Some Monotonicity Results Related to the Fermat Point of a Triangle

Transcription

1 Forum Geometricorum Volume FORUM GEOM ISSN Some Monotonicity Results Related to the Fermat Point of a Triangle Toufik Mansour and Mark Shattuck Astract. In this paper, we estalish a Fermat point analogue of the Steiner- Lehmus theorem y proving a more general monotonicity property. We also consider four pairs of segment lengths determined y intersecting Fermat point cevians to unequal sides of a triangle and determine comparale properties for these segments. Our proofs make use of trigonometric inequalities.. Introduction Starting with an isosceles triangle, if one were to draw the same type of segment to the congruent sides from the opposite angles, then the segments so otained are always congruent, y symmetry. Conversely, one might wonder when congruence of some particular pair of internal segments within a triangle implies congruence of the two corresponding sides. For example, it is easy to show that congruence of medians or altitudes to two sides implies congruence of the sides. On the other hand, congruence of two angle isectors implying congruence of the sides to which they are drawn is more difficult to show and is the content of the well-known Steiner-Lehmus theorem. The reader is referred to [,, 4, 8] for various proofs of this theorem and to [5] for stronger versions of it. Results such as these are frequently particular cases of more general monotonic ehavior. For example, a median, altitude or angle isector to a longer side is always shorter, and conversely. On the other hand, congruence of some particular pair of segments that are congruent within an isosceles triangle due to symmetry need not imply that the triangle within which they lie is isosceles. See, e.g., [7] for such an example. Here, we consider a variant of the Steiner-Lehmus result involving the Fermat point. The external Fermat point of a triangle ABC is otained as follows. On sides AB, AC and BC, construct externally equilateral triangles ABH, ACG and BCI. Then the segments CH, BG and AI are concurrent at a point F, known as the Fermat point, see [, p. 8]. See Figures - elow. It is well-known [9] that the Fermat point provides a minimum when computing the sum of the distances from an aritrary point to the vertices of triangle ABC. We will refer to the portion of either segment CH, BG or AI lying etween the respective vertex of Pulication Date: Novemer 8, 06. Communicating Editor: Paul Yiu.

2 56 T. Mansour and M. Shattuck triangle ABC and the point where the segment intersects the opposite side possily extended of triangle ABC as a Fermat point cevian. For instance, segments BD and CE in Figures - are Fermat cevians. Note that in the second figure, the side AC is extended, while oth sides AB and AC are extended in the third. H A G E D F B C Figure. Position of F when all angles of triangle ABC are less than 0. H A E C B D F G Figure. Position of F relative to BC when C >0. H G F D A E B C Figure. Position of F relative to BC when A >0.

3 Some monotonicity results related to the Fermat point of a triangle 57 In the next section, we show that a longer side always has a shorter Fermat cevian terminating on it, and conversely. In particular, if two Fermat cevians are congruent, then the sides to which they are drawn are congruent. In terms of the figures aove, we have if AB AC, then BD CE, with equality holding if and only AB = AC. Note that this stands in contrast to the lengths of the segments BG and CH which are always equal. Our result here extends earlier work concerning a Gergonne point analogue of Steiner-Lehmus done in [0], where the question of looking at analogues involving other kinds of centers is mentioned. See also [, 6] for related results involving extensions of the angle isector and other types of cevians. In the third section, we consider the same question for the two internal segments of a Fermat cevian determined y F as well as for the segments otained y the intersection of a cevian with the opposite side. That is, we consider the segments BF and FDon line BD in Fig s -, and segments AD and CD on side AC possily extended in these figures, and compare each of these four segment lengths with the analogous segments involving CE and AB. In two of the cases, one gets the same type of monotonicity as that demonstrated y the Fermat cevians themselves see Theorems 4 and 9 elow. In the other cases, the monotonicity demonstrated depends upon the measure of the vertex angle. In our proofs, we consider cases ased off of the figures aove. That is, we consider separately the cases when i all angles in triangle ABC are less than 0, ii one of the ase angles, B or C from which cevians are drawn, is greater than 0, or iii the vertex angle, A, exceeds 0. Note that the case when one of the angles of triangle ABC is exactly 0 can e degenerate and thus is sometimes excluded from consideration. Our proofs are trigonometric in nature and involve estalishing certain inequalities see, for example, Lemmas 5 and 8. At times, we will make use of the following result see, e.g., [, p. 65] concerning the Fermat point. Lemma. The segments BG and CH in Figures - are always congruent and meet at a 60 angle. In the proofs that follow, we will always take BC =for convenience, which can e done without loss of generality.. A Fermat analogue of the Steiner-Lehmus result In this section, we prove a version of the Steiner-Lehmus result for the Fermat point. In the theorem that follows and susequent theorems, the segment lengths refer to those defined in Figures - aove. Theorem. If AB > AC in triangle ABC, then BD > CE. Proof. Let x = CBF and y = BCF. First assume all angles of triangle ABC are less than. By the law of sines applied to triangle BCD in Figure aove, we have sinx + C sin x cos C +cosxsin C = = =sinxcot C +cosx. BD sin C sin C

