The Pons Asinorum for Tetrahedra

J. Geom. Online First c 2009 Birkh¨ auser Verlag Basel/Switzerland DOI 10.1007/s00022-009-1972-4

Journal of Geometry

The Pons Asinorum for Tetrahedra Mowaffaq Hajja Abstract. Proposition 5 of Book I of Euclid’s Elements, better known as the Pons Asinorum or the Asses’ Bridge, and its converse, Proposition 6, state that two sides of a triangle are equal if and only if the opposite angles are equal. A generalization of this statement to higher dimensional d-simplices is considered in [16], where it is shown that such a generalization holds only if the underlying d-simplex is orthocentric. In this paper, it is shown that a reasonably satisfactory generalization holds for all, not necessarily orthocentric, tetrahedra. It is also shown that a stronger statement than that given in [16] holds for orthocentric tetrahedra. Mathematics Subject Classification (2000). 51M04, 52B10, 52B15, 51F99. Keywords. Bridge of asses, content of a solid angle, orthocentric tetrahedron, polar sine, pons asinorum, sine of a solid angle, solid angle, tetrahedron, trihedral angle.

1. Introduction and summary of the main results Proposition 5 of Book I of Euclid’s Elements, better known as the Pons Asinorum or the Asses’ Bridge, and its converse, Proposition 6, state that two sides of a triangle are equal if and only if the opposite angles are equal. We shall refer to these two propositions as the pons asinorum theorems. Speculations regarding the reasons behind such an unusual name and regarding the unduly difficult proof given by Euclid can be found in [7, p. 6], [23, p. 54], and [17, pp. 415–416]. A generalization of the pons asinorum theorems to higher dimensional d-simplices is considered in [16], where it is shown that a reasonable generalization holds only if the underlying d-simplex is orthocentric. Here, a d-simplex is orthocentric if its altitudes are concurrent. That includes rectangular d-simplices, i.e., d-simplices in which the edges emanating from one of the vertices are orthogonal to each This work is supported by a research grant from Yarmouk University.

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other. In this paper, we show that a stronger generalization holds when d = 3, i.e., for orthocentric tetrahedra. We also show that a reasonably satisfactory version holds for all tetrahedra. Specifically, we prove that two faces of a tetrahedron are congruent (respectively, have the same area, the same perimeter, or the same product of side-lengths) if and only if the opposite corner angles are congruent (respectively, have the same sine, the same content, or the same polar sine). For orthocentric tetrahedra, we prove that all of these eight properties are equivalent. Thus no matter how one interprets equality among faces and among corner angles of an orthocentric tetrahedron, one gets a valid pons asinorum theorem. The afore-mentioned facts are presented in Theorems 3 and 8. We find these two theorems remarkable in that they are new additions to a subject that has long been exhausted. This statement refers to the geometry of tetrahedra and finds evidence in the many comprehensive classics on the subject (such as [2, 24], and [6]) and in the tremendous number of articles and problems that have appeared in the Monthly and similar journals in the past century. Regarding this last statement, one cannot but mention the late great problemist Victor Thébault who, alone, contributed tens of such articles and hundreds of such problems between 1933 and 1960; see [20, pp. 82–83]. It must be mentioned that a weaker version of Theorem 8 holds for orthocentric d-simplices for all d as proved in [16, Theorem 2]. However, we do not have any higher dimensional analogue of Theorem 3. Thus the questions whether Theorem 3 and all of Theorem 8 remain true in higher dimensions remain open. The paper is self-contained and does not require [16] as a prerequisite.

