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Question Number 3348 by RasheedSindhi last updated on 11/Dec/15
Prove that the regular pentagon is possible with ruler and compass.
$${Prove}\:{that}\:{the}\:{regular}\:{pentagon} \\ $$$${is}\:{possible}\:{with}\:{ruler}\:{and}\:{compass}. \\ $$
Commented by Rasheed Soomro last updated on 11/Dec/15
How can we prove that angle 108° is possible?
$$\mathcal{H}{ow}\:{can}\:{we}\:{prove}\:{that}\:{angle}\:\mathrm{108}°\:{is} \\ $$$${possible}? \\ $$
Commented by prakash jain last updated on 11/Dec/15
sin 5x=sin (4x+x)= sin 4xcos x+cos 4xsin x =2sin 2xcos 2xcos x+(1−2sin^2 2x)sin x =4sin xcos^2 x(1−2sin^2 x)+(1−8sin^2 xcos^2 x)sin x =4sin x(1−sin^2 x)(1−2sin^2 x)+(1−8sin^2 x+8sin^4 x)sin x =4sinx −12sin^3 x+8sin^5 x+sin x−8sin^3 x+8sin^5 x =16sin^5 x−20sin^3 x+5sin x x=18° 5x=90 sin 18°=y 16y^5 −20y^3 +5y−1=0 y=1 satisfies 16y^5 −16y^4 +16y^4 −16y^3 −4y^3 +4y^2 −4y^2 +4y+y−1=0 (y−1)(16y^4 +16y^3 −4y^2 −4y+1)=0 (y−1)[4y^2 (4y^2 +4y+1)−8y^2 −4y+1]=0 (y−1)[4y^2 (2y+1)^2 −4y(2y+1)+1]=0 (y−1)(2y(2y+1)−1)^2 =0 (y−1)(4y^2 +2y−1)^2 =0 y=1 or y=((−2±(√(4+16)))/8)=((−1±(√5))/4) sin 18>0 so sin 18°=(((√5)−1)/4)
$$\mathrm{sin}\:\mathrm{5}{x}=\mathrm{sin}\:\left(\mathrm{4}{x}+{x}\right)= \\ $$$$\mathrm{sin}\:\mathrm{4}{x}\mathrm{cos}\:{x}+\mathrm{cos}\:\mathrm{4}{x}\mathrm{sin}\:{x} \\ $$$$=\mathrm{2sin}\:\mathrm{2}{x}\mathrm{cos}\:\mathrm{2}{x}\mathrm{cos}\:{x}+\left(\mathrm{1}−\mathrm{2sin}^{\mathrm{2}} \mathrm{2}{x}\right)\mathrm{sin}\:{x} \\ $$$$=\mathrm{4sin}\:{x}\mathrm{cos}^{\mathrm{2}} {x}\left(\mathrm{1}−\mathrm{2sin}^{\mathrm{2}} {x}\right)+\left(\mathrm{1}−\mathrm{8sin}^{\mathrm{2}} {x}\mathrm{cos}^{\mathrm{2}} {x}\right)\mathrm{sin}\:{x} \\ $$$$=\mathrm{4sin}\:{x}\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} {x}\right)\left(\mathrm{1}−\mathrm{2sin}^{\mathrm{2}} {x}\right)+\left(\mathrm{1}−\mathrm{8sin}^{\mathrm{2}} {x}+\mathrm{8sin}^{\mathrm{4}} {x}\right)\mathrm{sin}\:{x} \\ $$$$=\mathrm{4sin}{x}\:−\mathrm{12sin}^{\mathrm{3}} {x}+\mathrm{8sin}^{\mathrm{5}} {x}+\mathrm{sin}\:{x}−\mathrm{8sin}^{\mathrm{3}} {x}+\mathrm{8sin}^{\mathrm{5}} {x} \\ $$$$=\mathrm{16sin}^{\mathrm{5}} {x}−\mathrm{20sin}^{\mathrm{3}} {x}+\mathrm{5sin}\:{x} \\ $$$${x}=\mathrm{18}° \\ $$$$\mathrm{5}{x}=\mathrm{90} \\ $$$$\mathrm{sin}\:\mathrm{18}°={y} \\ $$$$\mathrm{16}{y}^{\mathrm{5}} −\mathrm{20}{y}^{\mathrm{3}} +\mathrm{5}{y}−\mathrm{1}=\mathrm{0} \\ $$$${y}=\mathrm{1}\:{satisfies} \\ $$$$\mathrm{16}{y}^{\mathrm{5}} −\mathrm{16}{y}^{\mathrm{4}} +\mathrm{16}{y}^{\mathrm{4}} −\mathrm{16}{y}^{\mathrm{3}} −\mathrm{4}{y}^{\mathrm{3}} +\mathrm{4}{y}^{\mathrm{2}} −\mathrm{4}{y}^{\mathrm{2}} +\mathrm{4}{y}+{y}−\mathrm{1}=\mathrm{0} \\ $$$$\left({y}−\mathrm{1}\right)\left(\mathrm{16}{y}^{\mathrm{4}} +\mathrm{16}{y}^{\mathrm{3}} −\mathrm{4}{y}^{\mathrm{2}} −\mathrm{4}{y}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\left({y}−\mathrm{1}\right)\left[\mathrm{4}{y}^{\mathrm{2}} \left(\mathrm{4}{y}^{\mathrm{2}} +\mathrm{4}{y}+\mathrm{1}\right)−\mathrm{8}{y}^{\mathrm{2}} −\mathrm{4}{y}+\mathrm{1}\right]=\mathrm{0} \\ $$$$\left({y}−\mathrm{1}\right)\left[\mathrm{4}{y}^{\mathrm{2}} \left(\mathrm{2}{y}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}{y}\left(\mathrm{2}{y}+\mathrm{1}\right)+\mathrm{1}\right]=\mathrm{0} \\ $$$$\left({y}−\mathrm{1}\right)\left(\mathrm{2}{y}\left(\mathrm{2}{y}+\mathrm{1}\right)−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\left({y}−\mathrm{1}\right)\left(\mathrm{4}{y}^{\mathrm{2}} +\mathrm{2}{y}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$${y}=\mathrm{1}\:{or}\:{y}=\frac{−\mathrm{2}\pm\sqrt{\mathrm{4}+\mathrm{16}}}{\mathrm{8}}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{4}} \\ $$$$\mathrm{sin}\:\mathrm{18}>\mathrm{0}\:\mathrm{so}\:\mathrm{sin}\:\mathrm{18}°=\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}} \\ $$
Commented by prakash jain last updated on 11/Dec/15
Angle=108°
$$\mathrm{Angle}=\mathrm{108}° \\ $$
Answered by prakash jain last updated on 11/Dec/15
As shown in comments sin 18°=(((√5)−1)/4) With ruler and compass: We can draw (√5). So we can draw (√5) −1 we can draw 4. we can draw right angle So we can create a right angle △ABC with AB=(√5) −1 ∠ABC=90 AC=4 Then sin∠ACB=(((√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.
$$\mathrm{As}\:\mathrm{shown}\:\mathrm{in}\:\mathrm{comments} \\ $$$$\mathrm{sin}\:\mathrm{18}°=\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}} \\ $$$$\mathrm{With}\:\mathrm{ruler}\:\mathrm{and}\:\mathrm{compass}: \\ $$$$\mathrm{We}\:\mathrm{can}\:\mathrm{draw}\:\sqrt{\mathrm{5}}.\:\mathrm{So}\:\mathrm{we}\:\mathrm{can}\:\mathrm{draw}\:\sqrt{\mathrm{5}}\:−\mathrm{1} \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{draw}\:\mathrm{4}. \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{draw}\:\mathrm{right}\:\mathrm{angle} \\ $$$$\mathrm{So}\:\mathrm{we}\:\mathrm{can}\:\mathrm{create}\:\mathrm{a}\:\mathrm{right}\:\mathrm{angle}\:\bigtriangleup\mathrm{ABC}\:\mathrm{with} \\ $$$$\mathrm{AB}=\sqrt{\mathrm{5}}\:−\mathrm{1}\:\angle\mathrm{ABC}=\mathrm{90}\:\mathrm{AC}=\mathrm{4} \\ $$$$\mathrm{Then}\:\mathrm{sin}\angle\mathrm{ACB}=\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}} \\ $$$$\mathrm{so}\:\angle\mathrm{C}=\mathrm{18}° \\ $$$$\mathrm{Since}\:\mathrm{we}\:\mathrm{draw}\:\mathrm{90}°\:\mathrm{and}\:\mathrm{18}°\:\mathrm{with}\:\mathrm{ruler}\:\mathrm{and} \\ $$$$\mathrm{compass}\:\mathrm{hence}\:\mathrm{we}\:\mathrm{can}\:\mathrm{draw}\:\mathrm{required}\:\mathrm{108}°. \\ $$$$\mathrm{So}\:\mathrm{a}\:\mathrm{regular}\:\mathrm{pentagon}\:\mathrm{can}\:\mathrm{be}\:\mathrm{drawn}\:\mathrm{with} \\ $$$$\mathrm{ruler}\:\mathrm{and}\:\mathrm{compass}. \\ $$
Commented by prakash jain last updated on 11/Dec/15
sin 18=(((√5)−1)/4),sin^2 18=((6−2(√5))/(16))=((3−(√5))/8) cos^2 18=1−sin^2 18=1−((3−(√5))/8)=((5+(√5))/8) cos 36=cos^2 18−sin^2 18=((2+2(√5))/8)=((1+(√5))/4) For actual construction drawing 36° might be easier. Suppose you want draw pentagon of side 1. draw AB=1 Draw BX =(√5) so AX=1+(√5) Draw ⊥^r to AX at X. Draw an arc with Radius 4cm 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.
