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

Corollary

$ABC$ is acute-angled triangle, becouse $l^2 = -a^2+b^2 +c^2 >0, m^2 = a^2-b^2+c^2 > 0, n^2 = a^2+b^2-c^2 > 0.$

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 $180^\circ.$

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 $180^\circ$ and any two opposite edges are equal.

Proof

Let the sums of the plane angles at vertices $A$ and $B$ be equal to $180^\circ.$

Consider the development of the tetrahedron on the plane of face $ABC.$

The triples of points $A_1, B, C_1$ and $B_1, A, C_1$ are collinear. $A_1C = B_1C = CD \implies C$ lies on bisector of segment $A_1B_1.$

Case 1. Let the edges $AB$ and $CD$ are equal.

$AB$ is the midsegment $\triangle A_1C_1B_1 \implies A_1B_1 = 2AB.$

$CD = AB \implies C$ is the midpoint of the segment $A_1B_1 \implies$

the sum of the plane angles at vertices $C$ is equal to $180^\circ \implies ABCD$ is disphenoid.

Case 2. The edges other than $AB$ and $CD$ are equal. WLOG, $AC = BD.$ $A_1B = C_1B = BD = AC = \frac{A_1C_1}{2}.$ Note that in the process of constructing the development onto the plane $ABC,$ the image of the face $ABD$ and the face $ABC$ are in different half-planes of the line $AB.$

Accordingly, the image of the vertex $D$ of the face $ABD$ and the vertex $C$ are located on different sides of the line $AB.$

There are only two points on bisector of segment $A_1B_1$ such that distance from $A$ is equal to $BD.$

One of them is designated as $X$ on the diagram. It lies at the same semiplane $AB$ as $C_1$ which is impossible.

The second is the midpoint of segment $A_1B_1 \implies$

the sum of the plane angles at vertices $C$ is equal to $180^\circ \implies ABCD$ 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 $180^\circ$ and two pare of opposite edges are equal.

Proof

WLOG, the sum of the plane angles at vertex $A$ is equal to $180^\circ, AB = CD, AD = BC.$

Consider the development of the tetrahedron on the plane $ABC.$

$AB = B_1C, BC = AB_1 \implies ABCB_1$ is a parallelogram $\implies BC = AC_1, BC||AC_1 \implies BCAC_1$ is a parallelogram.

Therefore, $BC_1 = AC = BD\implies ABCD$ is disphenoid.

Equal circumradii of faces

Prove that if the radii of the circumcircles of the faces are equal, then tetrahedron $ABCD$ is a disphenoid.

Proof

If $\angle ADB$ and $\angle ACB$ are both subtended by equal chords in equal circles, then $\angle ADB = \angle ACB,$ or $\angle ADB + \angle ACB = 180^\circ.$

Consider the development of the tetrahedron on the plane $ABC.$

Case 1. The circumcircles of all faces are coincide (case is shown in the right of upper diagram).

Images of the point $D$ lies on the $\odot ABC.$ $\angle ADB + \angle CDB + \angle ADC = \angle AFB + \angle BGC + \angle AEC =$ $=180^\circ - \angle ACB +180^\circ - \angle BAC + 180^\circ - \angle CAB = 3 \cdot 180^\circ - 180^\circ = 360^\circ.$

The sum of the plane angles at the vertex $D$ is equal to 360^\circ. It is impossible because points $A,B,C,D$ are in the same plane.

Case 2. The circumcircles of $\triangle ABC$ and two another faces are coincide (case is shown in the left of upper diagram).

Images of the point $D$ are $D', E, F, BD' = BF \implies \triangle ABD' = \triangle ABF \implies AE = AD' = AF , CE = CF \implies \angle AEC = \angle AFC = 90^\circ.$ $\angle BFC = \angle ADB + \angle AEC.$ It is impossible because points $A, B, C, D$ are in the same plane.

Case 3. The circumcircles of $\triangle ABC$ and $\triangle FBC$ is coincide (case is shown in the right of down diagram).

$D'B = FB, CF = CE, \angle AEC = \angle AFC,$ $\angle AFB = \angle AD'B \implies$ $=\angle BFC = \angle AD'B + \angle AEC.$ Points $A, B, C, D$ are in the same plane.

Case 4. The development of the tetrahedron on the plane $ABC$ has no coincide circles (case is shown in the left of down diagram).

$\angle AEC = \angle ABC \implies AE||BC, \angle ACB = \angle AD'B \implies AD' || BC \implies$ points $D', A, E$ are collinear, $AD' = AE.$

Similarly, $CE = CF,$ points $C,E,F$ are collinear, $BD' = FB,$ points $D', B, F$ are collinear.

