Disphenoid

Disphenoid is a tetrahedron whose four faces are congruent acute-angled triangles.

Main

a) A tetrahedron $ABCD$ is a disphenoid iff $AB = CD, AC = BD, AD = BC.$

b) A tetrahedron is a disphenoid iff its circumscribed parallelepiped is right-angled.

c) Let $AB = a, AC = b, AD = c.$ The squares of the lengths of sides its circumscribed parallelepiped and the bimedians are: $AB'^2 = l^2 = \frac {a^2-b^2+ c^2}{2}, AC'^2 = m^2 = \frac {a^{2}-b^{2}+c^{2}}{2},$ $AD'^2 = n^2 = \frac {-a^{2}+b^{2}+c^{2}}{2}$ The circumscribed sphere has radius (the circumradius): $R=\sqrt {\frac {a^2+b^2+c^2}{8}}.$

The volume of a disphenoid is: $V= \frac {lmn}{3} = \sqrt {\frac {(a^2+b^2-c^2)(a^2-b^2+c^2)(-a^{2}+b^{2}+c^{2})}{72}}$ Each height of disphenoid $ABCD$ is $h=\frac {3V}{[ABC]},$ the inscribed sphere has radius: $r=\frac {3V}{4[ABC]},$ where $[ABC]$ is the area of $\triangle ABC.$

Proof

a) $AB \ne BC, AB \ne AD, \triangle ABC = \triangle BAD = \triangle CDA = \triangle DCB.$

$AB \ne AD, AB \ne BD$ because in $\triangle ABD$ there is no equal sides.

Let consider $\triangle BCD.$

$BD \ne AB, BC \ne AB,$ but one of sides need be equal $AB,$ so $AB = CD \implies AC = BD, AD = BC.$

b) Any tetrahedron can be assigned a parallelepiped by drawing a plane through each edge of the tetrahedron parallel to the opposite edge.

$AB = CD = C'D' \implies AD'BC'$ is parallelogram with equal diagonals, i.e. rectangle.

Similarly, $AB'CD'$ and $AB'DC'$ are rectangles.

If $AD'BC'$ is rectangle, then $AB = C'D' = CD.$

Similarly, $AC = BD, AD = BC \implies ABCD$ is a disphenoid.

c) $AB^2 = a^2 = AC'^2 + BC'^2 = m^2 + AD'^2 = m^2 + n^2.$

Similarly, $AC^2 = b^2 = l^2 + n^2, AD^2 = c^2 = l^2 + m^2 \implies 2(l^2 + m^2 + n^2) = a^2 + b^2 + c^2 \implies$ $l^2 = l^2 + m^2 + n^2 - (m^2 + n^2) = \frac {a^2 + b^2 + c^2}{2} - a^2 = \frac {-a^2 + b^2 + c^2}{2}.$

Similarly, $m^2 = \frac {a^2 - b^2 + c^2}{2}, n^2 = \frac {a^2 + b^2 - c^2}{2}.$

Let $E$ be the midpoint $AC$ , $E'$ be the midpoint $BD \implies$

$EE' = AC' = m = \sqrt {\frac {a^2 - b^2 + c^2}{2}}$ is the bimedian of $AC$ and $BD.$ $EE' || AC' \implies EE' \perp AC, EE' \perp BD.$

The circumscribed sphere of $ABCD$ is the circumscribed sphere of $AB'CD'C'DA'C,$ so it is $\frac {AA'}{2} = \sqrt {\frac {a^2+b^2+c^2}{8}}.$

The volume of a disphenoid is third part of the volume of $AB'CD'C'DA'C,$ so: $V= \frac {l \cdot m \cdot n}{3} = \sqrt {\frac {(a^2+b^2-c^2)(a^2-b^2+c^2)(-a^{2}+b^{2}+c^{2})}{72}}.$ The volume of a disphenoid is $V = \frac {1}{3} h [ABC] \implies h = \frac{3V}{[ABC]},$ where $h$ is any height.

The inscribed sphere has radius $r = \frac{h}{4}.$ $72 V^2 = (a^2+b^2-c^2) \cdot (a^2-b^2+c^2) \cdot (-a^2+b^2+c^2) =-a^6+a^4 b^2+a^4 c^2+a^2 b^4+a^2 c^4-b^6+b^4 c^2+b^2 c^4-c^6 -2a^2 b^2 c^2,$ $16 [ABC]^2 = (a+b+c)(a+b-c)(a-b+c)(-a+b+c) = -a^4+2a^2b^2+2a^2c^2-b^4+2b^2c^2-c^4,$ $8 R^2 = a^2+b^2+c^2,$ $128 R^2 [ABC]^2 = -a^6+a^4 b^2+a^4 c^2+a^2 b^4 +a^2 c^4-b^6+b^4 c^2+b^2 c^4-c^6 + 6a^2 b^2 c^2.$

Therefore $16 R^2 [ABC]^2 = a^2 b^2 c^2 + 9 V^2.$

vladimir.shelomovskii@gmail.com, vvsss

Constructing

Let triangle $ABC$ be given. Сonstruct the disphenoid $ABCD.$

Solution

Let $\triangle A_1B_1C_1$ be the anticomplementary triangle of $\triangle ABC, M$ be the midpoint $BC.$

Then $M$ is the midpoint of segment $AA_1 \implies$

$2MA' = AD, MA' || AD \implies A'$ is the midpoint $A_1D.$

Similarly, $B'$ is the midpoint $B_1D, C'$ is the midpoint $C_1D.$

So, $\angle A_1DB_1 = \angle B_1DC_1 = \angle C_1DA_1 = 90^\circ.$

Let $A_1A_0, B_1B_0, C_1C_0$ be the altitudes of $\triangle A_1B_1C_1, H$ be the orthocenter of $\triangle A_1B_1C_1 \implies DH \perp ABC.$

To construct the disphenoid $ABCD$ using given triangle $ABC$ we need:

1) Construct $\triangle A_1B_1C_1,$ the anticomplementary triangle of $\triangle ABC,$

2) Find the orthocenter $H$ of $\triangle A_1B_1C_1.$

3) Construct the perpendicular from point $H$ to plane $ABC.$

4) Find the point $D$ in this perpendicular such that $AD = BC.$

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 $180^\circ$ 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 $180^\circ.$

The development of the tetrahedron $ABCD$ on the plane $ABC$ is a hexagon $A_1CB_1AC_1B.$

a) If the angular defect of vertex $C$ is $180^\circ,$ then angle $\angle A_1CB_1 = \angle A_1CB + \angle BCA + \angle ACB_1 = \angle DCB + \angle BCA + \angle ACD = 180^\circ,$ so points $A_1, C,$ and $B_1$ are collinear.

Similarly, triples of points $A_1, B, C_1$ and $B_1, A, C_1$ are collinear.

The hexagon $A_1CB_1AC_1B$ is the triangle, where the points $A, B,$ and $C$ are the midpoints of sides $B_1C_1,A_1C_1,$ and $A_1B_1,$ respectively.

Consequently, $\triangle ABC = \triangle A_1CB = \triangle DCB.$

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 $180^\circ.$

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Disphenoid

Contents

Main

Constructing

Properties and signs of disphenoid

Three sums of the plane angles