2. The content, sine, and polar sine of a solid angle A (non-degenerate) solid (or trihedral) angle θ is defined to be any 4-tuple θ = D; A, B, C of affinely independent points (or position vectors) in Euclidean 3space. The point D is called its vertex and DA, DB, and DC are called its arms. Its content μ(θ) is defined to be the area of the spherical triangle whose vertices are the traces of the arms DA, DB, and DC on the unit sphere centered at D; see [12]. This content μ(θ) is given in [12] by μ(θ) = tan 2

1 − cos2 α − cos2 β − cos2 γ + 2 cos α cos β cos γ , 1 + cos α + cos β + cos γ

where α = ∠BDC , β = ∠ADC ,

γ = ∠BDA ,

(1)

The Pons Asinorum for Tetrahedra

3

D β α γ

X

A

Z

Y

B

z

y

γ

x

β α

C Figure 1. A typical tetrahedron.

as shown in Figure 1. Also, the sine and polar sine of θ are defined as in [11] by

2 2 area (DAB) area (DBC) area (DCA) sin θ vol ([D, A, B, C]) = , 3 DA DB DC polsin θ . vol ([D, A, B, C]) = 6

(2) (3)

Here, vol ([D, A, B, C]) is the volume of the tetrahedron [D, A, B, C]. Two angles α = A; A1 , A2 , A3 and β = B; B1 , B2 , B3 are said to be congruent (and we write α ≡ β) if there is an isometry τ of the underlying Euclidean 3-space E and a permutation σ of {1, 2, 3} such that τ (A) = B and τ (Ai −A) = ci (Bσ(i) −B) for positive constants ci , i = 1, 2, 3. It is easy to see that α ≡ β if and only if there exists a permutation σ of {1, 2, 3} such that ∠Ai AAj = ∠Bσ(i) BBσ(j) for all i, j = 1, 2, 3; see [16, Section 2]. If A = [A, B, C, D] is a (non-degenerate) tetrahedron, then the corner angle of A at the vertex D is the solid angle D; A, B, C. When no confusion may arise, we denote this angle by A, in accordance with Euler’s notation for the angles of a triangle. For an explanation of why we don’t simply call corner angles vertex angles, see [16, Note 7]. We say that the two tetrahedra A = [A1 , A2 , A3 , A4 ] and B = [B1 , B2 , B3 , B4 ] are congruent or isometric (and we write A ≡ B) if there is an isometry that takes A1 , A2 , A3 , A4 to B1 , B2 , B3 , B4 in some order. It is well-known that A ≡ B

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if and only if there exists a permutation σ of {1, 2, 3, 4} such that Ai − Aj = Bσ(i) − Bσ(j) for all i and j; see [3, Proposition 9.7.1, p. 236].

3. The pons asinorum for general tetrahedra To prove the first main theorem of this section, we need the following simple lemma. Lemma 1. Let α, β, and γ be the 2-subangles of a corner angle θ of a tetrahedron. Then 1 − cos2 α − cos2 β − cos2 γ + 2 cos α cos β cos γ = polsin θ . (4) Proof. Let θ be the corner angle D; A, B, C of the tetrahedron T = [A, B, C, D] and let the 2-subangles α, β, γ of θ and the side-lengths x, X, y, Y , z, Z of T be as shown in Figure 1. Let V be the volume of T . Using the Cayley-Menger determinant, we see that 144V 2 = (x2 X 2 + y 2 Y 2 + z 2 Z 2 )(x2 + X 2 + y 2 + Y 2 + z 2 + Z 2 ) − (x2 y 2 z 2 + x2 Y 2 Z 2 + X 2 y 2 Z 2 + X 2 Y 2 z 2 ) − 2(x4 X + X 4 x + y 4 Y + Y 4 y + z 4 Z + Z 4 z) ;

(5)

see [3, 9.7.3.1, p. 238] or [4, 9.1, p. 145]. We use the Law of Cosines to express cos α, cos β, and cos γ in terms of the side lengths, and we substitute these in the expression 1 − cos2 α − cos2 β − cos2 γ + 2 cos α cos β cos γ. Using Maple to factor the resulting polynomial, and using (5), we obtain 2 6V 2 2 2 1 − cos α − cos β − cos γ + 2 cos α cos β cos γ = . (6) XY Z The rest follows from the definition 6V = XY Z polsin θ; see (3).

