2) A _{xz} = (xz)/(2) A _{yz} = (yz)/(2)

Total area of a Tri-Rectangular Tetrahedron

A _{tet} = (xy + z l [x + y r ] + \sqrt{x ^{2y} ^{2+z} ^{2l[x} ^{2+y} ^{2r]})/(}{ 2 })

Example dimensions

a =14 . 4 b =10 c =12

x =9 . 037698822156002781875821128798

y =11 . 210709165793214885544254958172

z =4 . 2801869118065393196504607419769

The old way. Pythagorean's Theorem TWO dimensional. Then Modified for 3 dimensions BUT ONLY for Tri-Rectangular Tetrahedrons.

Given l [x,y,z r ] a = \sqrt{ x ^{2+y} ^{2 } b=\sqrt{ x} ^{2+z} ^{2 } c=\sqrt{ y} ^{2+z} ^{2 }} Given l [a,b,c r ] x = \sqrt{ (a ^{2+b} ^{2-c} ^{2})/(} y = \sqrt{ 2) (a ^{2+c} ^{2-b} ^{2})/(} z = \sqrt{ 2) (b ^{2+c} ^{2-a} ^{2})/(2) }

De Gua's theorem

See main article: De Gua's theorem. If the area of the base is

T _{0} and the areas of the three other (right-angled) faces are T _{1} , T _{2} and T _{3} , thenT ^{2 .} _{3}

This is a generalization of the Pythagorean theorem to a tetrahedron.

Integer solution

Perfect body

The area of the base (a,b,c) is always (Gua) an irrational number. Thus a trirectangular tetrahedron with integer edges is never a perfect body. The trirectangular bipyramid (6 faces, 9 edges, 5 vertices) built from these trirectangular tetrahedrons and the related left-handed ones connected on their bases have rational edges, faces and volume, but the inner space-diagonal between the two trirectangular vertices is still irrational. The later one is the double of the altitude of the trirectangular tetrahedron and a rational part of the (proved)^[3] irrational space-diagonal of the related Euler-brick (bc, ca, ab).

Integer edges

Trirectangular tetrahedrons with integer legs

a,b,c and sides d = \sqrt{b ^{2+c} ^{2},} e = \sqrt{a ^{2+c} ^{2},} f = \sqrt{a ^{2+b} ^{2}} of the base triangle exist, e.g. a =240 ,b =117 ,c =44 ,d =125 ,e =244 ,f =267 (discovered 1719 by Halcke). Here are a few more examples with integer legs and sides. a b c d e f ---- 240 117 44 125 244 267 275 252 240 348 365 373 480 234 88 250 488 534 550 504 480 696 730 746 693 480 140 500 707 843 720 351 132 375 732 801 720 132 85 157 725 732 792 231 160 281 808 825 825 756 720 1044 1095 1119 960 468 176 500 976 1068 1100 1008 960 1392 1460 1492 1155 1100 1008 1492 1533 1595 1200 585 220 625 1220 1335 1375 1260 1200 1740 1825 1865 1386 960 280 1000 1414 1686 1440 702 264 750 1464 1602 1440 264 170 314 1450 1464Notice that some of these are multiples of smaller ones. Note also .

Integer faces

Trirectangular tetrahedrons with integer faces

T _{c,} T _{a,} T _{b,} T _{0} and altitude h exist, e.g. a =42 ,b =28 ,c =14 ,T _{c=588,T} _{a=196,T} _{b=294,T} _{0=686,h=12} without or a =156 ,b =80 ,c =65 ,T _{c=6240,T} _{a=2600,T} _{b=5070,T} _{0=8450,h=48} with coprime a,b,c .

See also

• Irregular tetrahedra
• Standard simplex
• Euler Brick

Notes and References

1. Eves, Howard Whitley, "Great moments in mathematics (before 1650)", Mathematical Association of America, 1983, p. 41.
2. http://gogeometry.com/pythagoras/right_triangle_formulas_facts.htm Gutierrez, Antonio, "Right Triangle Formulas"
3. Walter Wyss, "No Perfect Cuboid",