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Topic: Help on tetrahedrons
Replies: 1   Last Post: Jun 15, 1996 12:03 PM

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Vincent R. Johns

Posts: 131
Registered: 12/6/04
Re: Help on tetrahedrons
Posted: Jun 15, 1996 12:03 PM
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Maciej Kosinski wrote:
>
> How to prove that the angule of the regular tetrahedron is 109
> degrees (or was it 109,5 degrees) in NH3 molecule?
> One of my ideas is to use inscribed and outscribed
> spheres which will give the answer: 2*arctan(r/R), where r is
> radius of the inscribed sphere and R is the radius of the out
> scribed sphere, but I don't know the relation of the r and R.
> Please help.



OK, each face of the tetrahedron is an equilateral triangle. Let's
consider the side (edge of tetrahedron) to have length 1.

B
e
e be
e
e b e
e
e b e
e
e b e
e
e b e
e F
e f b e
e f
e f b e
e f d d C
D d d a e
e a
A e
e
e e
e
e e
e

In the diagram, showing one face, the edges are identified as
"e". The center of the bottom face (not shown) is C. The
perpendicular bisector of an edge of the bottom face is "a",
and point A is the midpoint of the edge (not drawn to scale).
"d" is an angle bisector of the bottom face, intersecting
line "a" at point C. (I know it's crooked in my drawing;
please imagine a straight line there.) "b" is an altitude of
the tetrahedron. A, B, C, D, and F are points. Each of the
edges "e" has length 1.

BD = 1

The altitude of the triangle with a side of 1 is sin(60 deg.)
= distance AB (line segment not shown here; the diagram
gets messy fast as one adds lines)
AB = sqrt(3)/2

AD = 1/2
AC = length of line segment "a"
= AD*tan(30 deg.)
= (1/2)*tan(30 deg.)
= (1/2)*(1/sqrt(3))
= sqrt(3)/6
DC = length of line segment "d"
= (1/2)/cos(30 deg.)
= (1/2)*(2/sqrt(3))
= sqrt(3)/3

The angle that I think you want is the angle as measured at the
center of the tetrahedron to (any) two vertices.

We need a point on line "b" equidistant from B and D.

CD = sqrt(3)/3
BD = 1
BC= sqrt(1 - 1/3)
= sqrt(2/3)

Common distance FD must satisfy
DC^2 + (BC - FD)^2 = FD^2
1/3 + (sqrt(2/3) - FD)^2 = FD^2
1/3 + 2/3 - 2*sqrt(2/3)*FD + FD^2 = FD^2
FD*2*sqrt(2/3) = 1
FD = sqrt(3)/(2*sqrt(2))


Look at DC/FD -- we want the supplement of angle CFD.

arcsin(DC/FD) = arcsin((sqrt(3)/3)/(sqrt(3)/(2*sqrt(2))))
= arcsin((1/3)/(1/(2*sqrt(2))))
= arcsin(2*sqrt(2)/3)

... or about 109.4712 degrees.

-- Vincent Johns







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