Remark 2. Formula (4) appears with historical remarks in [11, p. 77]. See also [22]. Theorem 3 (The pons asinorum for general tetrahedra). For a general tetrahedron, the following statements hold: (a) Two corner angles have the same content μ if and only if the opposite faces have the same perimeter. (b) Two corner angles have the same sine if and only if the opposite faces have the same area. (c) Two corner angles have the same polar sine if and only if the opposite faces have the same product of side-lengths. (d) Two corner angles are congruent if and only if the opposite faces are congruent.


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Proof. We refer to Figure 1. Part (c) follows immediately from the definition as follows: polsin D = polsin C ⇐⇒ XY Z = xyZ ⇐⇒ XY z = xyz . Part (b) follows from (2). To prove (a), we use (1) and (6) to see that μ(D) = μ(C) 6V 6V = XY Z(1 + cos α + cos β + cos γ) xyZ(1 + cos α + cos β + cos γ ) XY (1 + cos α + cos β + cos γ) = xy(1 + cos α + cos β + cos γ ) . (7)

⇐⇒ ⇐⇒

Using the Law of Cosines, we express the terms of (7) in terms of the side-lengths, and we factor the resulting polynomial using Maple to obtain (7)

⇐⇒ (X + y + Z)(x + Y + Z)(x + y − X − Y ) = 0 ⇐⇒ x + y = X + Y ⇐⇒ triangles ABC and ABD have the same perimeter ,

as claimed in (a). It remains to prove (d). If C ≡ D, then the ordered triple (α, β, γ) is equal to (α , β , γ ), (β , α , γ ), or (β , γ , α ). In the first case, the triangles BAC and BAD are congruent by ASA. In the second case, the triangles BAC and ABD are congruent by ASA. It remains to deal with the last case. By Part (a), we have X +Y = x+y.

(8)

From cos α = cos β and the Law of Cosines, we see that Z 2 + Y 2 − x2 Z 2 + y2 − X 2 = . 2ZY 2Zy Therefore 0 = y(Z 2 + Y 2 − x2 ) − Y (Z 2 + y 2 − X 2 ) = (y − Y )Z 2 + y(Y − x)(Y + x) − Y (y − X)(y + X) = (y − Y )Z 2 + y(y − X)(Y + x) − Y (y − X)(y + X) , = (y − Y )Z + (y − X)(yx − Y X) = (y − Y )Z 2 + (y − X) (y − Y )x + Y (x − X) = (y − Y )Z 2 + (y − X) (y − Y )x + Y (Y − y) ,

by (8)

2

= (y − Y )Z + (y − Y )(y − X)(x − Y ) 2

= (y − Y )Z 2 + (y − Y )(y − X)(X − y) , = (y − Y ) Z 2 − (X − y)2 = (y − Y )(Z − X + y)(Z + X − y) .

by (8)

by (8)

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Therefore Y = y, and the triangles BAC and ABD are congruent by SAS. For the converse, suppose that the triangles BAC and ABD are congruent in some order. Then the pair (α, β) is equal to one of the pairs (α , β ) and (β , α ). Therefore (X, Y ) is equal to (x, y) or (y, x), and the triangle DAB is congruent to either CAB or CBA. Hence γ = γ and C ≡ D, as desired. Remark 4. It easily follows from (a), (b), and (c) of Theorem 3 that if θ and θ are two corner angles of a tetrahedron, then none of the equalities μ(θ) = μ(θ ), sin θ = sin θ , and polsin θ = polsin θ implies any one of the others. This is equivalent to saying that if ABCD is a convex quadrilateral with DA = X, DB = Y , CB = x, and CA = y, then none of the statements X + Y = x + y, XY = xy, and area (DAB) = area (CAB) implies any one of the others. Of course, (d) implies all of them. Remark 5. A tetrahedron whose faces are congruent is called equifacial. It is easy to see that if the faces of a tetrahedron have the same perimeter or the same product of side-lengths, then it is equifacial. A theorem of Bang states that the same holds if the faces have the same area; see [18, pp. 90–97] and [13]. Other characterizations of equifacial tetrahedra are given in [14, Theorem 4] and [15]. The next corollary, which follows immediately from Theorem 3, provides other characterizations. Corollary 6. For a general tetrahedron, the following statements are equivalent: (a) (b) (c) (d) (e)

The The The The The

tetrahedron four corner four corner four corner four corner

is equifacial. angles have the same content. angles have the same sine. angles have the same polar sine. angles are congruent.

