Difference between revisions of "Disphenoid"
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a) A tetrahedron <math>ABCD</math> is a disphenoid iff <math>AB = CD, AC = BD, AD = BC.</math> | a) A tetrahedron <math>ABCD</math> is a disphenoid iff <math>AB = CD, AC = BD, AD = BC.</math> | ||
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b) A tetrahedron is a disphenoid iff its circumscribed parallelepiped is right-angled. | b) A tetrahedron is a disphenoid iff its circumscribed parallelepiped is right-angled. | ||
+ | |||
c) Let <math>AB = a, AC = b, AD = c.</math> The squares of the lengths of sides its circumscribed parallelepiped and the bimedians are: | c) Let <math>AB = a, AC = b, AD = c.</math> The squares of the lengths of sides its circumscribed parallelepiped and the bimedians are: | ||
<cmath>AB'^2 = l^2 = \frac {a^2-b^2+ c^2}{2}, AC'^2 = m^2 = \frac {a^{2}-b^{2}+c^{2}}{2},</cmath> | <cmath>AB'^2 = l^2 = \frac {a^2-b^2+ c^2}{2}, AC'^2 = m^2 = \frac {a^{2}-b^{2}+c^{2}}{2},</cmath> | ||
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The volume of a disphenoid is: | The volume of a disphenoid is: | ||
<cmath>V= \frac {lmn}{3} = \sqrt {\frac {(a^2+b^2-c^2)(a^2-b^2+c^2)(-a^{2}+b^{2}+c^{2})}{72}}</cmath> | <cmath>V= \frac {lmn}{3} = \sqrt {\frac {(a^2+b^2-c^2)(a^2-b^2+c^2)(-a^{2}+b^{2}+c^{2})}{72}}</cmath> | ||
− | Each height of disphenoid <math>ABCD</math> is <math>h=\frac {3V}{[ABC]},</math> the inscribed sphere has radius: <math>r=\frac {3V}{4[ABC]}.</math> | + | Each height of disphenoid <math>ABCD</math> is <math>h=\frac {3V}{[ABC]},</math> the inscribed sphere has radius: <math>r=\frac {3V}{4[ABC]},</math> where <math>[ABC]</math> is the area of <math>\triangle ABC.</math> |
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+ | <i><b>Proof</b></i> | ||
+ | |||
+ | a) <math>AB \ne BC, AB \ne AD, \triangle ABC = \triangle BAD = \triangle CDA = \triangle DCB.</math> | ||
+ | |||
+ | <math>AB \ne AD, AB \ne BD</math> because in <math>\triangle ABD</math> there is no equal sides. | ||
+ | |||
+ | Let consider <math>\triangle BCD.</math> | ||
+ | |||
+ | <math>BD \ne AB, BC \ne AB,</math> but one of sides need be equal <math>AB,</math> so <math>AB = CD \implies AC = BD, AD = BC.</math> | ||
+ | |||
+ | b) Any tetrahedron can be assigned a parallelepiped by drawing a plane through each edge of the tetrahedron parallel to the opposite edge. | ||
+ | |||
+ | <math>AB = CD = C'D' \implies AD'BC'</math> is parallelogram with equal diagonals, i.e. rectangle. | ||
+ | |||
+ | Similarly, <math>AB'CD'</math> and <math>AB'DC'</math> are rectangles. | ||
+ | |||
+ | If <math>AD'BC'</math> is rectangle, then <math>AB = C'D' = CD.</math> | ||
+ | |||
+ | Similarly, <math>AC = BD, AD = BC \implies ABCD</math> is a disphenoid. | ||
+ | |||
+ | c) <math>AB^2 = a^2 = AC'^2 + BC'^2 = m^2 + AD'^2 = m^2 + n^2.</math> | ||
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+ | Similarly, <math>AC^2 = b^2 = l^2 + n^2, AD^2 = c^2 = l^2 + m^2 \implies 2(l^2 + m^2 + n^2) = a^2 + b^2 + c^2 \implies</math> | ||
