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Related-to-Q16140-What-if-the-three-lines-d-1-d-2-d-3-are-not-parallel-but-concurrent-

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Question Number 16302 by mrW1 last updated on 20/Jun/17
Related to Q16140 What if the three lines d_1 ,d_2 ,d_3 are not parallel, but concurrent?
$$\mathrm{Related}\:\mathrm{to}\:\mathrm{Q16140} \\ $$$$\mathrm{What}\:\mathrm{if}\:\mathrm{the}\:\mathrm{three}\:\mathrm{lines}\:\mathrm{d}_{\mathrm{1}} ,\mathrm{d}_{\mathrm{2}} ,\mathrm{d}_{\mathrm{3}} \:\mathrm{are} \\ $$$$\mathrm{not}\:\mathrm{parallel},\:\mathrm{but}\:\mathrm{concurrent}? \\ $$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 20/Jun/17
where is ′b′?
$${where}\:{is}\:'{b}'? \\ $$
Commented by mrW1 last updated on 20/Jun/17
Commented by mrW1 last updated on 20/Jun/17
positions of lines are given by α and β. position of point A is given by a. position of B and C is to determine. side length of equilateral triangle ΔABC is to determine.
$$\mathrm{positions}\:\mathrm{of}\:\mathrm{lines}\:\mathrm{are}\:\mathrm{given}\:\mathrm{by}\:\alpha\:\mathrm{and}\:\beta. \\ $$$$\mathrm{position}\:\mathrm{of}\:\mathrm{point}\:\mathrm{A}\:\mathrm{is}\:\mathrm{given}\:\mathrm{by}\:\mathrm{a}. \\ $$$$\mathrm{position}\:\mathrm{of}\:\mathrm{B}\:\mathrm{and}\:\mathrm{C}\:\mathrm{is}\:\mathrm{to}\:\mathrm{determine}. \\ $$$$\mathrm{side}\:\mathrm{length}\:\mathrm{of}\:\mathrm{equilateral}\:\mathrm{triangle} \\ $$$$\Delta\mathrm{ABC}\:\mathrm{is}\:\mathrm{to}\:\mathrm{determine}. \\ $$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 20/Jun/17
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 20/Jun/17
Answered by ajfour last updated on 21/Jun/17
side length of equilateral ΔABC is given by (where m_1 =tan 𝛂 , and m_2 =tan 𝛃 ) s=((2a(√(m_1 ^2 m_2 ^2 +m_1 ^2 +m_2 ^2 +m_1 m_2 −(√3)m_1 m_2 (m_1 +m_2 ))))/( (√3)(1−m_1 m_2 )+(m_1 +m_2 ))) .