$$\mathrm{sin}\:\mathrm{18}=\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}},\mathrm{sin}^{\mathrm{2}} \mathrm{18}=\frac{\mathrm{6}−\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{16}}=\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{8}} \\ $$$$\mathrm{cos}^{\mathrm{2}} \:\mathrm{18}=\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \mathrm{18}=\mathrm{1}−\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{8}}=\frac{\mathrm{5}+\sqrt{\mathrm{5}}}{\mathrm{8}} \\ $$$$\mathrm{cos}\:\mathrm{36}=\mathrm{cos}^{\mathrm{2}} \mathrm{18}−\mathrm{sin}^{\mathrm{2}} \mathrm{18}=\frac{\mathrm{2}+\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{8}}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}} \\ $$$$\mathrm{For}\:\mathrm{actual}\:\mathrm{construction}\:\mathrm{drawing}\:\mathrm{36}°\:\mathrm{might}\:\mathrm{be} \\ $$$$\mathrm{easier}. \\ $$$$\mathrm{Suppose}\:\mathrm{you}\:\mathrm{want}\:\mathrm{draw}\:\mathrm{pentagon}\:\mathrm{of}\:\mathrm{side}\:\mathrm{1}. \\ $$$$\mathrm{draw}\:\mathrm{AB}=\mathrm{1} \\ $$$$\mathrm{Draw}\:\mathrm{BX}\:=\sqrt{\mathrm{5}}\:\mathrm{so}\:\mathrm{AX}=\mathrm{1}+\sqrt{\mathrm{5}} \\ $$$$\mathrm{Draw}\:\bot^{{r}} \:\mathrm{to}\:\mathrm{AX}\:\mathrm{at}\:\mathrm{X}. \\ $$$$\mathrm{Draw}\:\mathrm{an}\:\mathrm{arc}\:\mathrm{with}\:\mathrm{Radius}\:\mathrm{4cm}\:\mathrm{with}\:\mathrm{center}\:\mathrm{at} \\ $$$$\mathrm{A}.\:\mathrm{Say}\:\mathrm{it}\:\mathrm{cuts}\:\mathrm{AX}\:\mathrm{at}\:\mathrm{Y}. \\ $$$$\angle\mathrm{YAX}=\mathrm{36}° \\ $$$$\mathrm{draw}\:\mathrm{line}\:\mathrm{joining}\:\mathrm{AY} \\ $$$$\mathrm{Draw}\:\mathrm{an}\:\mathrm{Arc}\:\mathrm{of}\:\mathrm{radius}\:\mathrm{1}\:\mathrm{from}\:\mathrm{B}.\:\mathrm{Say}\:\mathrm{it}\:\mathrm{cuts} \\ $$$$\mathrm{AY}\:\mathrm{at}\:\mathrm{C}. \\ $$$$\mathrm{join}\:\mathrm{BC}.\:\angle\mathrm{ABC}=\mathrm{108}° \\ $$$$\mathrm{AB}\:\mathrm{and}\:\mathrm{BC}\:\mathrm{are}\:\mathrm{2}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{pentagon}. \\ $$$$\mathrm{Rest}\:\mathrm{of}\:\mathrm{sides}\:\mathrm{can}\:\mathrm{be}\:\mathrm{drawn}\:\mathrm{by}\:\mathrm{repeating}\:\mathrm{ABC} \\ $$$$\mathrm{since}\:\mathrm{all}\:\mathrm{3}\:\mathrm{sides}\:\mathrm{are}\:\mathrm{known}.\: \\ $$
Commented by Rasheed Soomro last updated on 13/Dec/15
You interest and have knowledge in so many areas of mathematics !!! GREAT !
$$\boldsymbol{\mathrm{You}}\:\boldsymbol{\mathrm{interest}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{have}}\:\boldsymbol{\mathrm{knowledge}}\:\boldsymbol{\mathrm{in}}\:\boldsymbol{\mathrm{so}}\:\boldsymbol{\mathrm{many}}\: \\ $$$$\boldsymbol{\mathrm{areas}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{mathematics}}\:!!!\: \\ $$$$\mathcal{GREAT}\:! \\ $$

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