Non Euclidean versions of some classical triangle inequalities

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1 Non Euclidean versions of some classical triangle inequalities Dragutin Svrtan, Department of Mathematics, University of Zagreb, Bijeniča cesta 0, Zagreb, Croatia Daro Veljan, Department of Mathematics, University of Zagreb, Bijeniča cesta 0, Zagreb, Croatia September 0, 007 Abstract In this paper we recall and provide short proofs of some classical triangle inequalities and prove corresponding non Euclidean, i.e. spherical and hyperbolic versions of these inequalities. Among them are well nown Euler s inequality, Rouché s inequality (also called fundamental triangle inequality ), Finsler Hadwiger s inequality, isoperimetric inequality and others. 1 Introduction As it is well nown, the Euclid s Fifth Postulate (through any point in a plane outside of a given line there is only one line parallel to that line) has many equivalent formulations. Recall some of them: sum of the angles of a triangle is π (or 180 ), there are similar (non congruent) triangles, there is the area function (with usual properties), every triangle has unique circumcircle, Pythagoras theorem and it s equivalent theorems such as the law of cosines, the law of sines, heron s formula and many more. The negations of the Fifth Postulate lead to spherical and hyperbolical geometries. So, negations of some equalities characteristic for the Euclidean geometry lead to inequalities specific for either spherical or hyperbolic geometry. For example, for a triangle in the Euclidean plane we have the law of cosines c = a + b ab cos C, where we stic with standard notations (that is a, b and c are the side lengths and A, B and C are the angles opposite, respectively to the sides a, b and c). 1

2 It can be proved that in the spherical geometry one has the corresponding inequality c < a + b ab cos C, and in the hyperbolic geometry the opposite inequality c > a + b ab cos C. In fact, we have a + b ab cos C < c < a + b + ab cos(a + B). See [1] for details. On the other hand, there are plenty of interesting inequalities in the (ordinary or Euclidean) triangle geometry relating various triangle elements. In this paper we shall prove some of their counterparts in non Euclidean cases. Let us fix (mostly standard) notations. For a given triangle ABC, let a, b, c denote the side lengths (a opposite to the vertex A, etc.), A, B, C corresponding angles, s = a+b+c the perimeter of the triangle, S its area R the circumradius, r the in radius, r a, r b, r c the excircle radii. We use the symbols of cyclic sums and products such as: f(a) = f(a) + f(b) + f(c) f(a) = f(a) + f(b) + f(c) f(a, b) = f(a, b) + f(b, c) + f(c, a) f(a) = f(a)f(b)f(c) f(x) = f(x)f(y)f(z) Euler s inequality In Euler proved that the triangle s circumradius R is at least twice as big as its inradius r, i.e. R r, with equality iff the triangle is equilateral.here is a short proof. R r abc 4S S s sabc 8S = 8s (s a) (s b) (s c) (s x) 8 x }{{}}{{}}{{} =x =y =z s xy x 8 x x xy 9 x x y 6 x A G x y 6( x y) 1 6 = 6 x. (Yet another way to prove the last inequality: x y+yz = y(x +z ) xyz, and add such three similar inequalities.) The equality case is clear. The inequality 8S sabc (equivalent to Euler s) can also be easily obtained as a consequence (via A G) of the isoperimetric triangle inequality : S 4 (abc),

3 which we shall prove in 4. The Euler inequality has been improved and generalized (e.g. for simplices) many times. A recent and so far the best improvement of Euler s inequality is given by (see [15]) (and it improves [16]): R r abc + a + b + c abc ( a b + b c + c a ) a b + b c + c a 1 Now we turn to the non Euclidean versions of Euler s inequality. Let be the (constant) of the hyperbolic plane in which a hyperbolic triangle ABC sits. let ε = π (A + B + C) be the excess (or defect ). The area of the hyperbolic triangle is given by S = ε. Theorem.1 (Hyperbolic Euler s inequality) Suppose a hyperbolic triangle has a circumcircle and let R be its radius. Let r be the radius of the triangle s incircle. Then tanh R tanh r (.1) The equality is achieved for an equilateral triangle for any fixed excess. Proof. Recall that the radius R of the circumcircle of a hyperbolic triangle (if it exists) is given by (e.g. see [6], [7], [5], [8]): tanh R = sin ε a sinh a = (.) cosh a sinh s s a sinh Also the radius if the incircle (radius of the inscribed circle) r of the hyperbolic triangle is given by (same references): tanh r sinh s a = sinh s (.) We can tae = 1 in the above formulas. Then it is easy to see that (.1) is equivalent to sinh(s a) sinh a, or, by putting (as in the Euclidean case) x = s a, y = s b, z = s c, to s x sinh x sinh. (.4) By writing x instead of x etc., (.4) becomes sinh x sinh(s x) = sinh(y + z).

