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A Strong Triangle Inequality in Hyperbolic Geometry

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A Strong Triangle Inequality in Hyperbolic Geometry Forum Geometricorum Volume 16 (2016) 99–114. FORUM GEOM ISSN 1534-1178 A Strong Triangle Inequality in Hyperbolic Geometry Csaba Biro´ and Robert C. Powers Abstract. For a triangle in the hyperbolic plane, let α, β, γ denote the angles opposite the sides a, b, c, respectively. Also, let h be the height of the altitude to side c. Under the assumption that α, β, γ can be chosen uniformly in the interval (0,π) and it is given that α + β + γ<π, we show that the strong triangle inequality a + b>c+ h holds approximately 79% of the time. To accomplish this, we prove a number of theoretical results to make sure that the probability can be computed to an arbitrary precision, and the error can be bounded. 1. Introduction It is well known that the Euclidean and hyperbolic planes satisfy the triangle inequality. What is less known is that in many cases a stronger triangle inequality holds. Specifically, a + b>c+ h (1) where a, b, c are the lengths of the three sides of the triangle and h is the height of the altitude to side c. We refer to inequality (1) as the strong triangle inequality and note that this inequality depends on which side of the triangle is labeled c. The strong triangle inequality was first introduced for the Euclidean plane by Bailey and Bannister in [1]. They proved, see also Klamkin [4], that inequality (1) 24 holds for all Euclidean triangles if γ<arctan 7 where γ is the angle opposite side c. Bailey and Bannister also showed that a + b = c + h for any Euclidean γ =arctan 24 γ isosceles triangle such that 7 and is the unique largest angle of the 24 triangle. We let B =arctan 7 and refer to B as the Bailey-Bannister bound. In 2007, Baker and Powers [2] showed that the strong triangle inequality holds for any hyperbolic triangle if γ ≤ Γ where Γ is the unique root of the function γ f(γ)=−1 − cos γ +sinγ +sin sin γ 2 π ◦ ◦ in the interval [0, 2 ]. It turns out that B ≈ 74 and Γ ≈ 66 leading to roughly an 8◦ difference between the Euclidean and hyperbolic bounds. It appears that the strong triangle inequality holds more often in the Euclidean plane than in the hyperbolic plane. Let α and β denote the angles opposite the sides a and b, respectively. Under the assumption that the angles α and β can be chosen uniformly in the interval (0,π) and α + β<π,Fa˘ıziev et al. [3] showed the strong triangle inequality holds in the Publication Date: April 1, 2016. Communicating Editor: Paul Yiu. 100 C. Biro´ and R. C. Powers γ a b h c α β Figure 1. A hyperbolic triangle Euclidean plane approximately 69% of the time. In addition, they asked how this percentage will change when working with triangles in the hyperbolic plane. In this paper, we answer this question by showing that the strong triangle inequality holds approximately 79% of the time. Moreover, we show that the stated probability can be computed to an arbitrary precision and that the error can be bounded. Unless otherwise noted, all geometric notions in this paper are on the hyper- bolic plane. Since our problem is invariant under scaling, we will assume that the Gaussian curvature of the plane is −1. We will use the notations a, b, c, h, α, β, γ for sides, height, and angles of a given triangle. (See Figure 1.) We will exten- sively use hyperbolic trigonometric formulas such as the law of sines and the two versions of the law of cosines. We refer the reader to Chapter 8 in [5] for a list of these various formulas. 