4. The pons asinorum for orthocentric tetrahedra We now turn to orthocentric tetrahedra. We use the simple facts that if ABCD is a tetrahedron, then ABCD is orthocentric ⇐⇒ AB2 + CD2 = AC2 + BD2

(9)

⇐⇒ (A − D) · (B − D) = (B − D) · (C − D) = (C − D) · (A − D) . (10) This follows immediately from [16, (14)] and has also appeared in [5] and [21]. We start with a simple lemma that describes a typical non-rectangular orthocentric tetrahedron. It also shows that the shape of a non-rectangular orthocentric tetrahedron is completely determined by any of its corner angles.


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Lemma 7. Let T = [A, B, C, D] be an orthocentric tetrahedron, and let α = ∠BDC ,

β = ∠ADC ,

γ = ∠ADB

(11)

α = ∠BCD ,

β = ∠ACD ,

γ = ∠ACB

(12)

a = | cos α| ,

b = | cos β| ,

c = | cos γ|

(13)

a = cos α ,

b = cos β ,

c = cos γ .

Q = a + b + c − 2abc ; 2

2

2

(14) (15)

see Figure 2. Then, up to similarity, DA = a , CA = Q − b2 ,

DB = b , CB = Q − a2 ,

c − ab , a = Q − a2

DC = c , AB = Q − c2 ,

c − ab b = , Q − b2

c =

c(c − ab) (Q − a2 )(Q − b2 )

(16) (17) . (18)

Also, ⇐⇒ ABCD is degenerate .

Q=1

(19)

Proof. We refer to Figure 2. Without loss of generality, we assume that D is the origin. Since ABCD is orthocentric, it follows from (10) that A·B = B·C = C ·A. From A · B = A B cos γ, etc. and (13), it follows that A : B : C = cos α : cos β : cos γ = a : b : c . Hence we may assume that A = a, B = b, and C = c, thus proving (16). Using the Law of Cosines and (16), we obtain (17). Finally, (18) follows from (16) and (17) using the Law of Cosines again. Note that cos α, cos β, and cos γ have the same sign, and therefore a = cos α ,

b = cos β ,

c = cos γ ,

where = ±1. To prove the last statement, we use the following equivalences. Q = 1 ⇐⇒ ⇐⇒ ⇐⇒ ⇐⇒

c2 − (2ab)c + (a2 + b2 − 1) = 0 c = ab ± a2 b2 − (a2 + b2 − 1) c = ab ± 1 − a2 1 − b2 cos γ = cos α cos β ± sin α sin β

⇐⇒ cos γ = cos(α ± β) ⇐⇒ γ = α ± β or γ = β ± α ⇐⇒ ABCD is degenerate .

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D β α γ a = cos α c = cos γ

√

A

√

b = cos β

a2 + b2 − 2abc

B

√

γ

c2 + a2 − 2abc

b2 + c2 − 2abc

β α

C Figure 2. A typical non-rectangular orthocentric tetrahedron. The last statements follows from the fact that in a non-degenerate trihedral angle, the sum of any two ordinary subangles is greater than the third [19, Theorem 1]. The next theorem is the main theorem of this section. We use the symbols per, rad, and var for perimeter, circumradius, and variance, respectively. Here, the variance of a triangle is the sum of squares of its side-lengths. Theorem 8 (The pons asinorum for orthocentric tetrahedra). Let θ and θ be any two corner angles of an orthocentric tetrahedron, and let F and F be the opposite faces. Then the conditions μ(θ) = μ(θ ) ,

sin θ = sin θ ,

area (F ) = area (F ) , var (F ) = var (F ) ,

polsin θ = polsin θ ,

θ ≡ θ ,

per (F ) = per (F ) , rad (F ) = rad (F ) ,

F ≡ F

are all equivalent. Each is equivalent to the condition cos γ = 2 cos α cos β , where α, β, and γ are the 2-subangles of θ (or of θ ). Proof. In view of [16, Theorem 2], it remains to show that μ(θ) = μ(θ ) ⇐⇒ rad (F ) = rad (F ) ⇐⇒ c = 2ab ⇐⇒ θ ≡ θ .


