Other Versions of the Steiner-Lehmus Theorem

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as the set of all complex straight lines through the origin of Cn . The fixed points of this mapping (if there are any) are exactly the A-invariant 1-dimensional subspaces.

has a fixed point. Hence, the characteristic polynomial of A has a zero if and only if A Now, given the polynomial p(z) of degree n ≥ 1, which we may assume to be monic, let A be any linear operator on Cn whose characteristic polynomial is precisely p(z) [1, pp. 69–70] and consider the linear homotopy At = t A + (1 − t)I , where t ∈ [0, 1] and I denotes the identity mapping on Cn . If det At = 0 for some t, then t = 0 and we are done, for 1 − t −1 turns out to be a zero of p(z). Otherwise, At is a linear isomorphism for all values of t, inducing thus,

and the identity mapping on CP(n − 1). In particular,

t between A the homotopy A these two mappings have the same Lefschetz number λ = n (the Euler characteristic

and, consequently, of of CP(n − 1)). This ensures the existence of a fixed point of A a zero of p(z). REFERENCES 1. William C. Brown, Matrices over Commutative Rings, Marcel Dekker, Inc., New York, 1993. 2. Edwin H. Spanier, Algebraic Topology, Springer-Verlag, New York, 1966. Instituto de Matemática, UFRJ, Departamento de Matemática Aplicada, Ilha do Fundão, Rio de Janeiro, Brazil [email protected]

Other Versions of the Steiner-Lehmus Theorem Mowaffaq Hajja The Steiner-Lehmus Theorem Inside-out was the name given in [1] to the following theorem: Theorem 1. If B B and CC are the cevians of the acute-angled triangle ABC through its circumcenter O (Figure 1) and if AB = AC , then AB = AC.

Figure 1.

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c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 108

Figure 2.

A cevian is a line segment that joins a vertex to a point on the opposite side. The Gergonne center G of a triangle ABC is the intersection of the cevians A A , B B , CC defined by AB = AC ,

BC = B A ,

C A = C B ;

see Figure 2. If the Gergonne center coincides with the circumcenter, Theorem 1 implies that the triangle is equilateral. As far as this implication is concerned, the assumption in Theorem 1 that ABC is acute-angled is harmless since the Gergonne center is always interior while the circumcenter is interior for acute-angled triangles only. Is Theorem 1 valid if the assumption that ABC is acute-angled is removed? Theorem 1 was so-named because of its elegant equivalence with the well-known Steiner-Lehmus Theorem: If two internal angle bisectors of a triangle are equal then the triangle is isosceles. Figure 3 exhibits both the proof (of Theorem 1) and that equivalence: From O and A we draw lines parallel to AB and CC , respectively, and we let H

Figure 3.

October 2001]

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be their intersection point. We extend AH and O B to meet at D. Then E H and AB bisect DO A and D AO, respectively, internally. They are also equal because O H = AC . Thus D AO = DO A. Since D AO = 180◦ − 2B and DO A = 180◦ − 2C, it follows that B = C and AB = AC. Let us now drop the assumption that ABC is acute-angled. The several possibilities for the configuration A, B, C, O, A , B , C are exhibited in Figures 4(i)–(v). If we follow word by word our treatment of the acute-angled case, we end up again with a triangle AO D (with angles that can be easily computed from those of ABC) and equal angle bisectors AB and O H . However, these bisectors can be external, as shown in Figures 4(i)–(v). Is there a Steiner-Lehmus Theorem that deals with the external

Figure 4(i). C > 90, A > C − 90

Figure 4(ii). C > 90, A < C − 90

Figure 4(iii). A > 90, B > A − 90, C > A − 90

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Figure 4(iv). A > 90, B > A − 90, C < A − 90

Figure 4(v). A > 90, B < A − 90, C < A − 90

bisectors? Before revealing our answer in Theorem 4, we summarize in the following theorem what we have learned from Figures 3 and 4. Here the angles of AO D are denoted by U , V , W . Theorem 2. Let O be the circumcenter of the triangle ABC and let A A , B B , and CC be the cevians through O. Suppose that O has the Gergonne property AB = AC . October 2001]

