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A Vector Proof of a Theorem of Bang Mowaffaq Hajja Bang’s theorem states that if the faces of a tetrahedron have equal areas, then they are congruent. An elegant geometric proof is given in [4, pp. 90–97] and a trigonometry proof is given in [3]. Here I give another proof using vectors. Like that in [3], this proof uses the trigonometric identity sin 2A + sin 2B + sin 2C − 4 sin A sin B sin C = 2 sin(A + B + C) + 8 cos s sin(s − A) sin(s − B) sin(s − C)
(1)
where s = (A + B + C)/2. Let T be a tetrahedron whose faces have equal areas. Place one of the vertices O at the origin and place the other vertices at the points U = xU , V = yV , and W = zW , where x, y, z > 0 and where U, V, W are unit vectors. Let A, B, and C be the trihedral angles at O; see Figure 1. Since x y sin C = yz sin A = zx sin B, it follows that x : y : z = sin A : sin B : sin C. Without loss of generality, we assume that x = sin A, y = sin B, z = sin C. We also set X = cos A, Y = cos B, Z = cos C. Thus x = sin A = |V × W |, X = cos A = V · W,
y = sin B = |W × U |, Y = cos B = W · U,
z = sin C = |U × V | Z = cos C = U · V.
(2) (3)
Figure 1.
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Equating the area of the base U V W with that of a lateral face, we obtain x yz = |(xU − yV ) × (yV − zW )| = |xU × yV + yV × zW + zW × xU |. Dividing by x yz, we obtain 1 = |u + v + w|,
(4)
where u, v, w are the unit vectors defined by w=
U ×V , z
u=
V ×W , x
v=
W ×U . y
(5)
Squaring (4) and recalling that |u| = |v| = |w| = 1, we obtain u · v + v · w + w · u = −1.
(6)
From (5) and the vector identity (α × β) · (γ × δ) = (α · γ )(β · δ) − (α · δ)(β · γ ), we obtain u·v =
XY − Z , xy
v·w =
YZ − X , yz
w·u =
ZX −Y , xz
(7)
where x, y, z, X, Y , Z are as defined in (2) and (3). Substituting (7) in (6) and multiplying by x yz, we obtain x X + yY + z Z − (xY Z + y Z X + z XY ) = x yz. Replacing x X by 12 sin 2A, etc., and using the identity in [1, formula 674, p. 165], which, in view of (2), (3), reads sin(A + B + C) = xY Z + y Z X + z XY − x yz, we see that sin 2A + sin 2B + sin 2C − 2 sin(A + B + C) − 4 sin A sin B sin C = 0. In view of (1), this implies that cos s sin(s − A) sin(s − B) sin(s − C) = 0. Since 0 < A, B, C < π and A + B + C < 2π, it follows that A + B + C = π, A = B + C, B = A + C
or C = A + B.
(8)
None of the last three conditions in (8) holds except when the edges through O are coplanar [5, Theorem 1, p. 1051]. Thus A + B + C = π. The rest is traditional: if we take another vertex (say U ) and cut along the edges that pass through U and open T flat (as shown in Figure 2), then the points U1 , O, U2 are collinear. Similarly for U2 , V , U3 and U3 , W , U1 . Now it is quite transparent why the four triangles U1 OW , U2 O V , U3 W V , and O V W are congruent. This completes the proof. There is a yet more transparent interpretation of (6). If a denotes the area of each face of T , then applying the law of cosines [2] to our tetrahedron, we obtain a 2 = a 2 + a 2 + a 2 − 2a 2 (cos α + cos β + cos γ ), June/July 2001]
NOTES
563
Figure 2.
where α, β, γ are the dihedral angles between the faces at O. Their cosines are given by −(v · w), −(w · u), −(u · v) because u, v, w are the unit normals to those faces. REFERENCES 1. George S. Carr, Formulas and Theorems in Pure Mathematics, 2nd ed., Chelsea, New York, 1970. 2. R. J. Gregorac, A Law of Cosines in R n , Nieuw. Archief voor Wiskunde 3 (1991) 267–269. 3. Mowaffaq Hajja, A Trigonometric Identity that G. S. Carr Missed and a Trigonometric Proof of a Theorem of Bang, Math. Gazette (to appear). 4. Ross Honsberger, Mathematical Gems II, Doliciani Mathematical Expositions No. 2, Mathematical Association of America, Washington, DC, 1976. 5. Murray S. Klamkin, Vector proofs in solid geometry, Amer. Math. Monthly 77 (1970) 1051–1065. University of Sharjah, P. O. Box 27272, Sharjah, United Arab Emirates
[email protected]
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