Prove that ratio of a regular pentagon diagonal to its side is (((√5)+1)/2).


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Question Number 3584 by prakash jain last updated on 16/Dec/15

$Prove that ratio of a regular pentagon diagonal$ $to its side is \frac{\sqrt{5} + 1}{2} .$
Commented by Yozzii last updated on 15/Dec/15

$I t s s i d e i s o n e o f t h e 5 e d g e s ?$ $a b$ $c e$ $d$ $d i a g o n a l s = b d, a d, e c, a e, b c ?$
Commented by prakash jain last updated on 15/Dec/15

$y e s . Also it \frac{\sqrt{5} + 1}{2}$
Commented by Yozzii last updated on 16/Dec/15

$C o u l d a g e n e r a l f o r m u l a f o r t h e$ $v a l u e o f s u c h a r a t i o b e f o u n d i n t e r m s$ $o f n f o r a r e g u l a r n - g o n ?$
Commented by prakash jain last updated on 16/Dec/15

$A formula using \sin function is clear .$ $A general algebria formula will require solving$ $n^{t h} root of unity as an algebric formula .$
Commented by Rasheed Soomro last updated on 16/Dec/15

$W h a t i s t h e d i a g o n a l o f p o l y g o n ?$ $I f a r e g u l a r p o l y g o n h a s m o r e t h a n$ $o n e d i a g o n a l s o f d i g f e r e n t s i z e s t h e n$ $t h e r e m a y b e m o r e t h a n o n e s u c h r a t i o s$ $f o r e x a m p l e \frac{D_{1}}{s}, \frac{D_{2}}{s}, . . .$
Commented by prakash jain last updated on 16/Dec/15

