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


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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

	$You interest and have knowledge in so many$ $areas of mathematics!!!$ $GREAT!$
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