2. Simple observations In this section we mention a few simple, but important observations about the main question. Proposition 1. If γ is not the unique greatest angle in a triangle, then the strong triangle inequality holds. Proof. Suppose that γ is not the greatest angle, say, α ≥ γ. Then a ≥ c,so a + b ≥ c + b ≥ c + h. Equality could only hold, if α = γ = π/2, which is impossible. Proposition 2. If γ ≥ π/2, then the strong triangle inequality does not hold. We will start with a lemma that is interesting in its own right. Lemma 3. In every triangle the following equation holds. sinh c sinh h =sinha sinh b sin γ Note that in Euclidean geometry the analogous theorem would be the statement that ch = ab sin γ, which is true by the fact that both sides of the equation represent twice the area of the triangle. Interestingly, in hyperbolic geometry, the sides of the corresponding equation do not represent the area of the triangle. A strong triangle inequality in hyperbolic geometry 101 Proof. By the law of sines, sinh b sinh c = , sin β sin γ so sinh b sin γ sinh c = . sin β By right triangle trigonometry, sinh h =sinha sin β. Multiplying these equations, the result follows. Proof of Proposition 2. Note that a + b>c+ h if and only if cosh(a + b) > cosh(c + h). Using the addition formula for cosh, then the fact that cosh h ≥ 1 and cosh c ≥ 0, and then the law of cosines, in this order, we get cos(c + h)= coshc cosh h +sinhc sinh h ≥ cosh c +sinhc sinh h = cosh a cosh b − sinh a sinh b cos γ +sinha sinh b sin γ = cosh a cosh b +sinha sinh b(sin γ − cos γ) Notice that sin γ − cos γ ≥ 1 if π/2 ≤ γ ≤ π.So cosh a cosh b +sinha sinh b(sin γ − cos γ) ≥ cosh a cosh b +sinha sinh b =cosh(a + b). 3. Converting angles to lengths Since the angles of a hyperbolic triangle uniquely determine the triangle, it is possible to rephrase the condition a + b>c+ h with α, β, γ. In what follows, our goal is find a function f(α, β, γ), as simple as possible, such that a + b>c+ h if and only if f(α, β, γ) > 0. Following Proposition 1 and Proposition 2, in the rest of the section we will assume that γ<π/2 is the greatest angle of the triangle. The following lemma is implicit in [2]. We include the proof for completeness. Lemma 4. A triangle satisfies the strong triangle inequality if and only if cos α cos β +cosγ − 1 < cosh h. cos γ +1− sin γ Furthermore, the formula holds with equality if and only if a + b = c + h. Proof. Recall that a + b>c+ h if and only if cosh(a + b) − cosh(c + h) > 0. Using the cosh addition formula and the law of cosines on cosh c,wehave cosh(a + b) − cosh(c + h) = cosh a cosh b +sinha sinh b − cosh c cosh h − sinh c sinh h = cosh c +sinha sinh b cos γ +sinha sinh b − cosh c cosh h − sinh c sinh h = cosh c(1 − cosh h)+sinha sinh b(cos γ +1)− sinh c sinh h 102 C. Biro´ and R. C. Powers By Lemma 3, cosh c(1 − cosh h)+sinha sinh b(cos γ +1)− sinh c sinh h = cosh c(1 − cosh h)+sinha sinh b(cos γ +1)− sinh a sinh b sin γ sinh a sinh b 1 − cosh2 h = cosh c +(1+coshh)(cos γ +1− sin γ) 1+coshh sinh a sinh b By the fact that sinh h =sinhb sin α =sinha sin β,wehave 1 − cosh2 h − sinh2 h − sinh b sin α · sinh a sin β = = = − sin α sin β, sinh a sinh b sinh a sinh b sinh a sinh b so, using the dual form of the law of cosines, cos(a + b) − cosh(c + h) sinh a sinh b = [− cosh c sin α sin β +(1+coshh)(cos γ +1− sin γ)] 1+coshh sinh a sinh b = [−(cos α cos β +cosγ) + (1 + cosh h)(cos γ +1− sin γ)] . 1+coshh sinh a sinh b Since 1+cosh h > 0, we have the strong triangle inequality holds, if and only if cos α cos β +cosγ<(1 + cosh h)(cos γ +1− sin γ), and the result follows. A minor variation of the proof shows the case of equality. Lemma 5. For all triangles with γ>max{α, β}, cos α cos β +cosγ > 1. cos γ +1− sin γ Proof. Without loss of generality, 0 <α≤ β<γ<π/2. Then 0 > sin γ(sin γ − 1) = sin2 γ − sin γ>sin2 β − sin γ cos2 β>cos2 β +sin2 β − sin γ =1− sin γ cos2 β +cosγ>1 − sin γ +cosγ, so cos2 β +cosγ > 1. 1 − sin γ +cosγ Since 0 < cos β ≤ cos α,wehave cos2 β +cosγ cos α cos β +cosγ ≤ , 1 − sin γ +cosγ 1 − sin γ +cosγ and the results follows. By Lemma 4 and Lemma 5, we can conclude that the strong triangle inequality holds if and only if cos α cos β +cosγ 2 − 1 < cosh2 h. cos γ +1− sin γ (2) A strong triangle inequality in hyperbolic geometry 103 Using the law of cosines, cosh2 h ==sinh2 h +1=sin2 β sinh2 a +1=sin2 β(cosh2 a − 1) + 1 cos β cos γ +cosα 2 =sin2 β − sin2 β +1 sin β sin γ cos β cos γ +cosα 2 =cos2 β + . sin γ (3) Equations (2) and (3) together imply the following statement.
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