4 58 T. Mansour and M. Shattuck By the law of sines in triangle BCG,wehave = CG = sin C x =sin sin x C cot x cos C and sin x = BG sin C +. Therefore, y and, we get =sinxcot C +cotx =sinx cot C + +cos C BD sin C = sin C + cot C + +cos C BG sin C = sin C cos C +cos C sin C + sin C BG sin C = sin + sin C = +sinc, BG sin C BG sinc where we have made use of the sine-of-sum formula and the fact sin z =sin z. Noting that a comparale formula holds for CE, it follows that BD > CE if and only if +sinc < c +sinb. BG sinc CH sinb Since BG = CH y Lemma, this inequality reduces to sin C < c sin B, i.e., <c, y the law of sines. Since the last inequality is true y assumption, this completes the proof of the theorem in the case when all angles of triangle ABC are less than. If C > or if A >, then one may verify that the same formulas hold for BD and CE and the result follows as efore. If C =, then the result follows from the fact that BEC > BCE in the cyclic quadrilateral ACBH. On the other hand, if A =, then the result is tautological as D and E oth coincide with A in this case. Corollary. If BD = CE, then AB = AC.. Related monotonicity results In this section, we prove further monotonicity results for four additional segments determined y a given pair of intersecting Fermat point cevians. The first two results elow concern how a pair of cevians to sides of unequal length divide one another. Theorem 4. If AB > AC in triangle ABC, then BF > CF. Proof. Let x = CBF and y = BCF. Let l =. First assume C <. By the law of sines in triangles BCG and BCH considering separately the cases when A < or A CGsin BCG, we have sin x = BG = sinc+l BG and

5 Some monotonicity results related to the Fermat point of a triangle 59 BH sin CBH sin y = CH CH. By the law of sines in triangles BCF and ABC and since BG = CH, we then have = c sinb+l BF CF = sin y c sin B + l sin C sin B + l = = sin x sin C + l sin B sin C + l. To complete the proof in this case, note that C > B implies sinc B > 0 so that cosc B+ sinc B > cosb C+ sinb C, i.e., cos C B l > cos B C l. But then cos C B l cos C + B + l > cos B C l cos B + C + l, whence sin C sinb + l > sin B sinc + l, which implies BF > CF in this case. Now suppose C. Since the case when C = is seen to e trivial, assume C >. Then to show BF > CF in this case is to show sin C sin B + l > sin B sin C l, i.e., cosc B l cosc + B + l > cosb C +l cosb + C l. This may e rewritten as cosc B l cosc B l > cosb + C + l, i.e., sin C B sin 6 > cosb+c+l. To show sin C B < cosb + C + l where C >l, simply oserve that sin C B < < cosb+c +l since <B+C +l < 4 implies < cosb+c +l <, which completes the proof. For the next result, we will need the following inequality. Lemma 5. Suppose B < C < in triangle ABC. Then sinb +sinb + CsinC sinc +sinb + CsinB sin C > sin B if B + C >, with the inequality reversed if B + C <. Proof. Let l =. Consider the function hu, v = sinu +sinu + vsinvsinl u