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(a) If ABC is acute-angled and if U V W is a triangle whose angles are U = 180◦ − 2A,

V = 180◦ − 2B,

W = 180◦ − 2C,

then the internal angle bisectors of V and W are equal. (b) If A is obtuse and if U V W is a triangle whose angles are U = 2A − 180◦ ,

V = 2B,

W = 2C,

then the external angle bisectors of V and W are equal. (c) If C is obtuse and if U V W is a triangle whose angles are U = 2A,

V = 2B,

W = 2C − 180◦ ,

then the external angle bisector of V and the internal angle bisector of W are equal. If V D is the internal angle bisector of the angle V of U V W (Figure 5), then U D/DW = U V /W V and therefore UD =

wv , w+u

WD =

uv , w+u

where u, v, w are the sides of the triangle U V W . Using the Law of Cosines on the triangles U DV and W DV , and eliminating cos W DV , we see that 1 v | V D |2 = uvw . (1) − v (u + w)2 In a similar way, we find that if V D ∗ is the external bisector of V , then | V D ∗ |2 is obtained from (1) by replacing w by −w (or equivalently, by replacing u by −u). Thus the conclusions in (a), (b), (c) in Theorem 2 can be restated as algebraic relations among u, v, and w as follows: (a )

1 1 v w = − − v (u + w)2 w (u + v)2

(b )

1 1 v w = − − 2 v (u − w) w (u − v)2

(c )

1 −1 v w = . − + 2 v (u − w) w (u + v)2

Figure 5.

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Condition (a ) implies that v = w because the assumption v > w would lead to 1 1 < , v w

w v > , 2 (u + w) (u + v)2

contradicting (a ). More directly, (a ) simplifies to the equation (w − v)P = 0, where P is the positive polynomial P = (u + v)2 (u + w)2 + vw(u 2 + v 2 + w2 + vw + 2uv + 2uw). In fact, P can be factored as P = (u + v + w)(u 3 + u 2 (v + w) + 3uvw + vw(v + w)). (b ) does not lead to a similar contradiction. Since it is obtained from (a ) by replacing u by −u, it follows that if a triangle satisfies (b ), then either v = w or Q(u, v, w) = u 3 − u 2 (v + w) + 3uvw − vw(v + w) = 0.

(2)

The existence of a triangle satisfying (2) was first noted by Steiner. The elegant example (U, V, W ) = (36◦ , 12◦ , 132◦ ) of such a triangle is given in [2], which discusses pseudo-isosceles triangles in detail and notes that (2) can be reduced to the much more elegant relation sin2

U V W = sin sin ; 2 2 2

(3)

see Note 1 for a proof. In view of the relations U = 2A − 180◦ , V = 2B, W = 2C, (3) in turn takes the form cos2 A = sin B sin C

(4)

The identity (c ) is obtained from (b ) by replacing v by −v. Thus it is equivalent to the relation Q(u, −v, w) = u 3 + u 2 (v − w) − 3uvw − vw(v − w) = 0.

(5)

In Note 1 we show that (5) is equivalent to cos2

U V W = sin cos , 2 2 2

(6)

which, in view of (c ), surprisingly yields the same relation (4). Thus (b ) and (c ) translate into the same condition (4) about A, B, C. Because of the relation cos A = − cos(B + C) = sin B sin C − cos B cos C, (4) implies that cos B cos C = cos2 A − cos A = cos A(cos A − 1),

(7)

from which it follows that cos A, cos B, and cos C cannot all be positive. Thus (4) implies that ABC is obtuse-angled. This allows us to combine (a), (b), and (c) of Theorem 3 in the following generalization of Theorem 1: October 2001]

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Theorem 3. Let O be the circumcenter of the triangle ABC and let A A , B B , and CC be the cevians through O. Then O has the Gergonne property AB = AC if and only if either AB = AC or cos2 A = sin B sin C The family of triangles given by (4) can now be added to the so-called balanced families encountered in [1]. It follows from (4) and (7) that cos(B − C) = 2 cos2 A − cos A

(8)

But 2 cos2 A − cos A ∈ [−1, 1] if and only if cos A ∈ [−1/2, 1], i.e., if and only if A ∈ [0, 120◦ ]. Conversely, every A ∈ [0, 120◦ ] gives the unique value for | B − C | defined by (8). Thus the family of triangles that (4) defines is precisely those triangles ABC with A ∈ [0, 120◦ ] and with B, C defined by the equations B + C = π A,

| B − C |= cos−1 (2 cos2 A − cos A).

