Difference between revisions of "Disphenoid"
(Created page with "Disphenoid is a tetrahedron whose four faces are congruent acute-angled triangles. ==Main== 390px|right a) A tetrahedron <math>ABCD</math> is...") |
(→Main) |
||
(One intermediate revision by the same user not shown) | |||
Line 2: | Line 2: | ||
==Main== | ==Main== | ||
− | [[File: | + | [[File:Disphenoid -parallelepiped.png|390px|right]] |
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> | ||
+ | |||
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> | ||
Line 13: | Line 15: | ||
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> |
+ | |||
+ | <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> | ||
+ | |||
+ | 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> | ||
+ | |||
+ | 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''' | ||
+ | ==Constructing== | ||
+ | [[File:Disphenoid -parallelepiped A.png|390px|right]] | ||
+ | Let triangle <math>ABC</math> be given. Сonstruct the disphenoid <math>ABCD.</math> | ||
+ | |||
+ | <i><b>Solution</b></i> | ||
+ | |||
+ | Let <math>\triangle A_1B_1C_1</math> be the anticomplementary triangle of <math>\triangle ABC, M</math> be the midpoint <math>BC.</math> | ||
+ | |||
+ | Then <math>M</math> is the midpoint of segment <math>AA_1 \implies</math> | ||
+ | |||
+ | <math>2MA' = AD, MA' || AD \implies A'</math> is the midpoint <math>A_1D.</math> | ||
+ | |||
+ | Similarly, <math>B'</math> is the midpoint <math>B_1D, C'</math> is the midpoint <math>C_1D.</math> | ||
+ | |||
+ | So, <math>\angle A_1DB_1 = \angle B_1DC_1 = \angle C_1DA_1 = 90^\circ.</math> | ||
+ | |||
+ | Let <math>A_1A_0, B_1B_0, C_1C_0</math> be the altitudes of <math>\triangle A_1B_1C_1, H</math> be the orthocenter of <math>\triangle A_1B_1C_1 \implies DH \perp ABC.</math> | ||
+ | |||
+ | To construct the disphenoid <math>ABCD</math> using given triangle <math>ABC</math> we need: | ||
+ | |||
+ | 1) Construct <math>\triangle A_1B_1C_1,</math> the anticomplementary triangle of <math>\triangle ABC,</math> | ||
+ | |||
+ | 2) Find the orthocenter <math>H</math> of <math>\triangle A_1B_1C_1.</math> | ||
+ | |||
+ | 3) Construct the perpendicular from point <math>H</math> to plane <math>ABC.</math> | ||
+ | |||
+ | 4) Find the point <math>D</math> in this perpendicular such that <math>AD = BC.</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' |
Latest revision as of 20:07, 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
Constructing
Let triangle be given. Сonstruct the disphenoid
Solution
Let be the anticomplementary triangle of
be the midpoint
Then is the midpoint of segment
is the midpoint
Similarly, is the midpoint
is the midpoint
So,
Let be the altitudes of
be the orthocenter of
To construct the disphenoid using given triangle
we need:
1) Construct the anticomplementary triangle of
2) Find the orthocenter of
3) Construct the perpendicular from point to plane
4) Find the point in this perpendicular such that
vladimir.shelomovskii@gmail.com, vvsss