$${side}\:{length}\:{of}\:{equilateral}\:\Delta{ABC} \\ $$$${is}\:{given}\:{by}\:\left({where}\:\boldsymbol{{m}}_{\mathrm{1}} =\mathrm{tan}\:\boldsymbol{\alpha}\:,\right. \\ $$$$\left.\:{and}\:\:\boldsymbol{{m}}_{\mathrm{2}} =\mathrm{tan}\:\boldsymbol{\beta}\:\right) \\ $$$$\:\boldsymbol{{s}}=\frac{\mathrm{2}\boldsymbol{{a}}\sqrt{\boldsymbol{{m}}_{\mathrm{1}} ^{\mathrm{2}} \boldsymbol{{m}}_{\mathrm{2}} ^{\mathrm{2}} +\boldsymbol{{m}}_{\mathrm{1}} ^{\mathrm{2}} +\boldsymbol{{m}}_{\mathrm{2}} ^{\mathrm{2}} +\boldsymbol{{m}}_{\mathrm{1}} \boldsymbol{{m}}_{\mathrm{2}} −\sqrt{\mathrm{3}}\boldsymbol{{m}}_{\mathrm{1}} \boldsymbol{{m}}_{\mathrm{2}} \left(\boldsymbol{{m}}_{\mathrm{1}} +\boldsymbol{{m}}_{\mathrm{2}} \right)}}{\:\sqrt{\mathrm{3}}\left(\mathrm{1}−\boldsymbol{{m}}_{\mathrm{1}} \boldsymbol{{m}}_{\mathrm{2}} \right)+\left(\boldsymbol{{m}}_{\mathrm{1}} +\boldsymbol{{m}}_{\mathrm{2}} \right)}\:. \\ $$
Commented by ajfour last updated on 20/Jun/17
Commented by ajfour last updated on 20/Jun/17
A is taken on the right. ∠OAB =θ , ∠OAC = π/3−θ tan α = ((s sin θ)/(a−s cos θ)) = m_1 ...(i) tan β = ((s sin (π/3−θ))/(a−s cos (π/3−θ))) = m_2 ...(ii) from (i): s sin θ=am_1 −sm_1 cos θ s=((am_1 )/(sin θ+m_1 cos θ)) ...=(iii) from (ii): s=((am_2 )/(sin (π/3−θ)+m_2 cos (π/3−θ))) ....(iv) we shall find s from (iii) when we find 𝛉 . And to find 𝛉 we shall equate (iii) and (iv) : m_1 [sin (π/3−θ)+m_2 cos (π/3−θ)] =m_2 [sin θ+m_1 cos θ] m_1 [((√3)/2)cos θ−((sin θ)/2)]+m_1 m_2 [((cos θ)/2)+(((√3)sin θ)/2)] =m_2 sin θ+m_1 m_2 cos θ (cos θ)[(((√3)m_1 )/2) − ((m_1 m_2 )/2)]= (sin θ)[(m_1 /2)−((m_1 m_2 (√3))/2)+m_2 ] ((sin θ)/(cos θ))=((m_1 ((√3)−m_2 ))/(m_1 +2m_2 −(√3)m_1 m_2 )) = (p/b) From (iii) s=((am_1 )/(sin 𝛉+m_1 cos 𝛉)) = ((am_1 (√(p^2 +b^2 )))/(p+m_1 b)) =((am_1 (√(p^2 +b^2 )))/(m_1 ((√3)−m_2 )+m_1 (m_1 +2m_2 −(√3)m_1 m_2 ))) =((a (√(p^2 +b^2 )))/( (√3)(1−m_1 m_2 )+(m_1 +m_2 ))) (√(p^2 +b^2 )) comes out to be =2(√((m_1 ^2 m_2 ^2 +m_1 ^2 +m_2 ^2 +m_1 m_2 −(√3)m_1 m_2 (m_1 +m_2 ))) s =((2a(√(m_1 ^2 m_2 ^2 +m_1 ^2 +m_2 ^2 +m_1 m_2 −(√3)m_1 m_2 (m_1 +m_2 ))))/( (√3)(1−m_1 m_2 )+(m_1 +m_2 ))) To locate B and C we need only to draw AB and AC at angles 𝛉 and (𝛑/3−𝛉 ) with AO respectively from point A. Then we join BC. 𝛉=tan^(−1) (((m_1 ((√3)−m_2 ))/(m_1 +2m_2 −(√3)m_1 m_2 ))) The ΔABC is now in place and equilateral .