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However, we will give a proof that does not depend on [16]. Instead, we use Theorem 3. We refer to Figure 2, and we take the corner angles C and D for θ and θ . polsin D = polsin C

⇐⇒

CACBAB = DADBAB ,

⇐⇒

(c + a − 2abc)(c + b − 2abc) = a b

⇐⇒

(Q − b2 )(Q − a2 ) = a2 b2 ,

⇐⇒

where Q = a2 + b2 + c2 − 2abc , Q2 − (a2 + b2 )Q = 0

⇐⇒

Q = a2 + b2

⇐⇒ ⇐⇒

c2 = 2abc c = 2ab

=⇒ =⇒ =⇒

a = a , b = b , c = c , by (15) and (18), α = α , β = β , γ = γ D≡C.

2

2

2

2

by (3),

2 2

Thus we have proved that D ≡ C ⇐⇒ polsin D = polsin C ⇐⇒ c = 2ab .

(20)

Suppose next that μ(D) = μ(C). Then x + y = X + Y , by Theorem 3(a). Referring to Figure 1, it follows from (9) that x2 +X 2 = y 2 +Y 2 and therefore (x−y)(x+y) = (Y − X)(Y + X). Hence x − y = Y − X. From this and x + y = X + Y , we conclude that x = Y and y = X. By α = α and β = β , the triangles DBC and DAC are congruent, and so γ = γ . Thus we have proved that μ(D) = μ(C) ⇐⇒ D ≡ C . If area (DAC) = area (DBC), then ADCD sin β = BDCD sin α, and therefore cos α sin β = cos β sin α. Hence sin(β − α) = 0, and therefore α = β. From this, it follows that the triangles DAC and DBC are congruent. Therefore area (DAC) = area (DBC) ⇐⇒ triangles DAC and DBC are congruent . Alternatively, we use Heron’s formula area of x, y, z = 2(x2 y 2 + y 2 z 2 + z 2 x2 ) − (x4 + y 4 + z 4 ) , where x, y, z is the triangle whose side-lengths are x, y, and z; see [7, 1.53, p. 12]. If area (DAC) = area (DBC), then Heron’s formula implies that 2 16 (area (DAB))2 − area (CAB) = 4c(a2 + b2 − 2abc)(c − 2ab) . This shows that the areas are equal if and only if c = 2ab.

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Next, let R1 and R2 be the circumradii of triangles DAC and DBC. Then Q − a2 Q − b2 = R1 = R2 ⇐⇒ sin α sin β ⇐⇒ (Q − a2 ) sin2 β = (Q − b2 ) sin2 α ⇐⇒

(Q − a2 )(1 − b2 ) = (Q − b2 )(1 − a2 )

⇐⇒

(a2 − b2 )(Q − 1) = 0

⇐⇒ ⇐⇒

a = b , by (19) the triangles DAC and DBC are congruent .

Finally, var (ADC) = var (BDC)

⇐⇒ b2 + (Q − a2 ) = a2 + (Q − b2 ) ⇐⇒ a = b ⇐⇒ the triangles DAC and DBC are congruent .

This completes the proof.

Theorem 8 is, in a sense, the best possible generalization of the two-dimensional pons asinorum. It says that if two corner angles are equal in any reasonable sense (such as content, sine, polar sine), then the opposite faces are congruent, and if two faces are equal in any reasonable sense (such as area, perimeter, circumradius, variance), then the opposite angles are congruent. Remark 9. In the treatment of the Fermat–Torricelli point of a tetrahedron in [1], the average of the cosines of the (ordinary) 2-subangles α, β, and γ of a solid angle θ = D; A, B, C played a special role. Specifically, it was proved that the Fermat–Torricelli point of a tetrahedron is interior if and only if the average of the cosines of the 2-subangles of every corner angle is less than −1/3. A similar statement holds in all dimensions d, in which case −1/3 is replaced by −1/d. This gives rise to consider yet another measure of θ defined by cosD; A, B, C =

