Trirectangular tetrahedron

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 ${\displaystyle B_{3}}$ 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

${\displaystyle V={\frac {xyz}{6}}}$

The altitude h satisfies^[1]

${\displaystyle {\frac {1}{h^{2}}}={\frac {1}{x^{2}}}+{\frac {1}{y^{2}}}+{\frac {1}{z^{2}}}}$

The area ${\displaystyle T_{0}}$ of the base is given by^[2]

${\displaystyle T_{0}={\frac {xyz}{2h}}}$

Using the illustration in the upper right designate all 3 of the diagonals (sides) shown in the green triangle as |a,b,c|

Of the 3 sides in the green triangle the diagonal to the far right is "a", the bottom diagonal is "b" & the left diagonal is "c".

Also there are 3 axial aligned legs |x,y,z|. The rectangular leg pointing to the right is "x", the vertical pointing leg is "y" & the downward pointing leg is "z".

It does matter that each of the diagonal sides be matched with their corresponding rectangular legs. Each one of the 3 diagonals connects with 2 of the rectangular legs but not the 3rd. The 3rd leg IS the matching leg for the diagonal you chose. Side "a" mates with leg "z", side "b" mates with leg "y" & side "c" mates with leg "x".

Then the equations shown below work properly.

If you stare into the 1st equation presented here you can see Heron's constant [S] has been modified such each of the 3 diagonal elements |a,b,c| divided by 2 have been squared & have become |a²+b²+c²| divided by 2. Here [S] is replaced by [K]

And Pythagorean's theorem has been modified such that all three rectangular legs |x²,y²,z²| are used simultaneously instead of just 2 at a time & you don't take the square root.

Both of the modified versions of Heron's theorem & Pythagorean's Theorem also equal all 3 of the match pairs of the square of a diagonal + the square of its matching leg The interlocked mathematical structure of this is perfect. De Gua should have published this long ago & perhaps he did & if so it's another case of everybody in the universe could find it except me.

${\displaystyle K={\Biggl [}{\frac {a^{2}+b^{2}+c^{2}}{2}}{\Biggr ]}={\Bigl [}x^{2}+y^{2}+z^{2}{\Bigr ]}={\Bigl [}x^{2}+c^{2}{\Bigr ]}={\Bigl [}y^{2}+b^{2}{\Bigr ]}={\Bigl [}z^{2}+a^{2}{\Bigr ]}}$${\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Given{\Bigl [}a,b,c{\Bigr ]}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Given{\Bigl [}x,y,z{\Bigr ]}}$

${\displaystyle K={\Bigl [}Constant{\Bigr ]}\ \ \ \ \ \ \ \ \ K={\Biggl [}{\frac {a^{2}+b^{2}+c^{2}}{2}}{\Biggr ]}\ \ \ \ \ \ \ \ \ K={\Bigl [}x^{2}+y^{2}+z^{2}{\Bigr ]}}$

${\displaystyle Use\ {\Bigl [}x^{2}+c^{2}{\Bigr ]}\ \ \ \ \ \ \ \ x={\sqrt {_{\ }K-c^{2}\ \ }}\ \ \ \ \ \ \ \ \ \ \ c={\sqrt {_{\ }K-x^{2}\ \ }}}$

${\displaystyle Use\ {\Bigl [}y^{2}+b^{2}{\Bigr ]}\ \ \ \ \ \ \ \ \ y={\sqrt {_{\ }K-b^{2}\ \ }}\ \ \ \ \ \ \ \ \ \ \ b={\sqrt {_{\ }K-y^{2}\ \ }}}$

${\displaystyle Use\ {\Bigl [}z^{2}+a^{2}{\Bigr ]}\ \ \ \ \ \ \ \ \ z={\sqrt {_{\ }K-a^{2}\ \ }}\ \ \ \ \ \ \ \ \ \ \ a={\sqrt {_{\ }K-z^{2}\ \ }}}$

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

${\displaystyle V_{tet}={\frac {\ xyz\ }{6}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ V_{box}=xyz}$

