On an area property of the sum cota + cotb + cotγ in a triangle

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1 On an area property of the sum cota + cotb + cot in a triangle 1. Introduction Rummaging through an obscure trigonometry book published in Athens, Greece (and in the Greek language), and long out of print, I discovered the following, listed as an exercise (see reference [1]): Quote: Let be a triangle. If on the side AB we draw the line perpendicular to AB at the point B; on the side B the line perpendicular to it at the point ; and on A the perpendicular to it at A; a new triangle A Β is formed. Prove that E E (cotα + cotβ + cot), where E and E stand for the areas of the triangles Α Β Α and respectively. end of quote 1

2 Important Note Regarding Angle Notation In the interest of clarity and avoidance of confusion on the part of the reader, please note: 1) By refering to the angles A,B,, we always be refering to the internal triangles A,B,, in the context of a clearly defined triangle. In that context, angle A < BA < AB; or by using the alternative notation ^, the alternative notation ^, ^ ^ angle A BA AB. Similarly, angle B < AB < BA and ^ ^ angle AB BA. Consequently, all six angles A,B,,A,B,, referred to in Figures 1 through 4, are the internal angles of the triangles AB and A B. ) In any other context, we would use any of the two standard threeletter angle notations (mentioned above), with the vertex letter situated in the middle. However, in this paper, such other context, does not truly arise. 3) In a paper like this, where the internal angles A,B, are repeated very frequently throughout, using a three-letter notation (instead of a single letter one), would conceivably result in a significant increase in the length of the paper.

3 A first observation reveals that the sum cotα + cotβ + cot can be arbitrarily large. Indeed, if we restrict our attention only to those triangles in which none of the angles A, B, is obtuse; then each of the trigonometric numbers cotα, cotβ, cot is bounded below by zero, i.e. cota, cotb, cot 0. But say, angle can become arbitrarily small; in the language of calculus, 0 + ; so that cot +. A second observation shows that the triangles Figures, 3, 4). and Β Α are always similar (see Furthermore note that if we were to effect a similar construction of triangle Α Β from a given triangle Β Α but using an angle ϕ, 0 < ϕ 90 (instead of just ϕ 90 ), the same fact would emerge; that is, the two triangles are similar (see Figure 1). This paper has a two-fold aim: first establish that E E (cotα + cotβ + cot) ; and secondly that the minimum value of this area ratio is 3, attained precisely when the triangle is equilateral (and thus, by similarity, triangle Α Β as well). Even though, as we shall see, among all triangle pairs (, Α Β ), the ones with the smallest ratio ( 3) are those pairs in which both triangles are equilateral; if we restrict our search only to those pairs in which both and Β Α are right triangles, then, as we will show, the minimum value of the above area ratio is equal to 4; obtained precisely when both right triangles are isosceles.. Illustrations 3

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5 3. A few preliminaries The nine formulas listed below pertain to a triangle with sidelengths α Β, β Α, γ ΑΒ, angles Α, Β,, semiperimeter r (i.e. r α + β + γ) and area E. Of these nine formulas, the first five are well known to a wide mathematical audience, while the last four (Formulas 6, 7, 8, 9) are less so; for each of those formulas, Formulas 6, 7, 8, and 9 we offer a short proof in the next section. Formula 1 (Heron s Formula): E r (r - α) (r - β) (r - γ) Formula : cosθ cos θ sin θ for any angle θ, typically measured in degrees or radians Formula 3 (Law of Cosines): α β + γ - βγcosa, β α + γ - αγcosb, γ α + β - αβcos Formula 4 (summation formulas): For any angles θ and ω, cos (θ + ω) cosθcosω - sinθsinω, sin (θ + ω) sinθcosω + cosθsinω Formula 5: E 1 βγsina 1 αβsin 1 γαsinβ Formula 6: 16E (α β + β γ + γ α ) - (α 4 + β 4 + γ 4 ) Formula 7: For any angle θ which is not the form κπ or κπ + π, where κ is an integer, cotθ cot θ - 1 cotθ 5

6 Formula 8: cot A (r r (r - α), cot - β) (r - γ) B r (r - β), (r - α) (r - γ) cot (r r (r - γ) - β) (r - α) Formula 9: E α + β + γ 4 (cotα + cotb + cot) 4. Proofs of Formulas 6, 7, 8, and 9 a) Proof of Formula 6: From Formula 1 we have, E r (r - α) (r - β) (r - γ) 16E 16r (r - α) (r - β) (r - γ) ; 16E α + β + γ β + γ - α α + γ - β α + β - γ 16 (α + β + γ) (β + γ - α) [α - (β - γ)] (α + β - γ) [(β + γ) - α )] [α - (β - γ) ] α (β + γ) - α 4 - [(β + γ) (β - γ)] + α (β - γ) α [(β + γ) + (β - γ) ] - α 4 - [(β - γ )] α [ (β + γ )] - α 4 - β 4 - γ 4 + β γ (α β + β γ + γ α ) - (α 4 + β 4 + γ 4 ) b) Proof of Formula 7: For any two angles θ and ω such that θ + ω κπ, κ any integer (so that sin (ω + θ) 0) we have, in accordance with Formula 4, cot (θ + ω) cos (θ + ω) sin (θ + ω) cosθcosω - sinθsinω sinθcosω + sinωcosθ 6