+ | <cmath>l^2 = l^2 + m^2 + n^2 - (m^2 + n^2) = \frac {a^2 + b^2 + c^2}{2} - a^2 = \frac {-a^2 + b^2 + c^2}{2}.</cmath> | ||
+ | |||
+ | Similarly, <math>m^2 = \frac {a^2 - b^2 + c^2}{2}, n^2 = \frac {a^2 + b^2 - c^2}{2}.</math> | ||
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+ | Let <math>E</math> be the midpoint <math>AC</math>, <math>E'</math> be the midpoint <math>BD \implies</math> | ||
+ | |||
+ | <math>EE' = AC' = m = \sqrt {\frac {a^2 - b^2 + c^2}{2}}</math> is the bimedian of <math>AC</math> and <math>BD.</math> | ||
+ | <cmath>EE' || AC' \implies EE' \perp AC, EE' \perp BD.</cmath> | ||
+ | |||
+ | The circumscribed sphere of <math>ABCD</math> is the circumscribed sphere of <math>AB'CD'C'DA'C,</math> so it is | ||
+ | <cmath>\frac {AA'}{2} = \sqrt {\frac {a^2+b^2+c^2}{8}}.</cmath> | ||
+ | |||
+ | The volume of a disphenoid is third part of the volume of <math>AB'CD'C'DA'C,</math> so: | ||
+ | <cmath>V= \frac {l \cdot m \cdot n}{3} = \sqrt {\frac {(a^2+b^2-c^2)(a^2-b^2+c^2)(-a^{2}+b^{2}+c^{2})}{72}}.</cmath> | ||
+ | The volume of a disphenoid is <math>V = \frac {1}{3} h [ABC] \implies h = \frac{3V}{[ABC]},</math> where <math>h</math> is any height. | ||
+ | |||
+ | The inscribed sphere has radius <math>r = \frac{h}{4}.</math> | ||
+ | <cmath>72 V^2 = (a^2+b^2-c^2) \cdot (a^2-b^2+c^2) \cdot (-a^2+b^2+c^2) =-a^6+a^4 b^2+a^4 c^2+a^2 b^4+a^2 c^4-b^6+b^4 c^2+b^2 c^4-c^6 -2a^2 b^2 c^2,</cmath> | ||
+ | <cmath>16 [ABC]^2 = (a+b+c)(a+b-c)(a-b+c)(-a+b+c) = -a^4+2a^2b^2+2a^2c^2-b^4+2b^2c^2-c^4,</cmath> | ||
+ | <cmath>8 R^2 = a^2+b^2+c^2,</cmath> | ||
+ | <cmath>128 R^2 [ABC]^2 = -a^6+a^4 b^2+a^4 c^2+a^2 b^4 +a^2 c^4-b^6+b^4 c^2+b^2 c^4-c^6 + 6a^2 b^2 c^2.</cmath> | ||
+ | |||
+ | Therefore <math>16 R^2 [ABC]^2 = a^2 b^2 c^2 + 9 V^2.</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' |
Revision as of 20:01, 14 August 2024
Disphenoid is a tetrahedron whose four faces are congruent acute-angled triangles.
Main
a) A tetrahedron is a disphenoid iff
b) A tetrahedron is a disphenoid iff its circumscribed parallelepiped is right-angled.
c) Let The squares of the lengths of sides its circumscribed parallelepiped and the bimedians are:
The circumscribed sphere has radius (the circumradius):
The volume of a disphenoid is:
Each height of disphenoid
is
the inscribed sphere has radius:
where
is the area of
Proof
a)
because in
there is no equal sides.
Let consider
but one of sides need be equal
so
b) Any tetrahedron can be assigned a parallelepiped by drawing a plane through each edge of the tetrahedron parallel to the opposite edge.
is parallelogram with equal diagonals, i.e. rectangle.
Similarly, and
are rectangles.
If is rectangle, then
Similarly, is a disphenoid.
c)
Similarly,
Similarly,
Let be the midpoint
,
be the midpoint
is the bimedian of
and
The circumscribed sphere of is the circumscribed sphere of
so it is
The volume of a disphenoid is third part of the volume of so:
The volume of a disphenoid is
where
is any height.
The inscribed sphere has radius
Therefore
vladimir.shelomovskii@gmail.com, vvsss