Difference between revisions of "Disphenoid"
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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 | + | <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>AD'^2 = n^2 = \frac { | + | <cmath>AD'^2 = n^2 = \frac {a^{2}+b^{2}-c^{2}}{2}</cmath> |
The circumscribed sphere has radius (the circumradius): | The circumscribed sphere has radius (the circumradius): | ||
<math>R=\sqrt {\frac {a^2+b^2+c^2}{8}}.</math> | <math>R=\sqrt {\frac {a^2+b^2+c^2}{8}}.</math> | ||
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Therefore <math>16 R^2 [ABC]^2 = a^2 b^2 c^2 + 9 V^2.</math> | Therefore <math>16 R^2 [ABC]^2 = a^2 b^2 c^2 + 9 V^2.</math> | ||
+ | |||
+ | <i><b>Corollary</b></i> | ||
+ | |||
+ | <math>ABC</math> is acute-angled triangle, becouse <math>l^2 = -a^2+b^2 +c^2 >0, m^2 = a^2-b^2+c^2 > 0, n^2 = a^2+b^2-c^2 > 0.</math> | ||
'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
==Constructing== | ==Constructing== | ||
[[File:Disphenoid -parallelepiped A.png|390px|right]] | [[File:Disphenoid -parallelepiped A.png|390px|right]] |
Revision as of 05:00, 16 August 2024
Disphenoid is a tetrahedron whose four faces are congruent acute-angled triangles.
Contents
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
Corollary
is acute-angled triangle, becouse
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
Properties and signs of disphenoid
Three sums of the plane angles
The sums of the plane angles (the angular defects) at any three vertices of the tetrahedron are equal to iff the tetrahedron is disphenoid.
Proof
The sum of the all plane angles of the tetrahedron is the sum of plane angles of four triangles, so the sum of plane angles of fourth vertice is
The development of the tetrahedron on the plane is a hexagon
a) If the angular defect of vertex is then angle so points and are collinear.
Similarly, triples of points and are collinear.
The hexagon is the triangle, where the points and are the midpoints of sides and respectively.
Consequently,
Similarly, all faces of the tetrahedron are equal. The tetrahedron is disphenoid.
b) If the tetrahedron is disphenoid, then any two of its adjacent faces form a parallelogram when developed.
Consequently, the development of the tetrahedron is a triangle, i.e. the sums of the plane angles at the vertices of the tetrahedron are equal to
Angular defects at two vertices and pare of opposite edges
The tetrahedron is disphenoid if the sums of the plane angles (the angular defects) at any two vertices of the tetrahedron are equal to and any two opposite edges are equal.
Proof
Let the sums of the plane angles at vertices and be equal to
Consider the development of the tetrahedron on the plane of face
The triples of points and are collinear. lies on bisector of segment
1. Let the edges and are equal.
is the midsegment
is the midpoint of the segment
the sum of the plane angles at vertices is equal to is disphenoid.
2. The edges other than and are equal. WLOG, Note that in the process of constructing the development onto the plane the image of the face and the face are in different half-planes of the line
Accordingly, the image of the vertex of the face and the vertex are located on different sides of the line
There are only two points on bisector of segment such that distance from is equal to
One of them is designated as on the diagram. It lies at the same semiplane as which is impossible.
The second is the midpoint of segment
the sum of the plane angles at vertices is equal to is disphenoid.