The reader is invited to try A = 108◦ and obtain the elegant (66, 6)-triangle (A, B, C) = (108◦ , 66◦ , 6◦ ). Finally, we record the results concerning the equality of (external or internal) anglebisectors in a triangle. Theorem 4. (The Complete Steiner-Lehmus Theorem) Let U V W be a given triangle. (a) The internal angle bisectors of V and W are equal if and only if U V = U W . (b) The external angle bisectors of V and W are equal if and only if either U V = U W or sin2

U V W = sin sin . 2 2 2

(c) The external angle bisector of V and the internal angle bisector of W are equal if and only if cos2

U V W = sin cos . 2 2 2

Note 1. One way to show that (3) follows from (2) is to multiply (2) by u + v + w and go through the following sequence of reductions. Here, p stands for (u + v + w)/2. u 4 − u 2 (v + w)2 + 3u 2 vw + 3uvw(v + w) − vw(v + w) − vw(v + w)2 = 0 u 4 − u 2 (v 2 + w2 − vw) + 2uvw(v + w) − vw(v + w)2 = 0 u 2 (u 2 − v 2 − w2 + 2vw) − u 2 vw + 2uvw(v + w) − vw(v + w)2 = 0 u 2 (u 2 − (v − w)2 ) − vw(u 2 − 2uv − 2uw + v 2 + w2 + 2vw) = 0 u 2 (u + v − w)(u − v + w) − vw(−u + v + w)2 = 0 u 2 (2 p − 2w)(2 p − 2v) − vw(2 p − 2u)2 = 0 u 2 (2( p − w)( p − v))2 − vw(2( p − u)( p − v))(2( p − u)( p − w)) = 0 766


Now use the Law of Cosines in the form 2( p − w)( p − v) = wv(1 − cos U )

(9)

(see Note 2) to obtain the relation (1 − cos U )2 = (1 − cos W )(1 − cos V ).

(10)

The half-angle formula now gives the desired result (3). In a similar fashion, one deduces (6) from (5). It helps to use the Law of Cosines in the form 2 p( p − u) = wv(1 + cos U ).

(11)

Note 2. The forms (9) and (11) of the Law of Cosines follow immediately by adding ±2vw to both sides of u 2 − v 2 − w2 = −2vw cos U and then factoring the resulting left-hand side u 2 − (v ± w)2 as a difference of two squares. It is interesting to note that multiplying the resulting identities 2( p − v)( p − w) = vw(1 − cos U ),

2 p( p − u) = vw(1 + cos U )

and using the fact that the area of U V W is given by | U V W |= 12 vw sin U yields Heron’s Formula p( p − u)( p − v)( p − w) =| U V W |2 . REFERENCES 1. Mowaffaq Hajja, Triangle Centers: Some Questions in Euclidean Geometry, Internat. J. Math. Ed. Sci. Tech. 32 (2001) 21–37. 2. David L. MacKay, The Pseudo-Isosceles Triangle, School Science and Mathematics 40 (1940) 464–468. Mathematics, University of Sharjah, P. O. Box 27272, Sharjah, United Arab Emirates [email protected]

A Simpler ∞Proof of2 2 sin πz = πz k=1(1 − z /k ) Wladimir de Azevedo Pribitkin The identity in the title is fundamental and admits numerous simple proofs. But can we derive sine’s product without knowing cotangent’s sum? The answer yes is given in [1, pp. 204–205]. Here we provide a more basic proof, which uses only calculus and a modest dose of introductory complex analysis. 2 2 The function s(z) = π z ∞ k=1 (1 − z /k ) is entire and odd, and has simple zeros at all integers. By recoupling these roots and then rewriting (1 + z/k)[1 − z/(k + 1)], October 2001]

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