7 If, in addition to the above condition, we also have θ mπ and ω nπ (where m, n are any integers), then sinθsinω 0; and so, cos (θ + ω) cosθcosω sinθsinω - sinθsinω sinθsinω sinθcosω sinωcosθ - sinθsinω sinωsinθ cotθcotω - 1 cotω + cotθ cot θ - 1 Thus, by setting θ ω we obtain cotθ, under the cotθ conditions θ l π and θ l π; l any integer. Considering the cases l even and l odd, the condition can be simplified into θ κπ and θ κπ + π, κ any integer. c) Proof of Formula 8: We need only establish the first of the three (sub)formulas, the other two are established similarly. Indeed, if we apply the first of the two (sub)formulas in Formula with θ A ; in combination with the first (sub)formula in Formula 3 we obtain, β γ - α cos A 1+ βγ + cos A (β + γ) - α 4 βγ cos A (β + γ - α) (β + γ + α) 4 βγ (r - α) (r) 4 βγ (r - α) r βγ (1) Likewise, we apply cosθ 1 - sin θ with θ A in conjunction with the first (sub)formula in Formula 3 to obtain, β + γ - α sin A 1- βγ α - (β - γ) 4 βγ 7

8 sin A (α + γ - β) (α + β - γ) 4 βγ (r - β) (r - γ) 4 βγ (r - β) (r - γ) βγ () Equations (1) and () imply cot ( ) A cos A sin ( A ) r (r - α) (r - β) (r - γ) (3) Since A is a triangle angle, 0 < A < 90 ; which means, that in particular cot A > 0; thus, by virtue of (3) we conclude that cot A r (r - α) (r - β) (r - γ). d) Proof of Formula 9: We apply the first (sub)formula of Formula 5: E βγ ; we then substitute for βγ in the first (sub)formula of sina Formula 3 to obtain, α β + γ - 4E cosa α sina β + γ - 4Ε cota (4) Similarly, we also have, β α + γ - 4Ε cotβ (5) and, γ α + β - 4Ε cot (6) Adding equations (4), (5), and (6) memberwise and solving for the area E produces the desired result: E α + β + γ. 4 (cota + cotb + cot ) 8

9 5. Postulate 1 and its proof Postulate 1 Let be a triangle and Β Α be the triangle constructed from by drawing the following three straight lines: The perpendicular to AB at the point B, the perpendicular to B at, and the perpendicular to A at Α. If E and E are the areas of Α Β E respectively, (cota + cotb + cot). E and Proof: Without loss of generality, we may assume that B, are acute angles; 0 < B, < 90 ; while A maybe an acute, right, or obtuse angle. The key point in this proof is the observation or realization that, E Ε + γ cota + β cot + α cotb (7) To see why this is so let us first take a look at Figure. In that picture we see that lies in the interior of Β Α and that, E Ε + (area ) + (area B Α ) + (area Α Β ), from which (7) easily follows (look at the formulas listed in the space adjacent the illustration in Figure ; and just apply the area formula for a right triangle). Now, in the case of A 90 (Figure 3), (area A B ) 0 (since cota cot90 0); equation (7) still holds true with the term γ cota being zero. 9

10 Next, let us look at Figure 4. Here, triangle does not lie in the interior of Α Β ; however, as we show below, equation (7) still holds true; the piece of that lies outside the interior of triangle Β Α (that is, triangle M), is compensated by the fact that cota < 0, on account of 90 < A < 180. Indeed, E (area Ε (area B Α B Α ) + (area Α Μ ) + (area of the quadrilateral MΑ ); ) + Ε - (area Α Μ Β ) + (area Β Α ) - (area Μ ); Β and by virtue of (area ) (area Α Μ Β ) + (area Μ ), Β we obtain E Ε + (area B Α ) + (area Β Α ) - (area ) (8) We have, (area B Α ) β cot, (area Β Α ) α cotb, and (area ) γ Β ; and since (from triangle in Figure 4) we have Β γcot (180 - A) - γcota, we conclude that (area ) - γ cota ; and by (8) we arrive at equation (7); thus (7) holds true in all cases. Next note that, E (cota + cotb + cot) E (by Formula 9) E α + β + γ E 4Ε 16EE (α + β + γ ) (by (7)) 16E γ cota + β cot + α E + cotb (α + β + γ ) 10

