Disphenoid
Disphenoid is a tetrahedron whose four faces are congruent acute-angled triangles.
Contents
[hide]- 1 Main
- 2 Constructing
- 3 Properties and signs of disphenoid
- 3.1 Three sums of the plane angles
- 3.2 Angular defects at two vertices and pair of opposite edges
- 3.3 Angular defect at vertex and two pair of opposite edges
- 3.4 Equal circumradii of faces
- 3.5 Circumcenter and incenter coinside
- 3.6 Circumcenter and centroid coinside
- 3.7 Equal perimeters of faces
- 3.8 Equal areas of faces
- 3.9 Equal plane angles
- 4 Sphere of 12 points
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.
Equal circumradii of faces
Prove that if the radii of the circumcircles of the faces are equal, then tetrahedron is a disphenoid.
Proof
If and are both subtended by equal chords in equal circles, then or
Consider the development of the tetrahedron on the plane
Case 1. The circumcircles of all faces are coincide (case is shown in the right of upper diagram).
Images of the point lies on the
The sum of the plane angles at the vertex is equal to 360^\circ. It is impossible because points are in the same plane.
Case 2. The circumcircles of and two another faces are coincide (case is shown in the left of upper diagram).
Images of the point are It is impossible because points are in the same plane.
Case 3. The circumcircles of and is coincide (case is shown in the right of down diagram).
Points are in the same plane.
Case 4. The development of the tetrahedron on the plane has no coincide circles (case is shown in the left of down diagram).
points are collinear,
Similarly, points are collinear, points are collinear.
So tetrahedron is a 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 perimeters of faces
If the four faces of a tetrahedron have the same perimeter, then the tetrahedron is a disphenoid.
Proof
Denote tetrachedron Denote sides Similarly, is 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.
Equal plane angles
Prove that if then tetrahedron is a disphenoid.
Proof
Let be the midpoints of segments Denote
By applying the Law of Cosines to we get: Similarly, We add equations (1) + (3) and (2) + (4) and get: or
We add equations (1) + (4) and (2) + (3) and get: or
If then but
If
If
If
In any case, two pairs of opposite edges of a tetrahedron are equal.
WLOG,
is dishenoid.
vladimir.shelomovskii@gmail.com, vvsss
Sphere of 12 points
Prove that in a disphenoid the bases of the heights, the midpoints of the heights and the orthocenters of the faces lie on the sphere centered at the centroid of the disphenoid.
Proof
Consider the development of the disphenoid on the plane
Let be the base of the height of disphenoid. is the orthocenter
Let be the orthocenter
Let be the center of Euler circle. is the midpoint of and the center of
Let be the centroid of is the incenter and circumcenter of
Let be the midpoint of the height Therefore, is the midpoint of hypotenuse of right
All faces of are equal and all heights are equal, so all distances from to points similar to points and are the same.
vladimir.shelomovskii@gmail.com, vvsss