arXiv:1606.09244v1 [math.HO] 29 Jun 2016 ment a opr oamaueo od h eodw a aeapreci a name may we second into the line gold; of a measure of a division to the compare other, may the Pythagoras; of theorem orsodn etxo h yebl,nx,let next, hyperbola, the of vertex corresponding ON osrc nellipse an construct and rbe 2.1. Problem io axis, minor stesm-io xso h lis,and ellipse, the of axis semi-minor the is stegetatooe oansKpe ttd“emtyh stated,“Geometry Kepler Johannes astronomer great the As hn let Then, e’ osdra neetn rbe novn nellipse an involving problem interesting an consider Let’s Date OI ETO RBE NOVN H AIU GENERALIZED MAXIMUM THE INVOLVING PROBLEM SECTION CONIC A e n h ih directrix right the and Q P 2 2016.6. : ftehproastsytecondition the satisfy hyperbola the of triangle A BSTRACT epniua to perpendicular MN P If T etetprgtitreto on fteelpeadtehy the and ellipse the of point intersection top-right the be 2 N P h ao xs and axis, major the ssle,adtosml osrcin of constructions simple two and solved, is nitrsigcncscinpolmivligtemaximu the involving problem section conic interesting An . x a 2 2 k + KF F y b IGURE ON 2 x 2 2 htwill what , 1 = = n intersecting and a 1 ODNRGTTRIANGLE RIGHT GOLDEN .A 2. .A neetn oi eto problem section conic interesting An 1. n hyperbola a and ftehproa o,orpolmis: problem our now, hyperbola, the of O OI ETO PROBLEM SECTION CONIC .I 1. h rgn ihu oso eeaiy eset we generality, of loss without origin, the e 1 a , 2 e NTRODUCTION 2 = , U LI JUN ON e OQ b ON 1 2 1 F e + 2 1 x and c 2 2 and T 1 = c ttefoot the at 2 2 − uhta,teeccentricity the that, such , r shown. are HQ OQ y F b n h oio h lis eoe the becomes ellipse the of foci the and , 2 2 2 be? 1 = eoetefc fteellipse, the of foci the denote n yebl nFgr .Frt we First, 1. Figure in hyperbola a and xrm n enrto h rtwe first The ratio. mean and extreme Q where , stogettesrs n sthe is one treasures: great two as e eteitreto on of point intersection the be H let , u ee”1 .160]. p. jewel”[1, ous eeaie odnright golden generalized m a ebl,cntutaseg- a construct perbola, stesm-ao axis, semi-major the is e 1 c = fteellipse the of OF KL 2 1 = the . b 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]