The Triangle of Smallest Area Which Circumscribes a Semicircle 3

Home , Golden rectangle, Kepler triangle

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FIGURE 2. A sequence of generalized golden right triangles Tn

There is an identity (2.1) of the golden ratio[2] and Fibonacci numbers (see, e.g., [4, p. 78]). n+1 φ = Fn+1φ + Fn, (n = 0, 1, 2,...) (2.1) If we rewrite (2.1) in the form of (2.2), 2 2 n+1 φF +1 φ 1+ n = , (n = 1, 2, 3,...) (2.2) F F r n ! s n    THE TRIANGLE OF SMALLEST AREA WHICH CIRCUMSCRIBES A SEMICIRCLE 3

n+1 φFn+1 φ we will obtain a right triangles sequence Tn with sides (1, Fn , Fn ), see Figure 2. It’s easy to see that, the ﬁrst right triangle T1 with sides (1, √φq, φ) is theq Kepler triangle whose side lengths are in geometric progression, the second right triangle T2 is a (1, √2φ, φ√φ) triangle, and 2 furthermore, let n + , we ﬁnd that the limiting right triangle of Tn is just a (1, φ, 1+ φ ) triangle which forms→ half∞ of a golden rectangle. p Since there are only two golden (see [5, p. 73]) right triangles (one is the Kepler triangle with sides (1, √φ, φ), the other is the (1, φ, 1+ φ2) triangle), and they are both special cases of T , then, we can call T a generalized golden right triangles sequence. In addition, we have the n n p following simple area inequality (2.3) for Tn,

T1 Tn T2 (2.3) △ ≤△ ≤△ where Tn denoted as the area of Tn, and △ √φ F +1 T = n , (n = 1, 2, 3,...) (2.4) n 2 F △ r n in the inequality (2.3), we also notice that the second triangle T2 with sides (1, √2φ, φ√φ) is just the maximum area triangle in Tn, therefore, we can call T2 the maximum generalized golden right triangle. Interestingly enough, similar to the construction of the Kepler triangle T1 by ﬁrst creating a golden rectangle shown in [3], we can also construct the maximum generalized golden right triangle T2 by ﬁrst constructing a golden rectangle, see Figure 3.

FIGURE 3. An interesting construction of T2

Construction 2.1. A simple and interesting 3-step construction of T2: (1) construct a golden rectangle ABCD with BC = 1, AB = φ (see, e.g., [6, p. 118]) BO (2) construct O dividing BC in the golden ratio and OC = φ (3) draw an arc with the center at O and the radius of length AC, cutting the extension of BA at E, and join E to C Then EBC is just T2 having sides (1, √2φ, φ√φ). △ Proof. BO = BC = 1 , EO = AC = 1+ φ2, thus, BE = √EO2 BO2 = √2φ. φ φ − Last, back to Figure 1 again, we givep a problem to readers: 4 JUN LI

Problem 2.2. If the radius R and the two sides AB and AC of ABC can not be three edges of a triangle, determine ABC of smallest perimeter. And if ABC△ can not be an acute-angled triangle, determine ABC△ of smallest perimeter. △ △ REFERENCES [1] D. W. DeTemple, The triangle of smallest perimeter which circumscribes a semicircle, The Fibonacci Quarterly, 30.3 (1992) 274. [2] L. Mario, The Golden Ratio: The Story of Phi, The World's Most Astonishing Number, Broadway Books, New York, 2002. [3] Kepler triangle, Wikipedia, The Free Encyclopedia, https://en.wikipedia.org/wiki/Kepler_triangle. [4] T. Koshy, Fibonacci and Lucas Numbers with Applications, A Wiley-Interscience Publication, 2001. [5] M. Bicknell and V. E. Hoggatt, Golden triangles, rectangles, and cuboids, The Fibonacci Quarterly, 7.1 (1969) 73–91. [6] A. S. Posamentier and I. Lehmann, The Fabulous Fibonacci Numbers, Prometheus Books, 2007. Mathematics Subject Classiﬁcation (2010). 51M04, 51M15, 11B39 Keywords. Smallest Area, Triangle, Semicircle, Golden rectangle, Golden ratio, Fibonacci numbers, Kepler triangle, Golden right triangle, Maximum generalized golden right triangle

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