Locus of Points at Constant Distance from the Edges of a Trihedron

point A

1.047

point B

1.047

1.083

constant

3

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Choose a constant

a

and a trihedron with apex

O

and edges

OA

,

OB

,

OC

. How do you find the locus of points

P

such that the sum of the distances of

P

to the faces of the trihedron equals

a

?

First, construct the points

A'

,

B'

,

C'

, on the edges

OA

,

OB

,

OC

such that the distances of these points to the opposite faces are

a

.

By construction, each point on the triangle

A'

,

B'

,

C'

has this property, namely, with

P

on the triangle, the distances of this point to the faces of the trihedron are

a

1

,

a

2

,

a

3

.

Let

S

1

,

S

2

,

S

3

be the areas of the corresponding faces. Triple the volume of the tetrahedron

OA'B'C'

. The volume is now

V=a

S

i

/3

, so all the

S

i

are equal; let their common value be

S

. But the volume is also

(

a

1

S+

a

2

S+

a

3

S)/3

. So

a=

a

1

+

a

2

+

a

3

.

Let

A''

,

B''

,

C''

be the points symmetric to

A'

,

B'

,

C'

with respect to the point

O

, forming a second trihedron. Now each point lies inside one of the trihedral angles determined by the six edges of the trihedrons, so the general solution is the convex hull of the six points

A'

,

B'

,

C'

,

A''

,

B''

,

C''

, resembling a triangular antiprism.