Prove-that-ratio-of-a-regular-pentagon-diagonal-to-its-side-is-5-1-2-

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

Its side is one of the 5 edges? a b c e d diagonals=bd,ad,ec,ae,bc?

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

Could a general formula for the value of such a ratio be found in terms of n for a regular n−gon?

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^(th) root of unity as an algebric formula.

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

What is the diagonal of polygon? If a regular polygon has more than one diagonals of digferent sizes then there may be more than one such ratios for example (D_1 /s),(D_2 /s),...

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}, \dots

Commented by prakash jain last updated on 16/Dec/15

Yes. There will diagonals of different length.

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

Thus, in such a triangle, (s/2)=rsin36^° ⇒r=(s/(2sin36°)) . (∗) Now, draw a line joining two points such that a diagonal D of the pentagon is obtained (e.g AC). Draw a perpendicular bisector of AC so that another right−angled triangle is obtained as before. However, the angle at O in this triangle is 72°. Therefore, we get (D/2)=rsin72^° ⇒r=(D/(2sin72°)) . (∗∗) Equating (∗) and (∗∗) we have (D/(2sin72^° ))=(s/(2sin36°))⇒(D/s)=((sin72^° )/(sin36°))=2cos36°.

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

Let f=18°⇒ 5f=90°⇒2f=90°−3f. ∴ sin2f=sin(90°−3f) sin(90°−x)=cosx in general. ∴ sin2f=cos3f=4cos^3 f−3cosf But sin2f=2sinfcosf. ∴ 2sinfcosf=4cos^3 f−3cosf. Now, cosf≠0⇒2sinf=4cos^2 f−3 2sinf=4(1−sin^2 f)−3 2sinf=4−4sin^2 f−3 4sin^2 f+2sinf−1=0 ∴sinf=((−2±(√(4−4×4×(−1))))/8) sinf=((−2±2(√5))/8)=((−1±(√5))/4). 0<f<90°⇒sinf>0⇒sinf≠((−1−(√5))/4). ∴ sinf=((−1+(√5))/4) cos^2 f=1−sin^2 f cos^2 f=1−((((√5)−1)/4))^2 =((16−5−1+2(√5))/(16)) cos^2 f=((10+2(√5))/(16))=((5+(√5))/8) ⇒2cos^2 f=((5+(√5))/4)⇒2cos^2 f−1=((1+(√5))/4). But, cos2f=cos36°=2cos^2 f−1. ∴cos36°=((1+(√5))/4)⇒2cos36°=((1+(√5))/2). Hence (D/s)=((1+(√5))/2) ■

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

Let A,B,C are any three consecutive vertics, then AB^(−) ,BC^(−) are sides of pentagon and AC^(−) is its diagonal Now let mAB^(−) =mBC^(−) =s and mAC^(−) =d We have to prove d:s=(((√5)+1)/2):1 or (d/s)=(((√5)+1)/2) We know that m∠ABC=108° By cosine law d=(√(s^2 +s^2 −2(s)(s) cos 108°)) (d/s)=((√(s^2 +s^2 −2(s)(s) cos 108°))/s) =((s(√(2−2cos 108°)))/s)=(√(2−2cos 108°)) =(√(2−2(((1−(√5))/4))))=(√((8−2+2(√5))/4)) =((√(6+2(√5)))/2) Now we have to change the numerator into a+b(√5) form,where a ,b are rationals So let (√(6+2(√5))) =a+b(√5) 6+2(√5)=(a+b(√5))^2 =a^2 +5b^2 +2ab(√5) a^2 +5b^2 =6 ∧ 2ab=2⇒a=(1/b) a^2 +5b^2 =6⇒((1/b))^2 +5b^2 =6 ⇒1+5b^4 =6b^2 ⇒5b^4 −6b^2 +1=0 ⇒b^2 =((6±(√(36−20)))/(10))=((6±4)/(10))=1,(1/5) ⇒b=±1,±(1/( (√5))) [discarding as its irrational] b=±1⇒a=±1 a+b(√5)=1+(√5),1−(√5),−1+(√5),−1−(√5) Last three are extraneous(?) ((√(6+2(√5)))/2)=((1+(√5))/2)

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

Let A,B,C are three consecutive vertices of a pentagon.Then AB and BC are sides and AC is diagonal of pentagon. m∠ABC=108° △ABC is an issoscel triangle in which let AB=BC=s, AC=d m∠ABC=108° ,m∠CAB=m∠BCA=36° By sine law (d/(sin 108°))=(s/(36°))⇒(d/s)=((sin 108°)/(sin 36°)) =((((√(5+(√5))))/(((√(5−(√5)))))×((((√(5+(√5))))/(((√(5+(√5)))))=((5+(√5))/( (√(25−5))))=((5+(√5))/(2(√5))) =(5/(2(√5)))+((√5)/(2(√5)))=((√5)/2)+(1/2)= (d/s)=((1+(√5))/2)

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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