4 Now by the double formula and addition formula for sinh, after multiplication we get hence 8 sinh x cosh x sinh x sinh y cosh y cosh z+ + sinh x cosh x 6 sinh x cosh x sinh x sinh y cosh y cosh z (.5) However, (.5) is simply A G inequality for the six (nonnegative) numbers sinh x, cosh x,..., cosh z. The equality case follows easily. This proves the hyperbolic Euler s inequality. Note also that (.5) can be proved alternatively in the following way, using three times the simplest A G inequality sinh x sinh y cosh y cosh z + cosh x sinh y cosh y sinh z = = sinh y cosh y[(sinh x cosh z) + (cosh x sinh z) ] sinh y cosh y sinh x cosh z cosh x sinh z. In the spherical case the analogous formula to (.) and (.) and similar reasoning to the previous proof boils down to proving analogous inequality to (.4): sin x sin s x But (.6) follows in the same manner as above. So, we have the following. (.6) Theorem. (Spherical Euler s inequality) The circumradius R and the inradius r of a spherical triangle on a sphere of radius ρ are related by tan R ρ tan r ρ (.7) The equality is achieved for an equilateral triangle for any fixed spherical excess ε = (A + B + C) π. Remar. At present, we don t now how to improve these non Euclidean Euler s inequalities in the sense of the previous discussions in the Euclidean case. It would also be of interest to have the non Euclidean analogues of the Euler inequality R r for a tetrahedron (and similarly for any dimension n). Finsler Hadwiger s inequality In paper [] from 198. Finsler and Hadwiger proved the following sharp upper bound for the area S in terms of side lengths a, b, c of a triangle (improving Weitzenbrec s inequality): a (b c) + 4 S (.8) 4

5 Here are two short proofs of (.8). First proof (citejms): Start with the law of cosines a = b + c bc cos A, or equivalently a = (b c) + bc(1 cos A). From the area formula S = bc sin A, it then follows a = (b c) + 4S tan A. By adding all three such equalities we obtain a = (b c) + 4S tan A. By applying the Jensen inequality to the sum tan A (that is, using convexity of tan x, 0 < x < π) and the equality A + B + C = π, (.8) follows at once. Second proof ([7]): Put x = s a, y = s b, z = s c. Then [a (b c) ] = 4 xy. On the other hand, Heron s formula can be written as 4 S = 4 x x. Then (.8) is equivalent to x x xy, and this is equivalent to x yz (xy), which in turn is equivalent to [x(y z)] 0, and this is obvious. Remar.1 The seemingly weaer Weitzenbrec s inequality a 4 S is, in fact, equivalent to (.8) (see [16]). Also, there are various ways to write Finsler Hadwiger s inequality. For example, since [a (b c) ] = 4r(r + 4R), it follows that (.8) is equivalent to r(r + 4R) S, or, since S = rs, it is equivalent to s r + 4R. 5

6 There are many generalizations, improvements and sharpness of (.8) (see [4]). Let us mention here only two recent. One is (see [1]): (b + c) 1 b + c 10 r s [s + (r + 4R)], and the other one is (see [1]) a 4 S + (a b) + [ a(b + c a) b(c + a b)]. The opposite inequality is ([16]): a 4 S + (b c). Note that all these inequalities are sharp in the sense that equality hold iff the triangle is equilateral (regular). For the hyperbolic case, we need first an analogue of the area formula S = bc sin A. It is not common in the literature, so for reader s convenience we provide its short proof (see e.g. [5]). Lemma. (Cagnolli s first formula) The area S = ε of a hyperbolic triangle ABC is given by sin S = sinh a sinh b sin C cosh c (.9) Proof. From the well nown second (or polar ) law of cosines in elementary hyperbolic geometry we get cosh a cosh a = = cos A + cos B cos C sin B sin C sin ( ( ) B + ) ε sin C + ε, sinh a sin B sin C = sin ( ε ) ( ) sin A + ε sin B sin C By multiplying two expressions sinh a sinh b, and using (.10) we get sinh a sinh b = sin ε sin C cosh c. This implies (.9).. (.10) Theorem. (Hyperbolic Finsler Hadwiger s inequality) For a hyperbolic triangle ABC we have: a cosh cosh b c + 1 sin S cosh a cosh b cosh c tan π ε 6 (.11) The equality in (.11) holds iff for any fixed excess ε, the triangle is equilateral. 6