So $\triangle ABC = \triangle DCB = \triangle BAD = \triangle CDA,$ tetrahedron $ABCD$ 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 $I$ the incenter - circumcenter, $E$ and $F$ the points of tangency of the inscribed sphere with the faces $ABC$ and $ACD.$

$AF$ is circumradius of $\triangle ABC, AE$ is circumradius of $\triangle ACD.$

$E$ is circumcenter of $\triangle ABC, AF$ is circumradius of $\triangle ACD.$

$IF = IE, IE \perp ABC, IF \perp ACD \implies$ $\triangle AIE = \triangle AIF \implies AF = AE.$ $GF = GE, FG \perp AC, EG \perp AC, AF = AE \implies$ $\triangle AFG = \triangle AEG \implies \angle AFG = \angle AEG.$ $\angle ABC = \angle AEG = \angle AFG = \angle ADC.$ Similarly, $\angle ADB = \angle ACB, \angle BDC = \angle BAC \implies$ $\angle ADC + \angle CDB + \angle BDA = \angle ABC + \angle CAB + \angle BCA = 180^\circ.$ Similarly, the sums of the plane angles in vertices $A$ and $B$ are $180^\circ.$

Therefore, $ABCD$ 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 $E$ and $F$ be the midpoints of the edges $AC$ and $BD,$ $O$ be the centroid of the tetrahedron $ABCD.$ $O$ is the midpoint of $FE, FO = OE, OA = OB = OC = OD.$

The two sides and the median uniquely determine the third side, so $AC = BD.$

Similarly, $AB = CD, AD = BC \implies ABCD$ 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 $ABCD.$ Denote sides $AB = x, BC = y, CA = z, AD = a, BD = b, CD = c.$ $p = x+y+z = x + b +c = y + a + c = z + a + b \implies$ $4p = 2(x+y+z+a+b+c) \implies$ $2p = x+y+z+a+b+c = x+y+z + x + b +c \implies a = x.$ Similarly, $b = y, c = z \implies ABCD$ is disphenoid.

Equal areas of faces

Prove that $ABCD$ is a disphenoid if the areas of all faces of tetrahedron are equal.

Proof

Let's complete the tetrahedron $ABCD$ to a prism $ABCDEF$ $BE = CE = AD, BE || AD, CE || AD \implies$ $DF = AC, DE = AB, EF = BC, EF||BC.$ $\triangle ACD = \triangle FDC, \triangle ABD = \triangle EDB \implies$ $[ABC] = [BCD] = [CDF] = [DEF] = [BDE],$ where $[X]$ is the area of $X.$

Let $DH$ be the height of $DBCFE, DI \perp CE, DI' \perp BE \implies HI \perp CF, HI' \perp BE \implies$ points $I,H,I'$ are collinear.

$[CDF] =[BDE], BE = CF \implies DI = DI' \implies HI = HI'.$ Similarly, $HJ = HJ' \implies H = BF \cap CE \implies$ $CH = EH, BH = HF \implies$ $DB = DF = AC, DC = DE = AB.$ Similarly, $AD = CB \implies ABCD$ is disphenoid.

Equal plane angles

Prove that if $\angle BAC = \angle ABD = \angle ACD = \angle BDC < 90^\circ,$ then tetrahedron $ABCD$ is a disphenoid.

Proof

Let $S, E, F, H, G$ be the midpoints of segments $BC, AC, BD, CD, AB,$ $\angle BAC = \alpha, BC = 2a, AD = 2b.$ $FS || CD, SH || FD \implies \angle FSH = \angle BDC = \alpha.$ $GS ||AC,SE || AB \implies \angle GSE = \angle BAC = \alpha.$ $FS ||CD, GS || AC \implies \angle GSF = \angle ACD = \alpha.$ $HS ||FD, SE || AB \implies \angle HSE = \angle ACD = \alpha.$ Denote $FS = x, ES = y, HS = u, GS = v.$

By applying the Law of Cosines to $\triangle SFH,$ we get: $x^2 +u^2 -2 x u \cos \alpha = a^2. \hspace{10mm} (1)$ Similarly, $y^2 +v^2 - 2 y v \cos \alpha = a^2, \hspace{10mm} (2)$ $x^2 +v^2 - 2 x v \cos \alpha = b^2, \hspace{10mm} (3)$ $y^2 +u^2 - 2 y u \cos \alpha = b^2. \hspace{10mm} (4)$ We add equations (1) + (3) and (2) + (4) and get: $x^2 +u^2 - 2 x u \cos \alpha + x^2 +v^2 - 2 x v \cos \alpha = a^2 + b^2.$ $y^2 +v^2 - 2 y v \cos \alpha + y^2 +u^2 - 2 y u \cos \alpha = a^2 + b^2.$ $2(x^2 - y^2) = 2 \cos \alpha (x - y) (u +v) \implies x = y$ or $\cos \alpha = \frac {x + y}{u+v}.$

We add equations (1) + (4) and (2) + (3) and get: $x^2 +u^2 - 2 x u \cos \alpha + y^2 +u^2 - 2 y u \cos \alpha = a^2 + b^2.$ $y^2 +v^2 - 2 y v \cos \alpha + x^2 +v^2 - 2 x v \cos \alpha = a^2 + b^2.$ $2(u^2 - v^2) = 2 \cos \alpha (x + y) (u -v) \implies u = v$ or $\cos \alpha = \frac {u + v}{x+y}.$