11 16E + 8 (γ cota + β cot + α cotb) Ε - (α + β + γ ) 0 (9) Clearly, if we prove (9), the proof of Postulate 1 will be complete. First note that according to Formula 6 we have, 16E (α β + β γ + γ α ) - (α 4 + β 4 + γ 4 ) (10) Secondly, we compute the term 8Eγ cota in (9), in terms of α, β, γ; and by cyclicity of the letters α, β, γ, we correspondingly find similar formulas for the terms 8Eβ cot and 8Eα cotb. We start by applying the first (sub)formula of Formula 8 together with Formulas 7 and 1 to obtain, 8Eγ 1 r (r - α) cota 8 r (r - α) (r - β) (r - γ) - 1 (r - β) (r - γ) (r - β) (r - γ) r (r - α) 8Eγ cota 4γ [ r (r - α) - (r - β) (r - γ) ] 8Eγ cota 4γ α + β + γ β + γ - α α + γ - β α + β - γ) - 8Eγ cota γ [(β + γ) - α - {α - (β - γ) }] 8Eγ cota γ (β + γ - α ) (11) 11

12 Similarly, by cyclicity, 8Eβ cot β (α + β - γ ) and (1) 8Eα cotβ α (γ + α - β ) (13) Also, by expansion we have, - (α + β + γ ) - (α β + β γ + γ α ) - (α 4 + β 4 + γ 4 ) (14) If we add (10), (11), (1), (13), and (14) memberwise, the resulting equation is (9). 6. Two Lemmas from calculus and their conclusion Lemma 1: Over the open interval (0, + ), the function f (x) x - x x x has an absolute minimum at x 1 ; the 3 absolute minimum value being f Lemma : Let k be a positive constant. Over the open interval (0, π ), the function gk (x) cot x + kcotx + k cotx + k + 1 has an absolute minimum at x θ k ; where θ k is the unique real number in (0, π ) such that cotθ k - k + k + 1; the absolute minimum value being g k (θ k ) k - k k k (it can be easily verified that g k (θ k ) > 0.) 1

13 The proof to the first Lemma utilizes the standard calculus techniques and is left to the reader. The proof of the second Lemma can also be done by the use of the standard techniques. To the interested reader we point out that the derivative of the function g k is given by - csc x [cot x + kcotx - 1] g k (x) (cotx + k) - ccs x [cotx + k - (cotx + k) k + 1] [cotx + k + k + 1] Note that in the last fraction, only the factor cotx + k - k + 1 changes sign over the interval (0, π ); that happens at the only critical number θk this function has: the unique number θ k such that cotθ k - k + k + 1 (recall that cotx is a decreasing function on (0, π ) and whose range is (0, + )). Combining the two Lemmas we arrive at, The conclusion from the two Lemmas Among all functions g k in Lemma, the one with the smallest absolute minimum is the function g k with k 3 1. The smallest absolute minimum occurs at π x θ k (from cotθ ). 3 1 The smallest minimum is given by g

14 7. Postulates and 3 Now let us look at the sum cota + cotb + cot with 0 < B, < 90 and 0 < A < 180. Since A (B + ) we have, cota + cotb + cot cot [180 - (Β + )] + cotb + cot - cot (Β + ) + cotb + cot. Recall from the proof of Formula 7 that cot (Β + ) cotbcot - 1. cotb + cot cotbcot - 1 Therefore cota + cotb + cot - cotb + cot + cotb + cot; cota + cotb + cot cot B + cot cotb + cot cotb + cot + 1. If we hold the angle fixed, cot becomes a positive constant; cot k. And if we set B x, and let x vary in the open interval (0, π ), the above cotangent sum becomes simply one of the functions g k of Lemma! Hence, according to the Conclusion of the two Lemmas, the smallest value of the above sum is equal to 3 attained when k cot 1 and 3 B x 3 π ; in degrees, B 60 ; A 60 as well. Combining this with Postulate 1 leads to 14

15 Postulate The ratio E E has the minimum value 3. The minimum value is attained precisely among the triangle pairs (, Β ), with Α both triangles being equilateral. We finish by taking a look at the cases in which (and thus Α Β as well) is a right triangle: A 90 ; 90 - Β; so that cota + cotb + cot 0 + cotb + cot (90 - Β) cotb + tanb cosb sinb cos B + sin B + sinb cosb sinbcosb 1 sinbcosb sinbcosb sinb, since 0 < sinb 1, in view of 0 < B < 180. Thus the smallest value of (cota + cotb + cot) is 4; this happens when sinb 1; B 45. Postulate 3 Among all pairs (, Β ) in which both triangles are right Α ones the minimum value of E E is 4 attained when both triangles are isosceles. REFERENCES 15

16 [1] (transliterated from the Greek) Marinos Zevas, Trigonometry, (in Greek), Gutenberg Press, Athens, Greece, July 1973, no ISBN number. 16