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

${\displaystyle h_{tet}\ ={\frac {xyz}{\ \ {\sqrt {\ x^{2}y^{2}+z^{2}{\Bigl [}x^{2}+y^{2}{\Bigr ]}\ }}\ \ }}}$

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

${\displaystyle A_{abc}={\frac {xyz}{\ 2h_{tet}}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ A_{abc}={\frac {\ \ {\sqrt {x^{2}y^{2}+z^{2}{\Bigl [}x^{2}+y^{2}{\Bigr ]}\ }}\ \ \ }{2}}}$

Area of all 4 surfaces of a Tri-Rectangular Tetrahedron

${\displaystyle A_{abc}={\frac {xyz}{\ 2h_{tet}}}\ \ \ \ A_{xy}={\frac {\ xy\ }{2}}\ \ \ \ A_{xz}={\frac {\ xz\ }{2}}\ \ \ \ A_{yz}={\frac {\ yz\ }{2}}}$

Total area of a Tri-Rectangular Tetrahedron

${\displaystyle A_{tet}={\frac {\ xy+z{\Bigl [}x+y{\Bigr ]}+{\sqrt {x^{2}y^{2}+z^{2}{\Bigl [}x^{2}+y^{2}{\Bigr ]}\ }}\ \ }{2}}}$

${\displaystyle a=14.4\ \ \ \ \ b=10\ \ \ \ \ c=12}$

${\displaystyle x=9.037698822156002781875821128798}$

${\displaystyle y=11.210709165793214885544254958172}$

${\displaystyle z=4.2801869118065393196504607419769}$

Pythagorean's Theorem is TWO dimensional. Here Heron's Theorem is modified for 3 dimensions BUT ONLY for Tri-Rectangular Tetrahedrons. These work but are not as slick as the above interlocking equations.

${\displaystyle Given{\Bigl [}x,y,z{\Bigr ]}\ \ \ \ a={\sqrt {\ x^{2}+y^{2}\ }}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ b={\sqrt {\ x^{2}+z^{2}\ }}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ c={\sqrt {\ y^{2}+z^{2}\ }}}$${\displaystyle Given\ {\Bigl [}a,b,c{\Bigr ]}\ \ \ \ x={\sqrt {{\frac {\ a^{2}+b^{2}-c^{2}\ }{2}}\ }}\ \ \ \ y={\sqrt {{\frac {\ a^{2}+c^{2}-b^{2}\ }{2}}\ }}\ \ \ \ z={\sqrt {{\frac {\ b^{2}+c^{2}-a^{2}\ }{2}}\ }}}$

If the area of the base is ${\displaystyle T_{0}}$ and the areas of the three other (right-angled) faces are ${\displaystyle T_{1}}$, ${\displaystyle T_{2}}$ and ${\displaystyle T_{3}}$, then

${\displaystyle T_{0}^{2}=T_{1}^{2}+T_{2}^{2}+T_{3}^{2}.}$

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

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).

Trirectangular tetrahedrons with integer legs ${\displaystyle a,b,c}$ and sides ${\displaystyle d={\sqrt {b^{2}+c^{2}}},e={\sqrt {a^{2}+c^{2}}},f={\sqrt {a^{2}+b^{2}}}}$ of the base triangle exist, e.g. ${\displaystyle a=240,b=117,c=44,d=125,e=244,f=267}$ (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     1464

Notice that some of these are multiples of smaller ones. Note also A031173.

Trirectangular tetrahedrons with integer faces ${\displaystyle T_{c},T_{a},T_{b},T_{0}}$ and altitude h exist, e.g. ${\displaystyle a=42,b=28,c=14,T_{c}=588,T_{a}=196,T_{b}=294,T_{0}=686,h=12}$ without or ${\displaystyle a=156,b=80,c=65,T_{c}=6240,T_{a}=2600,T_{b}=5070,T_{0}=8450,h=48}$ with coprime ${\displaystyle a,b,c}$.

• Irregular tetrahedra
• Standard simplex
• Euler Brick

• Weisstein, Eric W. "Trirectangular tetrahedron". MathWorld.