Prove-that-the-regular-pentagon-is-possible-with-ruler-and-compass-

Question Number 3348 by RasheedSindhi last updated on 11/Dec/15

P r o v e t h a t t h e r e g u l a r p e n t a g o n

i s p o s s i b l e w i t h r u l e r a n d c o m p a s s .

Commented by Rasheed Soomro last updated on 11/Dec/15

H o w c a n w e p r o v e t h a t a n g l e 108 ° i s

p o s s i b l e ?

Commented by prakash jain last updated on 11/Dec/15

\sin 5 x = \sin (4 x + x) =

\sin 4 x \cos x + \cos 4 x \sin x

= 2 \sin 2 x \cos 2 x \cos x + (1 - {2 \sin}^{2} 2 x) \sin x

= 4 \sin x \cos^{2} x (1 - {2 \sin}^{2} x) + (1 - {8 \sin}^{2} x \cos^{2} x) \sin x

= 4 \sin x (1 - \sin^{2} x) (1 - {2 \sin}^{2} x) + (1 - {8 \sin}^{2} x + {8 \sin}^{4} x) \sin x

= 4 \sin x - {12 \sin}^{3} x + {8 \sin}^{5} x + \sin x - {8 \sin}^{3} x + {8 \sin}^{5} x

= {16 \sin}^{5} x - {20 \sin}^{3} x + 5 \sin x

x = 18 °

5 x = 90

\sin 18 ° = y

16 y^{5} - 20 y^{3} + 5 y - 1 = 0

y = 1 s a t i s f i e s

16 y^{5} - 16 y^{4} + 16 y^{4} - 16 y^{3} - 4 y^{3} + 4 y^{2} - 4 y^{2} + 4 y + y - 1 = 0

(y - 1) (16 y^{4} + 16 y^{3} - 4 y^{2} - 4 y + 1) = 0

(y - 1) [4 y^{2} (4 y^{2} + 4 y + 1) - 8 y^{2} - 4 y + 1] = 0

(y - 1) [4 y^{2} {(2 y + 1)}^{2} - 4 y (2 y + 1) + 1] = 0

(y - 1) {(2 y (2 y + 1) - 1)}^{2} = 0

(y - 1) {(4 y^{2} + 2 y - 1)}^{2} = 0

y = 1 o r y = \frac{- 2 \pm \sqrt{4 + 16}}{8} = \frac{- 1 \pm \sqrt{5}}{4}

\sin 18 > 0 so \sin 18 ° = \frac{\sqrt{5} - 1}{4}

Commented by prakash jain last updated on 11/Dec/15

Angle = 108 °

Answered by prakash jain last updated on 11/Dec/15

As shown in comments

\sin 18 ° = \frac{\sqrt{5} - 1}{4}

With ruler and compass :

We can draw \sqrt{5} . So we can draw \sqrt{5} - 1

we can draw 4 .

we can draw right angle

So we can create a right angle △ ABC with

AB = \sqrt{5} - 1 ∠ ABC = 90 AC = 4

Then \sin ∠ ACB = \frac{\sqrt{5} - 1}{4}

so ∠ C = 18 °

Since we draw 90 ° and 18 ° with ruler and

compass hence we can draw required 108 ° .

So a regular pentagon can be drawn with

ruler and compass .

Commented by prakash jain last updated on 11/Dec/15

\sin 18 = \frac{\sqrt{5} - 1}{4}, \sin^{2} 18 = \frac{6 - 2 \sqrt{5}}{16} = \frac{3 - \sqrt{5}}{8}

\cos^{2} 18 = 1 - \sin^{2} 18 = 1 - \frac{3 - \sqrt{5}}{8} = \frac{5 + \sqrt{5}}{8}

\cos 36 = \cos^{2} 18 - \sin^{2} 18 = \frac{2 + 2 \sqrt{5}}{8} = \frac{1 + \sqrt{5}}{4}

For actual construction drawing 36 ° might be

easier .

Suppose you want draw pentagon of side 1 .

draw AB = 1

Draw BX = \sqrt{5} so AX = 1 + \sqrt{5}

Draw ⊥^{r} to AX at X .

Draw an arc with Radius 4 cm with center at

A . Say it cuts AX at Y .

∠ YAX = 36 °

draw line joining AY

Draw an Arc of radius 1 from B . Say it cuts

AY at C .

join BC . ∠ ABC = 108 °

AB and BC are 2 sides of pentagon .

Rest of sides can be drawn by repeating ABC

since all 3 sides are known .

Commented by Rasheed Soomro last updated on 13/Dec/15

You interest and have knowledge in so many

areas of mathematics!!!

GREAT!