7 Proof. The idea is to try to mimic (as much as possible) the first proof of (.8). Start with the hyperbolic law of cosines cosh a = cosh b cosh c sinh b sinh c cos A. By adding and subtracting sinh b sinh c, we obtain cosh a = cosh b c = cosh b c = cosh b c + sinh b sinh c sinh b sinh c cos A = + sinh b sinh c sin A = + 4 sinh b sinh c cosh b cosh c sin A By Cagnolli s formula (.9), substitute here the part sinh b sinh c cosh a = cosh b c + 4 cosh a cosh b cosh c sin S to obtain tan A (.1) Apply to both sides of (.1) the cyclic sum operator, and (again) apply Jensen s inequality (i.e. convexity of tan x ): 1 ( A 1 tan tan ) A = tan π ε 6 This implies (.11). The equality claim is also clear from the above argument. The corresponding spherical inequality can be obtained by mutatis mutandis from the hyperbolic case. The area S of a spherical triangle ABC on a sphere of radius ρ is given by S = ρ ε, where ε = A+B +C π is the triangle s excess. The spherical Cagnolli s formula (lie.9) reads as follows: sin S ρ = sin a ρ sin b ρ sin C cos c. (.1) ρ So, starting with the spherical law of cosines, using (.1) and Jensen s inequality, one can show the following. Theorem.4 (Spherical Finsler Hadwiger s inequality) For a spherical triangle ABC on a sphere of radius ρ we have a cos ρ cos b c + 1 sin S ρ ρ cos a ρ cos b ρ cos c ρ tan ε π. (.14) 6 The equality in (.14) holds iff for any fixed ε, the triangle is equilateral. Remar.5 Note that both hyperbolic and spherical inequalities (.11) and (.14) reduce to Finsler Hadwiger s inequality (.8) when in (.5), or ρ in (.14). This is immediate from the power sum expansions of trigonometric or hyperbolic functions. 7

8 4 Isoperimetric triangle inequalities In the Euclidean case, if we multiply all the area formulas, one of which is S = 1 bc sin A, we obtain a symmetric formula for the triangle area S = 1 8 (abc) sin A sin B sin B. (4.15) By using A G inequality and concavity of the function sin x on [0, π] (or, Jensen s inequality again), we have: ( ) sin A + sin B + sin C sin A sin B sin C ( sin A + B + C ) = sin π = 8 This and (4.15) imply the so called isoperimetric inequality for a triangle: S 64 (abc), or in a more appropriate form S 4 (abc). (4.16) Inequality (4.16) and A G imply that S 6 (a + b + c), and this is why we call it isoperimetric inequality. By Heron s formula we have (4S) = sd (a, b, c), where s = a + b + c and d (a, b, c) := (a + b c)(b + c a)(c + a b). By [10], Cor. 6., we have a sharp inequality d (a, b, c) (abc) a + b + c + abc From Heron s formula and (4.17) it easily follows S 1 abc a + b + c a + b + c + abc (4.17) (4.18) namely, we claim 1 abc a + b + c a + b + c + abc (abc) (4.19) 4 But (4.19) is equivalent to ( a + b + c ) ( ) + abc a + b + c (abc). (4.0) 4 8

9 To prove (4.0) we can tae abc = 1 and prove a + b + c ( ) + 1 a + b + c. (4.1) 4 Instead, we prove even stronger inequality a + b + c a + b + c. (4.) Inequality (4.) is stronger than (4.1) because the means are increasing, i.e. M p (a, b, c) M q (a, b, c), for a, b, c > 0 and 0 p q, where M p (a, b, c) = [ ] (a p +b p +c p 1 ). To prove (4.), denote x = a + b + c and consider the function ( ) x + 1 f(x) = x 4. Since (by A G) x abc = 1, i.e. x, we consider f(x) only for x. Since f() = 0 and derivative f (x) 0 for x, we conclude f(x) 0 for x and hence prove (4.19). Putting all together, we finally have a chain of inequalities for the triangle area S symmetrically expressed in terms of the side lengths a, b, c. Theorem 4.1 (Improved Euclidean isoperimetric triangle inequalities) S 1 abc a + b + c a + b + c + abc 1 6 (a + b + c) (abc) 4 4 a + b + c + abc 4 (abc) (4.) We shall now mae an analogue of the isoperimetric inequality (4.16) in the hyperbolic case. Start with Cagnolli s formula (.9) and multiply all such three formulas to get (since S = ε ): sin ε = sinh a As in the Euclidean case we have sin A ( sin A + sin B + sin C a tanh sin A. (4.4) ) ( sin A + B + C ) ( = sin π ε ) So, this inequality together with (4.4) implies the following. Theorem 4. The area S = ε of a hyperbolic triangle with side lengths a, b, c satisfies the following inequality ( sin ε ) sin π ε sinh a tanh a. (4.5) For a regular triangle (a = b = c, A = B = C) and any fixed excess ε, the inequality (4.5) becomes an equality (by Cagnolli s formula (.9)). 9