$Yes . There will diagonals of different length .$
	Answered by Yozzii last updated on 16/Dec/15
	$All regular n−gons can be inscribed in a circle. PROOF: Consider the n roots of the equation z^n =re^(iθ) where i=(√(−1)),r>0,n∈N+{0},−π<θ≤π. Since e^(iθ) =e^(i(θ+2πt)) (t∈Z), because of the periodicity of the sine and cosine functions which arise from the form e^(iθ) =cosθ+isinθ, ⇒ z^n =re^(i(θ+2tπ)) ⇒z=r^(1/n) e^(i((θ+2tπ)/n)) . The n values of z are obtained for 0≤t≤n−1. Observe that each value of z has the common modulus r^(1/n) . On an Argand diagram, say the n values of z are plotted with lines joining their points to the origin. Suppose one point has argument α_t =((θ+2tπ)/n) and an adjacent point has argument α_(t+1) =((θ+2(t+1)π)/n). ⇒ α_(t+1) −α_t =((θ+2tπ+2π−θ−2tπ)/n)=((2π)/n). The angle between the lines joining consecutive points is a constant (((2π)/n)) independent of t and all of these n angular differences sum to 2π. Thus, the n points are regularly positioned about the origin and they may be joined by lines to form a regular n−gon figure. These n points are placed about the origin and they are at the equal distances (r^(1/n) ) from the origin. These points hence, by definition of a circle as the locus of points at a fixed distance from a fixef point, lie on a circle. Since these n points form the corners of a regular n−gon, it follows that such a figure can be inscribed in a circle. □ So, inscribe a pentagon ABCDE in a circle and let its radius be denoted by r. The angle at the centre O of the pentagon in each of the five sectors is ((2π)/5)≡72°. Denote s as the side length of the pentagon. Draw a perpendicular bisector of one side of the pentagon (e.g AB). In so doing, one of the sectors is symmetrically bisected to form two right−angled triangles, with hypothenuses being the radius r of the circle, and s is halved.$
	$A l l r e g u l a r n - g o n s c a n b e i n s c r i b e d$ $i n a c i r c l e .$ $PROOF :$ $C o n s i d e r t h e n r o o t s o f t h e e q u a t i o n$ $z^{n} = {r e}^{i θ}$ $w h e r e i = \sqrt{- 1}, r > 0, n \in N + {0}, - π < θ ⩽ π .$ $S i n c e e^{i θ} = e^{i (θ + 2 π t)} (t \in Z), b e c a u s e o f t h e p e r i o d i c i t y$ $o f t h e s i n e a n d c o s i n e f u n c t i o n s w h i c h$ $a r i s e f r o m t h e f o r m e^{i θ} = c o s θ + i s i n θ,$ $\Rightarrow z^{n} = {r e}^{i (θ + 2 t π)} \Rightarrow z = r^{1 / n} e^{i \frac{θ + 2 t π}{n}} .$ $T h e n v a l u e s o f z a r e o b t a i n e d f o r 0 ⩽ t ⩽ n - 1 .$ $O b s e r v e t h a t e a c h v a l u e o f z h a s t h e$ $c o m m o n m o d u l u s r^{1 / n} . O n a n A r g a n d$ $d i a g r a m, s a y t h e n v a l u e s o f z a r e p l o t t e d$ $w i t h l i n e s j o i n i n g t h e i r p o i n t s t o t h e$ $o r i g i n . S u p p o s e o n e p o i n t h a s a r g u m e n t$ $α_{t} = \frac{θ + 2 t π}{n} a n d a n a d j a c e n t p o i n t h a s$ $a r g u m e n t α_{t + 1} = \frac{θ + 2 (t + 1) π}{n} .$ $\Rightarrow α_{t + 1} - α_{t} = \frac{θ + 2 t π + 2 π - θ - 2 t π}{n} = \frac{2 π}{n} .$ $T h e a n g l e b e t w e e n t h e l i n e s j o i n i n g$ $c o n s e c u t i v e p o i n t s i s a c o n s t a n t (\frac{2 π}{n})$ $i n d e p e n d e n t o f t a n d a l l o f t h e s e n$ $a n g u l a r d i f f e r e n c e s s u m t o 2 π . T h u s, t h e$ $n p o i n t s a r e r e g u l a r l y p o s i t i o n e d a b o u t t h e$ $o r i g i n a n d t h e y m a y b e j o i n e d b y l i n e s$ $t o f o r m a r e g u l a r n - g o n f i g u r e .$ $T h e s e n p o i n t s a r e p l a c e d a b o u t t h e$ $o r i g i n a n d t h e y a r e a t t h e e q u a l d i s t a n c e s (r^{1 / n})$ $f r o m t h e o r i g i n . T h e s e p o i n t s h e n c e,$ $b y d e f i n i t i o n o f a c i r c l e a s t h e l o c u s o f$ $p o i n t s a t a f i x e d d i s t a n c e f r o m a f i x e f$ $p o i n t, l i e o n a c i r c l e . S i n c e t h e s e n p o i n t s$ $f o r m t h e c o r n e r s o f a r e g u l a r n - g o n,$ $i t f o l l o w s t h a t s u c h a f i g u r e c a n b e$ $i n s c r i b e d i n a c i r c l e . ◻$ $S o, i n s c r i b e a p e n t a g o n A B C D E i n a c i r c l e a n d$ $l e t i t s r a d i u s b e d e n o t e d b y r . T h e$ $a n g l e a t t h e c e n t r e O o f t h e p e n t a g o n i n$ $e a c h o f t h e f i v e s e c t o r s i s \frac{2 π}{5} \equiv 72 ° .$ $D e n o t e s a s t h e s i d e l e n g t h o f t h e p e n t a g o n .$ $D r a w a p e r p e n d i c u l a r b i s e c t o r o f o n e$ $s i d e o f t h e p e n t a g o n (e . g A B) . I n s o d o i n g, o n e$ $o f t h e s e c t o r s i s s y m m e t r i c a l l y b i s e c t e d$ $t o f o r m t w o r i g h t - a n g l e d t r i a n g l e s,$ $w i t h h y p o t h e n u s e s b e i n g t h e r a d i u s r o f$ $t h e c i r c l e, a n d s i s h a l v e d .$
	Commented by Yozzii last updated on 16/Dec/15