6 60 T. Mansour and M. Shattuck for 0 <u<v<min {l, u}. By the product-to-sum trigonometric identities, we get hu, v = sinusinl u+sinu + vsinl usinv = cosu l cos l + cosu + v l cosv + l sin v = cosu l cos l+ sinl u+sinu +v l sin l+sinv + l = cosu l cos l+ sinu +v l+sinl +sin u v cosu v l, where we have applied the sine sum-to-product identity in the last equality. To show the inequality in question, we compare the quantities hu, v and hv, u. Note that hu, v >hv, u if and only if cosu l sinu+v cosu v l > cosv l sinu+v cosv u l, which may e rewritten as cosu l cosv l > sinu + vcosu v l cosu v + l, i.e., sinv + u lsinv u > sinu + vsinu vsinl, 4 y the cosine difference-to-product formula. Since v>u, inequality 4 is immediate if u + v l for then the left side is non-negative, while the right is negative. If u + v < l, then 4 reduces to sinl u v < sinu+v, which holds if l <u+v <lfor then sinl u v < sin l < sinu+v in that case. The reverse inequality is seen to hold if u+v < l, which completes the proof. We now prove a monotonicity result concerning the segments EF and DF in Figures - aove. Note that when A =, the points D, E and F coincide and so we exclude this case from consideration. Theorem 6. Suppose B < C < A < and EF < DF if A >. in triangle ABC. Then EF > DF if Proof. Let x = CBD and y = BCE. First assume A <. Then DF = CF sinc y BF sinb x sinc+x and EF = sinb+y, y the law of sines in triangles CDF and BEF. Thus, we have EF sinc + xsinb x DF = sin y, 5 sin x sinb + ysinc y

7 Some monotonicity results related to the Fermat point of a triangle 6 y the law of sines in triangle BCF. Let l =. By the expressions for cot x and for cot x +cotc found in the proof of Theorem aove, and the law of sines applied to triangles BCG and ABC, we get sinc + xsinb x =sinccot x +coscsin B cos x cos B sin x sin x =sinccot x +cotcsinxsin B cot x cos B sin l + sin C CGsinC + l =sinc sin C sinl C BG sin B + cosl C cos B sinl C sinl C sinb +sinb + CsinC = sinb sinl C BG sinb + C+sinB + C l, where we have used the difference-of-sine formula in the last equality. Oserve that the quotient sinb+ysinc y sin y is given y a comparale formula involving CH in place of BG. By Lemma, and the law of sines in triangle ABC, it follows from 5 that EF sinb +sinb + CsinC DF = sinl B. sinl C sinc +sinb + CsinB 6 Since A < EF, it follow from Lemma 5 that DF > in this case. If A >, then one is led again to 6, and Lemma 5 implies EF DF <, which completes the proof. We then otain the following restricted Steiner-Lehmus type result. Corollary 7. If EF = DF in triangle ABC where B, C < in particular, if triangle ABC is acute, then AB = AC. Remark. The results of the preceding theorem and corollary may not hold if C > 5. For example, when B + C =, then EF > DF if C = 6, while EF < DF if C = 8 9. Note that EF > DF if and only if hu, v+hv, u > 0 where v = C > and h is as aove. Thus, y continuity, it is seen that when A =, there exists a triangle ABC where C> and thus AB AC such that EF = DF. Before proving our next result, we will need the following trigonometric inequality. Lemma 8. If 0 <x<ywith x + y<, then sin x sinx + y+ sin x sin y sin x < sin y sinx + y+ sin x sin y sin. 7 y

