The trigondodecahedron (also pyramid tetrahedron ) is a polyhedron with twelve congruent equilateral triangles as faces, 8 corners and 18 edges. Five edges adjoin four of the corners and four edges adjoin the other four corners.
It is a deltahedron and the Johnson body J 84 of 92, all named after the mathematician Norman Johnson .
Cartesian coordinates
The Cartesian coordinates of the corner points can be, with center at origin and edge length 2:
(
0
,
±
1
,
p
)
(
±
r
,
0
,
q
)
(
0
,
±
r
,
-
q
)
(
±
1
,
0
,
-
p
)
{\ displaystyle {\ begin {alignedat} {4} & (& 0, && \; \ pm 1, && \; p &) \\ & (& \ pm r, && \; 0, && \; q &) \\ & (& 0, && \; \ pm r, && \; - q &) \\ & (& \ pm 1, && \; 0, && \; - p &) \ end {alignedat}}}
According to the Pythagorean theorem:
r
2
+
(
p
-
q
)
2
=
3
{\ displaystyle r ^ {2} + (pq) ^ {2} = 3}
(
r
-
1
)
2
+
(
p
+
q
)
2
=
4th
{\ displaystyle (r-1) ^ {2} + (p + q) ^ {2} = 4}
r
2
+
2
q
2
=
2
{\ displaystyle r ^ {2} + 2q ^ {2} = 2}
It follows:
p
=
1
2
(
3
+
2
r
-
r
2
+
3
-
r
2
)
≈
1
,
57
{\ displaystyle p = {\ frac {1} {2}} ({\ sqrt {3 + 2r-r ^ {2}}} + {\ sqrt {3-r ^ {2}}}) \ approx 1 { ,} 57}
q
=
1
2
(
3
+
2
r
-
r
2
-
3
-
r
2
)
=
2
-
r
2
2
{\ displaystyle q = {\ frac {1} {2}} ({\ sqrt {3 + 2r-r ^ {2}}} - {\ sqrt {3-r ^ {2}}}) = {\ sqrt {\ frac {2-r ^ {2}} {2}}}}
q
=
x
≈
0.411
123
{\ displaystyle q = {\ sqrt {x}} \ approx 0 {,} 411123}
as one of three solutions to the equation .
2
x
3
+
11
x
2
+
4th
x
-
1
=
0
{\ displaystyle 2x ^ {3} + 11x ^ {2} + 4x-1 = 0}
r
≈
1.289
169
{\ displaystyle r \ approx 1 {,} 289169}
as one of three solutions to the equation .
r
3
-
3
r
2
-
4th
r
+
8th
=
0
{\ displaystyle r ^ {3} -3r ^ {2} -4r + 8 = 0}
The coordinates show that a minimal cuboid enclosing the body has the shape of a square column with a height of and a width of . All six faces of the column touch one edge of the trigonododecahedron. Due to the fact that the aforementioned casing body no dice is that trigondodecahedron neither has circumsphere still inscribed ball or midsphere .
2
p
{\ displaystyle 2p}
r
2
{\ displaystyle r {\ sqrt {2}}}
Formulas
Sizes of a trigonododecahedron with edge length a
volume
V
=
r
12
a
3
(
6th
3
-
r
2
+
r
4th
-
2
r
2
)
{\ displaystyle V = {\ frac {r} {12}} a ^ {3} \ left (6 {\ sqrt {3-r ^ {2}}} + r {\ sqrt {4-2r ^ {2} }} \ right)}
Surface area
A.
O
=
3
a
2
3
{\ displaystyle A_ {O} = 3a ^ {2} {\ sqrt {3}}}
1. Face angle ≈ 96.2 °
cos
α
1
=
1
3
(
3
-
2
r
2
)
{\ displaystyle \ cos \, \ alpha _ {1} = {\ frac {1} {3}} \ left (3-2r ^ {2} \ right)}
2. Face angle ≈ 121.74 °
cos
α
2
=
1
3
(
1
-
2
r
)
=
1
3
(
4th
q
2
-
1
)
{\ displaystyle \ cos \, \ alpha _ {2} = {\ frac {1} {3}} \ left (1-2r \ right) = {\ frac {1} {3}} \ left (4q ^ { 2} -1 \ right)}
3. Face angle ≈ 166.44 °
cos
α
3
=
2
3
(
1
-
p
2
)
{\ displaystyle \ cos \, \ alpha _ {3} = {\ frac {2} {3}} \ left (1-p ^ {2} \ right)}
See also
Web links
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