10 The corresponding isoperimetric inequality can be obtained for a spherical triangle: ( sin ε ) sin π+ε sin a ρ tan a ρ. (4.6) Remar 4. In the dimensional case we have a well nown upper bound of the volume V of a (Euclidean) tetrahedron in terms of product of lengths of its edges (lie (4.16)) : V abcdef 1 with equality iff the tetrahedron is regular (and similarly in any dimension), see [11]. Non Euclidean tetrahedra (and simplices) lac good volume formulas of Heron s type, except the Cayley Menger determinant formulas in all three geometries. Kahan s formula 1 for volume of a Euclidean tetrahedron is nown only for the Euclidean case. There are some volume formulas for tetrahedra in all three geometries now available on Internet, but they are rather involved. We don t now at present how to use them to obtain a good and simple enough upper bound. Un dimension two, Heron s formula in all three geometries can very easily be deduced. A very short proof of Heron s formula is as follows. Start with the triangle area (in fact, law of sines) and the law of cosines 4S = ab sin C, and a + b c = ab cos C. Now square them both and then add them. The result is (a form of) the Heron s formula (4S) + (a + b c ) = (ab). In a similar way one can get triangle area formulas in the non Euclidean case by starting with Cagnolli s formula ((.9) or (.1)) and the appropriate law of cosines. Remar 4.4 In order to improve the non Euclidean dimensional isoperimetric inequality analogous to (4.) we would need an analogue of the function d (a, b, c) and a corresponding inequality lie (4.17). This inequality was proved in [10] as a consequence of the inequality d (a, b, c ) d (a, b, c), and this follows from an identity expressing the difference d (a, b, c) d (a, b, c ) as the sum of four squares. But at present we don t now the right hyperbolic analogue d H (a, b, c) or spherical analogue d S (a, b, c) of d (a, b, c). 5 Rouche s inequality and Blundon s inequality The following inequality is a necessary and sufficient condition for the existence of an (Euclidean) triangle with elements R, r and s (see [4]): R +10Rr r (R r) R Rr s R +10Rr r +(R r) R Rr. 1 see

11 (5.7) This inequality (sometimes called the fundamental triangle inequality ) was first proved by É. Rouché in answering a question of Ramus. It was recently improved in [14]. A short proof of (5.7) is as follows. Let r a, r b, r c be excircle radii of the triangle ABC. Then it is well nown (and easy to chec) that r a = 4R + r, ra r b = s and r a r b r c = rs. Hence r a, r b, r c are the roots of the cubic x (4R + r)x + s x rs = 0. (5.8) Now consider the discriminant of this cubic, that is D = (r a r b ). In terms of elementary symmetric functions e 1, e, e (in variables r a, r b, r c ) the discriminant is given by D = e 1e 4e 4e 1e + 18e 1 e e 7e. (5.9) Since e 1 = r a = 4R + r, e = r a r b = s, e = r a = rs, we have D = s [(4R + r) s 4s 4 4(4R + r) r + 18(4R + r)rs 7r s ]. From D 0, (5.7 follows easily. In fact, the inequality D 0 reduces to the quadratic inequality in s : s 4 (R + 10Rr r )s + (4R + r) r 0 (5.0) The fundamental inequality (5.7) implies a sharp linear upper bound of s in terms of r and R, nown as Blundon s inequality []: s ( 4)r + R. (5.1) To prove (5.1), it is enough to prove that R + 10Rr r + (R r) R Rr [( 4)r + R]. A little computation shows that this is equivalent to the following cubic inequality (with x = R/r): f(x) := 4( 5)x 0(60 10)x +1(48 8)x+4(9 1 ) 0. By Euler s inequality x, f() 0 and clearly f(x) 0 for x. Yet another (standard) way to prove Blundon s inequality (5.1) is to use convexity of the biquadratic function on the left hand side of the inequality (5.0). Blundon s inequality is also sharp in the sense that equality holds in (5.1) iff the triangle is equilateral. (Recall by the way that a triangle is a right triangle if and only if s = r + R). 11