cos ∠ADB + cos ∠BDC + cos ∠CDA . 3

It is now natural to ask whether the condition cos θ = cos θ can be added to the list of conditions in Theorem 8. This is an open question. Referring to Figure 2, and letting Q = a2 + b2 + c2 − 2abc as before, we reduce this to the question of whether the equation c − ab c − ab c(c − ab) a+b+c= + + 2 2 Q−a Q−b Q − a2 Q − b2 has any other solution besides c = 2ab.


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References [1] S. Abu-Saymeh, M. Hajja, On the Fermat–Torricelli points of tetrahedra and of higher dimensional simplexes, Math. Mag. 70 (1997), 372–378. [2] N. Altshiller-Court, Modern Pure Solid Geometry, Chelsea Publishing Co., New York, 1964. [3] M. Berger, Geometry I, Springer-Verlag, Berlin, Germany, 1994. [4] C. J. Bradley, Challenges in Geometry, Oxford University Press, Oxford, 2005. [5] B. H. Brown, Isogonal tetrahedra, Amer. Math. Monthly 31 (1924), 371–375. [6] P. Couderc, A. Balliccioni, Premier Livre du T´ etraèdre, Gauthier-Villars, Paris, 1935. [7] H. S. M. Coxeter, Introduction to Geometry, John Wiley, Inc., New York, 1969. [8] A. L. Edmonds, M. Hajja, H. Martini, Coincidences of simplex centers and related facial structures, Beitr¨ age Algebra Geom. 46 (2005), 491–512. [9] A. L. Edmonds, M. Hajja, H. Martini, Orthocentric simplices and their centers, Results Math. 47 (2005), 266–295. [10] A. L. Edmonds, M. Hajja, H. Martini, Orthocentric simplices for which the incenter lies on the Euler line, submitted. [11] F. Ericksson, The law of sines for tetrahedra and n-simplices, Geom. Dedicata 7 (1978), 71–80. [12] F. Ericksson, On the measure of solid angles, Math. Mag. 63 (1990), 184–187. [13] M. Hajja, A vector proof of a theorem of Bang, Amer. Math. Monthly 108 (2001), 562–564. [14] M. Hajja, P. Walker, Equifacial tetrahedra, Internat. J. Math. Ed. Sci. Tech. 32 (2001), 501–508. [15] M. Hajja, P. Walker, Equifaciality of tetrahedra whose incenter and Fermat–Torricelli center coincide, J. Geom. Graph. 9 (2005), 37–41. [16] M. Hajja, The pons asinorum in higher dimensions, preprint. [17] T. L. Heath, Euclid – The Thirteen Books of the Elements, Vol. 1, Second edition, Dover Publications, Inc., New York, 1956. [18] R. Honsberger, Mathematical Gems II, Dolciani Mathematical Expositions No. 2, Mathematical Association of America, Washington, D. C., 1976. [19] M. S. Klamkin, Vector proofs in geometry, Amer. Math. Monthly 77 (1970), 1051– 1065. [20] C. S. Ogilvy, Tomorrow’s Math – Unsolved Problems for the Amateur, Second edition, Oxford University Press, New York, 1972. [21] J. Rosenbaum, Problem 3644, Amer. Math. Monthly 40 (1933), 561–562; Solution, ibid, 42 (1935), 51–53. [22] I. Sofair, Problem 1515, Math. Mag. 70 (1997), 64; Solution, ibid, 71 (1998), 69–70. [23] S. Stahl, Geometry – From Euclid to Knots, Prentice Hall, New Jersey, 2003. [24] V. Thébault, Géométrie dans l’espace (Géométrie du Tétraèdre). Libraire Vuibert, Paris, 1956.

12 Mowaffaq Hajja Mathematics Department Yarmouk University Irbid Jordan e-mail: [email protected] [email protected] Received: 20 October 2006. Revised: 12 April 2007.

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