8 6 T. Mansour and M. Shattuck Proof. Let l =. Equivalently, we show sin x sinx + ysinl y+ sin x sin y sinl y < sin y sinx + ysinl x+ sin x sin y sinl x, which can e rewritten as sin x sinx + ysinl y sinl x + sin x sin ysinl y sinl x < sin y sin xsinx + ysinl x. 8 If l y, then sinl y > sinl x since l x>l y. If l y<, then l y> l x since x + y<l =, y assumption, and thus sinl y > sin l x = sinl x. Dividing oth sides of 8 y sinl y sinl x > 0, we then must show sin x sinx+y+ sin x sin y sin xsinx + ysinl x y<sin. 9 sinl y sinl x Using sinl y sinl x =sin x y cos x+y l and the identity sin y sin x =siny xsiny + x, wehave sin y sin xsinx + ysinl x sinl y sinl x Thus, we need to show sin x sinx + y+ sin x sin y<cos = sin y sin xsinx + ysinl x sin x y cos x+y l = siny xsin x + ysinl x sin y x cos x+y+l = cos y x sin x + ysinl x cos. x+y+l y x sin x + ysinl x cos x+y+l. 0 Since the cosine function is decreasing in the first quadrant, to prove 0, it suffices to show x+y sin sin x + ysinl x x sinx + y+ sin x sin y<cos. cos x+y+l Dividing oth the numerator and denominator of the fraction in the last inequality y cos x+y gives sin x sinx + y+ sin x sin x + ysinl x y<sin tan x+y,

9 Some monotonicity results related to the Fermat point of a triangle 6 which can e rewritten as sin x sinx + y+ 4 cosx y cosx + y < sinl x sin x + y tan x+y. Since the right-hand side of depends on x and y only through the sum, our strategy is to fix u = x + y and consider the left-hand side as a function of x which we seek to maximize. Given 0 <u<, define the function hx for 0 <x< u y 4 hx = sin u sin x + cosu x cos u. sinl x Then h x > 0 if and only if sinl x sinusin x cos x + sinu x > cosl xsin u sin x + cosu x cos u. 4 Since sinl x > cosl x > 0 for 0 <x< 6 and since 0 <x< u,to show, it suffices to show sinusin x cos x>sin u sin x + cosu x cos u. 4 Note that 0 <x< 6 implies sin u sin x cos x>sin u sin x and sin u sin x cos x> cosu x cos u = sin x sin y, 4 as sin u =sinx + y > sin y and cos x>. Adding these inequalities then implies the desired inequality and shows that hx is increasing on the interval 0, u. Thus, to prove, it is enough to show that it holds when x = u where u = x + y. That is, we must show sin v sinv+ sin v tanv < 4sin vsinl v, 0 <v< 6, i.e., sinv+ tanv < 6 cos v sinl v. The last inequality holds since sinv+ tanv < < 6 cos v sinl v for 0 <v< 6. This implies, which completes the proof. Our final two results concern the monotonicity of the segments determined when a Fermat cevian intersects the opposite side. Theorem 9. If AB > AC in triangle ABC, then BE > CD.

10 64 T. Mansour and M. Shattuck Proof. First suppose all angles of triangle ABC are less than. In this case, y the law of sines in triangles BCE and BCD in Figure, we have BE = and CD = and only if sin x sinc+x sin y sinb+y, where x = CBD and y = BCE. Then BE > CD if sin Bcot y +cotb < sin Ccot x +cotc. By the formulas found previously for cot x +cotc and cot y +cotb and the law of sines in triangle ABC, this last inequality is equivalent to sin B sinb + C+ sin C < sin C sinb + C+ sin B, sinl BsinC sinl CsinB where l =. By Lemma 5 and since sin C>sin B>0, the preceding inequality holds, as desired. We now consider the remaining cases. If A = or C =, then the result is clear. If A >, then one may verify that BE > CD if and only if holds as efore; in which case, the result now follows from Lemma 8. If C >, then one may verify that BE > CD if and only if sin B sinb + C+ sin C < sin C sinb + C+ sin B. 4 sinl BsinC sinc lsinb To show 4, first note that sinc l < sinl B since C l < l <l B <l. sin B sinb+c sin C sinb+c Hence, sinl BsinC < sinc lsinb since sin B < sin C and sinl B < sinc l. Thus 4 follows from adding these two inequalities, which completes the proof. Theorem 0. Let AB > AC in triangle ABC where A. Then AE < AD if A < or A > and AE > AD if < A <, with AE = AD if A =. Proof. First assume that all angles in triangle ABC are less than. Using the formula for EB found in the proof of Theorem 9 and the fact that sin C = c sinb +