12 Let us turn to non Euclidean versions of the fundamental triangle inequality. Suppose a hyperbolic triangle has the circumscribed circle. As before, denote by R, r, and r a, r b, r c, respectively, the radii of the circumscribed, inscribed and excribed circles of the triangle. Then by (.) and (.) we now R and r, while r a (and similarly r b and r c ) is given by tanh r a = sinh s tan A, (5.) and by using tan A = sinh s b sinh A sinh s c. (5.) sinh s a The combination of these two expresses r a in terms of A, b, and c. In order to obtain the analogue of cubic equation (5.8) for the hyperbolic triangle and whose roots are x 1 = tanh ra, x = tanh r b, x = tanh rc, we have to compute the elementary symmetric functions e 1, e, e in the variables x 1, x, x. We compute first (the easiest) e. Equations 5., 5. and. yield e = tanh r a = sinh s tanh r. (5.4) Next, by (5.) and (5.): e = tanh r a tanh r b = s A sinh tan tan B = sinh s s a sinh Using the identity sinh(x+y +z) (sinh x+sinh y +sinh z) = 4 sinh y + z applied to x = s a, y = s b, z = s c, it follows that sinh s sinh s a = 4 sinh a And now from (.) and (.) we get ( e = sinh s 1 tanh r tanh R ) Finally, to compute e 1, we use the identity tan(x + y + z) = By (5.), e 1 = sinh s tan A tan A = tan A+B+C tan A+B+C = tan π ε = cot ε. sinh z + x tan x + tan y + tan z tan x tan y tan z 1 tan x tan y tan y tan z tan z tan x. Now from (5.7): ( 1 tan A tan B 1 ) + tan A, sinh x + y, (5.5) (5.6) (5.7)

13 From (.), we have A tan = tanh r sinh s. By (5.), (5.5) and (.) and (.) it follows easily 1 tan A tan B = tanh r tanh R sinh s. Finally, putting all together yields e 1 = tanh r ( 1 + tanh R sinh s cot ε ). (5.8) Equations (5.4), (5.6) and (5.8) yield via x e 1 x + e x e = 0 the cubic equation ( x tanh r 1 + tanh R sinh s cot ) ε x + + sinh ( s 1 tanh r tanh ) R x sinh s tanh r = 0 (5.9) This cubic (with roots tanh ra etc.) by letting l reduces to the cubic (5.8). This follow from the identity sinh s tanh r sin ε = cosh a If, then the right hand side tends to and therefore the coefficient by x in (5.9) goes to r + 4R which appears in (5.8). And similarly the other coefficients. Consider the discriminant of (5.9) D = ( tanh r a tanh r ) b. Now, by applying (5.9) and (5.4), (5.6) and (5.8) we obtain the four degree polynomial (in fact six degree polynomial) in sinh s for an expression D. By the following legend r tanh r R tanh R ε cot ε (5.40) s sinh s we can write D as follows (after some computation); note that it has almost double number of terms than the corresponding Euclidean discriminant D = s [(r R ε + 4r 4 R 4 ε 4r R ε 1 + 6rR 1r R + 8r R )s 4 +r Rε(1 4rR + 4r R ε 8r R ε + 9ε + 18rRε)s +r (r R 10rR 1r R ε )s 6r 4 Rεs r 4 ]. 1

14 (5.41) By definition D 0, so degree four polynomial in s (in fact in sinh s ), i.e. polynomial in bracets in (5.41) is 0. So the hyperbolic analogue of the fundamental triangle inequality (5.7), or rather degree four polynomial inequality (5.0) is the quadratic (in s) polynomial inequality D s 0. Theorem 5.1 (Hyperbolic fundamental triangle inequality ) For a hyperbolic triangle that has the circumference of radius R, the incircle of radius r semiperimeter s and excess ε, we have D s 0 (5.4) where D is given by (5.41) together with the legend (5.40). When, (5.4) reduces to (5.0). Blundon s hyperbolic inequality can also be derived as a Corollary of theorem 5.1. The spherical version of the fundamental inequality as well as the corresponding spherical Blundon s inequality can also be obtained, but we shall omit them here. In conclusion, we may say that all these triangle inequalities give more information and better insight to the geometry of and manifolds. 14

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