Unexpectedly MARCUS BIZONY RATIO THE GOLDEN tan tan If 1), Figure lengths (see have 1 will is sides hypotenuse the the of length the if Clearly, A tan o lrt,w oeta eaeuighr h symbol the 1 here value using the are represent we that note we clarity, For aldteGle ai stercpoa fti number, this of also reciprocal is the which is Ratio Golden the called tan ϕ ϕ Keywords: Keplertriangle, goldenratio,incircle,geometric progression 2 θ =( 2 4 2 = θ

stelre ct nl ftetinl,then triangle, the of angle acute larger the is θ θ θ = 1 ϕ = = cos elrtriangle Kepler htissdsaei h ai 1 ratio the in are sides its requires that which Progression, Geometric in are sides + √ + sec tan θ ϕ √ .Teohrvlewihcudeulywellbe could equally which value other The 1. hsagei h ct ouintotheequation solution the acute is angle This . = 2 θ 5 ϕ θ ) since .(o,since (For, 1. / − + steGle Ratio. Golden the is 2 ,adwihi eegvntesymbol the given here is which and 1, B ,ie,tan i.e., 1, θ . 618 saue ec tan Hence acute. is θ ϕ 1 sargtage ragewhose triangle right-angled a is , ... iue1 Figure 1 1/ √ hc en that means which , 4 ϕ ϕ 1 θ 2 ϕ = = , At Right Angles |Vol.6,No. 1,March2017 sec ϕ 1 C A + 2 . : 1/ θ √ ,w get we 1, hc yields which , √ 2 ϕ ϕ θ : cos ϕ where θ = 1 Part 1 1.) ϕ to G. 29

ClassRoom A

AB = 1 = J BC G K AC = G 1/2 F H CD = G 3/2 AD = G DE = G 2 AF = G 2

D = 5/2 = 3 E EF G AJ G FH = G 3 HJ = G 7/2 JK = G 4

B C

Figure 2

1 1 1 1 1 1

Figure 3

Dropping perpendiculars progressively from the not equilateral but in the form of a double Kepler right-angle vertex to the hypotenuse and then triangle, produced by placing two Kepler triangles back to a leg (see Figure 2) produces lengths which alongside each other, with their longer legs are powers of G 1/2; continuing the process coinciding (see Figure 4). generates similar triangles, so any length of the n/2 To see why, we need first to understand something form G can be achieved in this manner. about the incentre of a triangle, which is at the Now consider an isosceles triangle whose equal same distance r from each of its sides (see sides have length 1. Clearly there are lots of Figure 5). The figure shows a triangle and its possible ‘shapes’ for it, and presumably therefore different areas for its incircle (see Figure 3).It A seems reasonable to ask which of these isosceles triangles has the largest incircle. An intuitive response might be that the required 1 1 isosceles triangle is going to be equilateral; 1/√ϕ certainly if that turned out to be the case, it would fit one’s sense of what is ‘right’. And, indeed, itis when the isosceles triangle is equilateral that the θ largest proportion of its area is included in its B / C / D incircle. But if the incircle itself should be as large 1 ϕ 1 ϕ as possible, we need to make the isosceles triangle Figure 4

30 At Right Angles | Vol. 6, No. 1, March 2017 A

c b r E r I V

r

a B C D

Figure 5 Figure 6 incircle, and also lines connecting the vertices of Setting the derivative of r equal to zero gives the triangle to the incentre; they demonstrate how (2 + 2 cos x)2 cos 2x sin 2x( 2 sin x)=0, the area of the triangle can be seen as the sum of − − 4(1 + cos x) 2 cos2 x 1 the areas of three smaller triangles. Of course the ∴ − radii are perpendicular to the sides and therefore + 4 sin2 x cos(x = 0, ) serve as heights on those bases for the smaller (1 + cos x) 2 cos2 x 1 triangles. ∴ − + cos x 1 cos2 x = 0, Thus the area of the whole triangle is −( ) ra/2 + rb/2 + rc/2 = rs where s is the ∴ 2 cos2 x( 1 + cos)x(1 cos x)=0, semi-perimeter of the triangle, and we deduce that − − since 1 + cos x = 0. The equation in the last line for any triangle the length r of its in-radius is given ̸ yields by the formula cos2 x + cos x 1 = 0, − area of triangle so that cos x = G. The verification that this indeed r = . yields a maximum value of r is left to the reader. semi-perimeter of triangle This particular isosceles triangle ( ABD, with △ In the case of an isosceles triangle whose equal AB : AD : BD = 1 : 1 : 2/ϕ) has the additional sides have length 1 and with base angle x, we have: interesting property that if AC is the altitude to its base, then the perpendicular to AC through the 1 incentre I of the triangle meets side AB at the foot area of triangle = cos x sin x = sin 2x, · 2 E of the perpendicular CE to that side from the perimeter of triangle = 2 + 2 cos x. midpoint C of the base BD; see Figure 6. Moreover, I is a Golden Point of AC (i.e., divides Hence the radius of the incircle is given by it in the Golden Ratio). Further, the point V where BI meets CE is a Golden Point of both CE sin 2x and BI. Readers might like to tackle these proofs r = . 2 + 2 cos x for themselves.

MARCUS BIZONY is now retired, having been Deputy Head (Academic) at Bishops in Cape Town, and before that Head of Mathematics; he taught at Bishops for over thirty years. He was Associate Editor of Learning and Teaching Mathematics, a journal published by the Association for Mathematics Educators of South Africa. He was also convenor of the panel that sets the South African Olympiad questions for juniors (Grades 8 and 9). He may be contacted at [email protected].

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