11 Some monotonicity results related to the Fermat point of a triangle 65 C, wehave AE = AB EB = c sin B sinb + C sin B sinb + C+ sin C sinb + C sin B +sin B + = c sin C sin B sinb + C+ sin C sinb + Csin B cos + =c sin C cos C cosc +B+ sinc = c cos C + cos C +B + sinc sin C + 6 cosc +B = c cos C cos C +B sin c sin B + C C + sin B 6 = cosc +B sin C + 6 cosc +B, where we have made use of several trigonometric identities. Noting the comparale expression for AD, it follows that AE < AD if and only if c sin B + C sin B sin C + 6 cosc +B < sin C + B sin C sin B + 6 cosb +C. Since sin B + C is positive as A < and since c sin B = sin C, this last inequality is equivalent to C cosc +B cosb +C < sin + sin B +, 6 6 i.e., C B B + C C B B + C sin sin < sin cos +. 6 Thus, we have AE < AD if B + C sin < sin B + C, 5 since AB > AC implies that the sin C B factor is positive. Note that AE > AD if 5 is reversed. We now consider cases on the sum B + C. First suppose B + C > and let u = B+C where 0 <u< 6. Then 5 holds if and only if sinu< sinu + =sinu =sinu 4sin u, i.e., 4sin u<, which is true since 0 <u< 6. Thus AE < AD if AB > AC in this case. If < B + C <, then let u = B+C where 0 <u< 6. In this case, the reverse of inequality 5 holds since sin u >sinu, whence AE > AD if AB > AC. If B + C =, then there is equality in 5. Comining the preceding cases thus implies the theorem whenever all angles in ABC are less than.

12 66 T. Mansour and M. Shattuck If A >, then AE = EB AB and one gets c sin AE = B C sin B sin C + 6 cosc +B in this case and a comparale expression for AD. Since B + C <, the factor of sin B C is positive and thus cancels out when comparing AE and AD, which leads to inequality 5 as efore. Let u = B+C,so 6 <u<. Then 5 is seen to hold in this case since sin u < sinu, whence AE < AD if AB > AC. Finally, if C, then one gets AD = AC + CD = + sin C sinb + C sin C sinb + C+ sin B, which is seen to yield the same formula as efore. Note that the expression for AE also does not change in this case and we are again led to inequality 5. Since 5 holds if B + C >,wehaveae < AD if AB > AC when C, which completes the proof. Corollary. If AE = AD in triangle ABC where A,, then AB = AC. References [] S. Au-Saymeh, M. Hajja, and H. A. ShahAli, Another variation on the Steiner-Lehmus theme, Forum Geom., [] H. S. M. Coxeter, Introduction to Geometry, John Wiley & Sons, Inc., 96. [] H. S. M. Coxeter and S. L. Greitzer, Geometry Revisited, Random House, Inc., 967. [4] M. Hajja, A short trigonometric proof of the Steiner-Lehmus theorem, Forum Geom., [5] M. Hajja, Stronger forms of the Steiner-Lehmus theorem, Forum Geom., [6] M. Hajja, Cyril F. Parry s variations on the Steiner-Lehmus theme, pre-print. [7] T. Mansour and M. Shattuck, Monotonicity results concerning certain lengths within a triangle, J. Adv. Math. Appl., [8] R. Oláh-Gál and J. Sándor, On trigonometric proofs of the Steiner-Lehmus theorem, Forum Geom., [9] L. Sanger and J. Wales, Wikipedia, under point. [0] K. R. S. Sastry, A Gergonne analogue of the Steiner-Lehmus theorem, Forum Geom., Toufik Mansour: Department of Mathematics, University of Haifa, Haifa, Israel address: tmansour@univ.haifa.ac.il Mark Shattuck: Department of Mathematics, University of Tennessee, Knoxville, Tennessee 7996, USA address: shattuck@math.utk.edu