Difference between revisions of "Disphenoid"
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Similarly, <math>AB = CD, AD = BC \implies ABCD</math> is the disphenoid. | Similarly, <math>AB = CD, AD = BC \implies ABCD</math> is the disphenoid. | ||
===Equal areas of faces=== | ===Equal areas of faces=== | ||
− | [[File:Areas are equal.png| | + | [[File:Areas are equal.png|430px|right]] |
Prove that <math>ABCD</math> is a disphenoid if the areas of all faces of tetrahedron are equal. | Prove that <math>ABCD</math> is a disphenoid if the areas of all faces of tetrahedron are equal. | ||
Latest revision as of 17:03, 18 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 pair 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
Case 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.
Case 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.
Angular defect at vertex and two pair of opposite edges
The tetrahedron is disphenoid if the sum of the plane angles (the angular defects) at one vertex of the tetrahedron is equal to and two pare of opposite edges are equal.
Proof
WLOG, the sum of the plane angles at vertex is equal to
Consider the development of the tetrahedron on the plane
is a parallelogram
is a parallelogram.
Therefore, is disphenoid.
Circumcenter and incenter coinside
A tetrahedron is a disphenoid if the centers of the circumscribed sphere and the inscribed sphere coincide.
Proof
Denote the incenter - circumcenter,
and
the points of tangency of the inscribed sphere with the faces
and
is circumradius of
is circumradius of
is circumcenter of
is circumradius of
Similarly,
Similarly, the sums of the plane angles in vertices
and
are
Therefore, is the disphenoid.
Circumcenter and centroid coinside
A tetrahedron is a disphenoid if the centers of the circumscribed sphere and the centroid coincide.
Proof
Let and
be the midpoints of the edges
and
be the centroid of the tetrahedron
is the midpoint of
The two sides and the median uniquely determine the third side, so
Similarly, is the disphenoid.
Equal areas of faces
Prove that is a disphenoid if the areas of all faces of tetrahedron are equal.
Proof
Let's complete the tetrahedron to a prism
where
is the area of
Let be the height of
points
are collinear.
Similarly,
Similarly,
is disphenoid.