	$T h u s, i n s u c h a t r i a n g l e,$ $\frac{s}{2} = r s i n 36^{°} \Rightarrow r = \frac{s}{2 s i n 36 °} . ()$ $N o w, d r a w a l i n e j o i n i n g t w o p o i n t s$ $s u c h t h a t a d i a g o n a l D o f t h e p e n t a g o n$ $i s o b t a i n e d (e . g A C) . D r a w a p e r p e n d i c u l a r$ $b i s e c t o r o f A C s o t h a t a n o t h e r r i g h t - a n g l e d$ $t r i a n g l e i s o b t a i n e d a s b e f o r e . H o w e v e r,$ $t h e a n g l e a t O i n t h i s t r i a n g l e i s 72 ° .$ $T h e r e f o r e, w e g e t$ $\frac{D}{2} = r s i n 72^{°} \Rightarrow r = \frac{D}{2 s i n 72 °} . ( )$ $E q u a t i n g () a n d (* *) w e h a v e$ $\frac{D}{2 s i n 72^{°}} = \frac{s}{2 s i n 36 °} \Rightarrow \frac{D}{s} = \frac{s i n 72^{°}}{s i n 36 °} = 2 c o s 36 ° .$
	Commented by Yozzii last updated on 16/Dec/15

	$L e t f = 18 ° \Rightarrow 5 f = 90 ° \Rightarrow 2 f = 90 ° - 3 f .$ $∴ s i n 2 f = s i n (90 ° - 3 f)$ $s i n (90 ° - x) = c o s x i n g e n e r a l .$ $∴ s i n 2 f = c o s 3 f = 4 {c o s}^{3} f - 3 c o s f$ $B u t s i n 2 f = 2 s i n f c o s f .$ $∴ 2 s i n f c o s f = 4 {c o s}^{3} f - 3 c o s f .$ $N o w, c o s f \neq 0 \Rightarrow 2 s i n f = 4 {c o s}^{2} f - 3$ $2 s i n f = 4 (1 - {s i n}^{2} f) - 3$ $2 s i n f = 4 - 4 {s i n}^{2} f - 3$ $4 {s i n}^{2} f + 2 s i n f - 1 = 0$ $∴ s i n f = \frac{- 2 \pm \sqrt{4 - 4 \times 4 \times (- 1)}}{8}$ $s i n f = \frac{- 2 \pm 2 \sqrt{5}}{8} = \frac{- 1 \pm \sqrt{5}}{4} .$ $0 < f < 90 ° \Rightarrow s i n f > 0 \Rightarrow s i n f \neq \frac{- 1 - \sqrt{5}}{4} .$ $∴ s i n f = \frac{- 1 + \sqrt{5}}{4}$ ${c o s}^{2} f = 1 - {s i n}^{2} f$ ${c o s}^{2} f = 1 - {(\frac{\sqrt{5} - 1}{4})}^{2} = \frac{16 - 5 - 1 + 2 \sqrt{5}}{16}$ ${c o s}^{2} f = \frac{10 + 2 \sqrt{5}}{16} = \frac{5 + \sqrt{5}}{8}$ $\Rightarrow 2 {c o s}^{2} f = \frac{5 + \sqrt{5}}{4} \Rightarrow 2 {c o s}^{2} f - 1 = \frac{1 + \sqrt{5}}{4} .$ $B u t, c o s 2 f = c o s 36 ° = 2 {c o s}^{2} f - 1 .$ $∴ c o s 36 ° = \frac{1 + \sqrt{5}}{4} \Rightarrow 2 c o s 36 ° = \frac{1 + \sqrt{5}}{2} .$ $H e n c e \frac{D}{s} = \frac{1 + \sqrt{5}}{2} ◼$
	Answered by Rasheed Soomro last updated on 16/Dec/15

