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A Conic Section Problem Involving the Maximum Generalized Goldenrighttriangle 3 A CONIC SECTION PROBLEM INVOLVING THE MAXIMUM GENERALIZED GOLDEN RIGHT TRIANGLE JUN LI ABSTRACT. An interesting conic section problem involving the maximum generalized golden right triangle T2 is solved, and two simple constructions of T2 are shown. 1. INTRODUCTION As the great astronomer Johannes Kepler stated,“Geometry has two great treasures: one is the theorem of Pythagoras; the other, the division of a line into extreme and mean ratio. The first we may compare to a measure of gold; the second we may name a precious jewel”[1, p. 160]. 2. A CONIC SECTION PROBLEM FIGURE 1. An interesting conic section problem arXiv:1606.09244v1 [math.HO] 29 Jun 2016 Let’s consider an interesting problem involving an ellipse and a hyperbola in Figure 1. First, we 2 2 x2 y x2 y construct an ellipse a2 + b2 = 1 and a hyperbola c2 b2 = 1, where a is the semi-major axis, b 2 2 2− is the semi-minor axis of the ellipse, and a = b + c , such that, the eccentricity e1 of the ellipse and e2 of the hyperbola satisfy the condition e1e2 = 1, and the foci of the ellipse becomes the corresponding vertex of the hyperbola, next, let F1 and F2 denote the foci of the ellipse, KL the minor axis, MN the major axis, and O the origin, without loss of generality, we set c = OF2 = 1. Then, let P be the top-right intersection point of the ellipse and the hyperbola, construct a seg- ment PQ perpendicular to ON and intersecting ON at the foot Q, let H be the intersection point of 1 ON and the right directrix x = a of the hyperbola, now, our problem is: Problem 2.1. If PN KF , what will e , e , ON and OQ be? k 2 1 2 OQ HQ Date: 2016.6. 1 2 JUN LI Proposition 2.2. If PN KF , then e = 1 , e = φ√φ and ON = OQ = φ, where φ = 1+√5 , k 2 1 φ√φ 2 OQ HQ 2 in other words, Q divides ON into the golden ratio, and H divides OQ into the golden ratio. Proof. First, solve the equation set (2.1) 2 x2 y 2 + 2 = 1 a b2 (2.1) 2 y ( x 2 = 1 − b 2 2 2a2 (a 1)b to get the coordinates of P , we get xP = OQ = a2+1 , yP = PQ = a2−+1 , and by PN KF , we have PQ = KO , and get q q k 2 QN OF2 (a2 1)b2 2a2 a b 2− = ( 2 ) (2.2) r a + 1 − ra + 1 and the final form (a2 1)(a4 4a2 1) = 0 (2.3) − − − c Since c = 1 and a > c, then we obtain the unique solution a = φ√φ from (2.3), hence e1 = a = 1 , e = φ√φ, then ON = φ√φ = φ, OH = 1 = 1 , HQ = OQ OH = 1 , thus, OQ = φ, φ√φ 2 OQ √φ a φ√φ − √φ HQ and we also get QN = ON OQ = 1 = HQ, which means Q is the midpoint of HN. − √φ 1 AC BD Next, we show an interesting property in the ellipse which has eccentricity φ√φ , let and denote the latus rectum of the ellipse, then we have Proposition 2.3. The rectangle ACDB is made up of 4 congruent right triangles similar to the F1F2 Kepler triangle[2, p. 149] and AF1 = √φ. Also, it is shown in [3] that, AF2C is just the kind of isosceles triangle of smallest perimeter which circumscribes a semicircle.△ Proof. F F , AF 2 , then F1F2 √φ, hence proved. 1 2 = 2 1 = √φ AF1 = Interestingly, we notice that the KOF in Figure 1 is just the maximum generalized golden △ 2 right triangle T2[4] which has sides (1, √2φ, φ√φ), and we’ve already got an interesting construc- tion in [4], next, we will show another two simple constructions of it, see Figure 2 and 3. FIGURE 2. A simple construction of T2 Construction 2.4. A 3-step construction of T2: (1) construct a Kepler triangle ABC with BC = 1, AB = √φ (see, e.g., [5]) (2) construct a square ABDE △externally on the side AB A CONIC SECTION PROBLEM INVOLVING THE MAXIMUM GENERALIZED GOLDENRIGHTTRIANGLE 3 (3) draw an arc with the center at B and the radius BE, cutting the extension of BA at F , and join F to C Then FBC is T . △ 2 Proof. BF = BE = √2AB = √2φ. FIGURE 3. Another simple construction of T2 Construction 2.5. Another 3-step construction of T2: BC (1) construct a segment AB = 1, construct point C on the extension of AB that BA = φ, and extend BC to D that CD = BC (2) a semicircle is drawn with its center at the midpoint O of AD, and the radius OA, naturally passing through D (3) through B, construct a perpendicular segment BE to AB, and intersecting the semicircle at point E, and join E to A Then ABE is T . △ 2 Proof. According to Thales’ theorem, AED is a right triangle, then we get BE = √AB BD = √2φ. △ · REFERENCES [1] H. S. M. Coxeter, Introduction to Geometry, John Wiley and Sons, Inc., New York, 1961. [2] L. Mario, The Golden Ratio: The Story of Phi, The World's Most Astonishing Number, Broadway Books, New York, 2002. [3] D. W. DeTemple, The triangle of smallest perimeter which circumscribes a semicircle, The Fibonacci Quarterly, 30.3 (1992) 274. [4] J. Li, The triangle of smallest area which circumscribes a semicircle (2016), http://arxiv.org/abs/1606.08804 [5] Kepler triangle, Wikipedia, The Free Encyclopedia, https://en.wikipedia.org/wiki/Kepler_triangle. Mathematics Subject Classification (2010). 51M04, 51M09, 51M15, 11B39 Keywords. Conic section problem, Ellipse, Hyperbola, Golden ratio, Fibonacci numbers, Kepler triangle, Golden right triangle, Maximum generalized golden right triangle SCHOOL OF SCIENCE, JIANGXI UNIVERSITY OF SCIENCE AND TECHNOLOGY,GANZHOU, 341000, CHINA. E-mail address: [email protected].
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