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Right-Angled Tetrahedron

Let
T=ABCD
be a tetrahedron with the three plane angles at
D
all right angles, that is,
∠ADB=∠BDC=∠CDA=90°
. (This is more explicitly known as a trirectangular tetrahedron.) Let
α=∠CAD
,
β=∠CBD
,
ϕ=∠ACB
. Then
cosϕ=sinαsinβ
. The lines that join the midpoints of opposite edges are equal and meet at a point. The proof, outlined in the Details, implies that these three lines are diagonals of a rectangular prism, intersecting at the center.

Details

Proof
Let
DC=c
. Then
AC=c/sinα
,
BC=c/sinβ
,
AD=ccotα
,
BD=ccotβ
. The length of
AB
can be evaluated from triangle
ABD
using the Pythagorean theorem,
2
AB
=
2
BD
+
2
AD
, and from triangle
ABC
using the law of cosines,
2
AB
=
2
BC
+
2
AC
-2BCACcosϕ
, giving
2
c
(
2
cot
α+
2
cot
β)=
2
c
2
sin
α
+
2
c
2
sin
β
-
2
2
c
sinαsinβ
cosϕ
.
Simplifying this identity, we find:
cosϕ=sinαsinβ
[1, pp. 102 and 117].

References

[1] V. V. Prasolov and I. F. Sharygin, Problems in Stereometry (in Russian), Moscow: Nauka, 1989.

External Links

Permanent Citation

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