	$L e t A, B, C a r e a n y t h r e e c o n s e c u t i v e$ $v e r t i c s, t h e n \overset{―}{A B}, \overset{―}{B C} a r e s i d e s o f$ $p e n t a g o n a n d \overset{―}{A C} i s i t s d i a g o n a l$ $N o w l e t m \overset{―}{A B} = m \overset{―}{B C} = s$ $a n d m \overset{―}{A C} = d$ $W e h a v e t o p r o v e d : s = \frac{\sqrt{5} + 1}{2} : 1$ $o r \frac{d}{s} = \frac{\sqrt{5} + 1}{2}$ $W e k n o w t h a t m ∠ A B C = 108 °$ $B y c o s i n e l a w$ $d = \sqrt{s^{2} + s^{2} - 2 (s) (s) \cos 108 °}$ $\frac{d}{s} = \frac{\sqrt{s^{2} + s^{2} - 2 (s) (s) \cos 108 °}}{s}$ $= \frac{s \sqrt{2 - 2 \cos 108 °}}{s} = \sqrt{2 - 2 \cos 108 °}$ $= \sqrt{2 - 2 (\frac{1 - \sqrt{5}}{4})} = \sqrt{\frac{8 - 2 + 2 \sqrt{5}}{4}}$ $= \frac{\sqrt{6 + 2 \sqrt{5}}}{2}$ $N o w w e h a v e t o c h a n g e t h e n u m e r a t o r i n t o$ $a + b \sqrt{5} f o r m, w h e r e a, b a r e r a t i o n a l s$ $S o l e t \sqrt{6 + 2 \sqrt{5}} = a + b \sqrt{5}$ $6 + 2 \sqrt{5} = {(a + b \sqrt{5})}^{2}$ $= a^{2} + 5 b^{2} + 2 a b \sqrt{5}$ $a^{2} + 5 b^{2} = 6 \land 2 a b = 2 \Rightarrow a = \frac{1}{b}$ $a^{2} + 5 b^{2} = 6 \Rightarrow {(\frac{1}{b})}^{2} + 5 b^{2} = 6$ $\Rightarrow 1 + 5 b^{4} = 6 b^{2} \Rightarrow 5 b^{4} - 6 b^{2} + 1 = 0$ $\Rightarrow b^{2} = \frac{6 \pm \sqrt{36 - 20}}{10} = \frac{6 \pm 4}{10} = 1, \frac{1}{5}$ $\Rightarrow b = \pm 1, \pm \frac{1}{\sqrt{5}} [d i s c a r d i n g a s i t s i r r a t i o n a l]$ $b = \pm 1 \Rightarrow a = \pm 1$ $a + b \sqrt{5} = 1 + \sqrt{5}, 1 - \sqrt{5}, - 1 + \sqrt{5}, - 1 - \sqrt{5}$ $L a s t t h r e e a r e e x t r a n e o u s (?)$ $\frac{\sqrt{6 + 2 \sqrt{5}}}{2} = \frac{1 + \sqrt{5}}{2}$
	Commented by prakash jain last updated on 16/Dec/15

	$\frac{d}{s} = \frac{\sqrt{6 + 2 \sqrt{5}}}{2} = \frac{1 + \sqrt{5}}{2}$
	Commented by Rasheed Soomro last updated on 16/Dec/15

	$TH α n k S!$
	Answered by Rasheed Soomro last updated on 16/Dec/15

	$L e t A, B, C a r e t h r e e c o n s e c u t i v e v e r t i c e s$ $o f a p e n t a g o n . T h e n A B a n d B C a r e s i d e s$ $a n d A C i s d i a g o n a l o f p e n t a g o n . m ∠ A B C = 108 °$ $△ A B C i s a n i s s o s c e l t r i a n g l e i n w h i c h$ $l e t A B = B C = s, A C = d$ $m ∠ A B C = 108 °, m ∠ C A B = m ∠ B C A = 36 °$ $B y s i n e l a w$ $\frac{d}{s i n 108 °} = \frac{s}{36 °} \Rightarrow \frac{d}{s} = \frac{s i n 108 °}{s i n 36 °}$ $= \frac{(\sqrt{5 + \sqrt{5}}}{(\sqrt{5 - \sqrt{5}}} \times \frac{(\sqrt{5 + \sqrt{5}}}{(\sqrt{5 + \sqrt{5}}} = \frac{5 + \sqrt{5}}{\sqrt{25 - 5}} = \frac{5 + \sqrt{5}}{2 \sqrt{5}}$ $= \frac{5}{2 \sqrt{5}} + \frac{\sqrt{5}}{2 \sqrt{5}} = \frac{\sqrt{5}}{2} + \frac{1}{2} =$ $\frac{d}{s} = \frac{1 + \sqrt{5}}{2}$
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