$${A}\:{is}\:{taken}\:{on}\:{the}\:{right}. \\ $$$$\:\angle{OAB}\:=\theta\:,\:\angle{OAC}\:=\:\pi/\mathrm{3}−\theta \\ $$$$\:\:\mathrm{tan}\:\alpha\:=\:\frac{{s}\:\mathrm{sin}\:\theta}{{a}−{s}\:\mathrm{cos}\:\theta}\:\:=\:{m}_{\mathrm{1}} \:\:\:\:…\left({i}\right) \\ $$$$\:\:\mathrm{tan}\:\beta\:=\:\frac{{s}\:\mathrm{sin}\:\left(\pi/\mathrm{3}−\theta\right)}{{a}−{s}\:\mathrm{cos}\:\left(\pi/\mathrm{3}−\theta\right)}\:=\:{m}_{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…\left({ii}\right) \\ $$$${from}\:\left({i}\right): \\ $$$$\:\:{s}\:\mathrm{sin}\:\theta={am}_{\mathrm{1}} −{sm}_{\mathrm{1}} \mathrm{cos}\:\theta \\ $$$$\:\:{s}=\frac{{am}_{\mathrm{1}} }{\mathrm{sin}\:\theta+{m}_{\mathrm{1}} \mathrm{cos}\:\theta}\:\:\:\:\:\:…=\left({iii}\right) \\ $$$$\:\:{from}\:\left({ii}\right): \\ $$$$\:\:{s}=\frac{{am}_{\mathrm{2}} }{\mathrm{sin}\:\left(\pi/\mathrm{3}−\theta\right)+{m}_{\mathrm{2}} \mathrm{cos}\:\left(\pi/\mathrm{3}−\theta\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:….\left({iv}\right) \\ $$$${we}\:{shall}\:{find}\:\boldsymbol{{s}}\:{from}\:\left({iii}\right)\:{when} \\ $$$${we}\:{find}\:\boldsymbol{\theta}\:.\:{And}\:{to}\:{find}\:\boldsymbol{\theta}\:{we}\:{shall} \\ $$$$\:{equate}\:\left({iii}\right)\:{and}\:\left({iv}\right)\:: \\ $$$${m}_{\mathrm{1}} \left[\mathrm{sin}\:\left(\pi/\mathrm{3}−\theta\right)+{m}_{\mathrm{2}} \mathrm{cos}\:\left(\pi/\mathrm{3}−\theta\right)\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={m}_{\mathrm{2}} \left[\mathrm{sin}\:\theta+{m}_{\mathrm{1}} \mathrm{cos}\:\theta\right] \\ $$$${m}_{\mathrm{1}} \left[\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{cos}\:\theta−\frac{\mathrm{sin}\:\theta}{\mathrm{2}}\right]+{m}_{\mathrm{1}} {m}_{\mathrm{2}} \left[\frac{\mathrm{cos}\:\theta}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}\mathrm{sin}\:\theta}{\mathrm{2}}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={m}_{\mathrm{2}} \mathrm{sin}\:\theta+{m}_{\mathrm{1}} {m}_{\mathrm{2}} \mathrm{cos}\:\theta \\ $$$$\left(\mathrm{cos}\:\theta\right)\left[\frac{\sqrt{\mathrm{3}}{m}_{\mathrm{1}} }{\mathrm{2}}\:−\:\frac{{m}_{\mathrm{1}} {m}_{\mathrm{2}} }{\mathrm{2}}\right]= \\ $$$$\:\:\:\:\:\:\:\:\:\left(\mathrm{sin}\:\theta\right)\left[\frac{{m}_{\mathrm{1}} }{\mathrm{2}}−\frac{{m}_{\mathrm{1}} {m}_{\mathrm{2}} \sqrt{\mathrm{3}}}{\mathrm{2}}+{m}_{\mathrm{2}} \right] \\ $$$$\frac{\mathrm{sin}\:\theta}{\mathrm{cos}\:\theta}=\frac{{m}_{\mathrm{1}} \left(\sqrt{\mathrm{3}}−{m}_{\mathrm{2}} \right)}{{m}_{\mathrm{1}} +\mathrm{2}{m}_{\mathrm{2}} −\sqrt{\mathrm{3}}{m}_{\mathrm{1}} {m}_{\mathrm{2}} }\:=\:\frac{\boldsymbol{{p}}}{\boldsymbol{{b}}} \\ $$$${From}\:\left({iii}\right) \\ $$$$\:\:\boldsymbol{{s}}=\frac{\boldsymbol{{am}}_{\mathrm{1}} }{\mathrm{sin}\:\boldsymbol{\theta}+\boldsymbol{{m}}_{\mathrm{1}} \mathrm{cos}\:\boldsymbol{\theta}} \\ $$$$\:\:\:\:=\:\frac{\boldsymbol{{am}}_{\mathrm{1}} \sqrt{\boldsymbol{{p}}^{\mathrm{2}} +\boldsymbol{{b}}^{\mathrm{2}} }}{\boldsymbol{{p}}+\boldsymbol{{m}}_{\mathrm{1}} \boldsymbol{{b}}} \\ $$$$\:\:\:=\frac{\boldsymbol{{am}}_{\mathrm{1}} \sqrt{\boldsymbol{{p}}^{\mathrm{2}} +\boldsymbol{{b}}^{\mathrm{2}} }}{\boldsymbol{{m}}_{\mathrm{1}} \left(\sqrt{\mathrm{3}}−\boldsymbol{{m}}_{\mathrm{2}} \right)+\boldsymbol{{m}}_{\mathrm{1}} \left(\boldsymbol{{m}}_{\mathrm{1}} +\mathrm{2}\boldsymbol{{m}}_{\mathrm{2}} −\sqrt{\mathrm{3}}\boldsymbol{{m}}_{\mathrm{1}} \boldsymbol{{m}}_{\mathrm{2}} \right)} \\ $$$$=\frac{{a}\:\sqrt{{p}^{\mathrm{2}} +{b}^{\mathrm{2}} }}{\:\sqrt{\mathrm{3}}\left(\mathrm{1}−{m}_{\mathrm{1}} {m}_{\mathrm{2}} \right)+\left({m}_{\mathrm{1}} +{m}_{\mathrm{2}} \right)} \\ $$$$\:\sqrt{\boldsymbol{{p}}^{\mathrm{2}} +\boldsymbol{{b}}^{\mathrm{2}} }\:{comes}\:{out}\:{to}\:{be} \\ $$$$ \\ $$$$\:\:\:\:\:\:=\mathrm{2}\sqrt{\left({m}_{\mathrm{1}} ^{\mathrm{2}} {m}_{\mathrm{2}} ^{\mathrm{2}} +{m}_{\mathrm{1}} ^{\mathrm{2}} +{m}_{\mathrm{2}} ^{\mathrm{2}} +{m}_{\mathrm{1}} {m}_{\mathrm{2}} −\sqrt{\mathrm{3}}{m}_{\mathrm{1}} {m}_{\mathrm{2}} \left({m}_{\mathrm{1}} +{m}_{\mathrm{2}} \right)\right.}\: \\ $$$$ \\ $$$$\:\boldsymbol{{s}}\:=\frac{\mathrm{2}\boldsymbol{{a}}\sqrt{\boldsymbol{{m}}_{\mathrm{1}} ^{\mathrm{2}} \boldsymbol{{m}}_{\mathrm{2}} ^{\mathrm{2}} +\boldsymbol{{m}}_{\mathrm{1}} ^{\mathrm{2}} +\boldsymbol{{m}}_{\mathrm{2}} ^{\mathrm{2}} +\boldsymbol{{m}}_{\mathrm{1}} \boldsymbol{{m}}_{\mathrm{2}} −\sqrt{\mathrm{3}}\boldsymbol{{m}}_{\mathrm{1}} \boldsymbol{{m}}_{\mathrm{2}} \left(\boldsymbol{{m}}_{\mathrm{1}} +\boldsymbol{{m}}_{\mathrm{2}} \right)}}{\:\sqrt{\mathrm{3}}\left(\mathrm{1}−\boldsymbol{{m}}_{\mathrm{1}} \boldsymbol{{m}}_{\mathrm{2}} \right)+\left(\boldsymbol{{m}}_{\mathrm{1}} +\boldsymbol{{m}}_{\mathrm{2}} \right)} \\ $$$$\boldsymbol{{T}}{o}\:{locate}\:{B}\:{and}\:{C}\:{we}\:{need}\:{only} \\ $$$${to}\:{draw}\:{AB}\:{and}\:{AC}\:{at}\:{angles} \\ $$$$\:\boldsymbol{\theta}\:{and}\:\left(\boldsymbol{\pi}/\mathrm{3}−\boldsymbol{\theta}\:\right)\:{with}\:{AO}\:{respectively} \\ $$$${from}\:{point}\:{A}.\:{Then}\:{we}\:{join}\:{BC}. \\ $$$$\:\boldsymbol{\theta}=\mathrm{tan}^{−\mathrm{1}} \left(\frac{\boldsymbol{{m}}_{\mathrm{1}} \left(\sqrt{\mathrm{3}}−\boldsymbol{{m}}_{\mathrm{2}} \right)}{\boldsymbol{{m}}_{\mathrm{1}} +\mathrm{2}\boldsymbol{{m}}_{\mathrm{2}} −\sqrt{\mathrm{3}}\boldsymbol{{m}}_{\mathrm{1}} \boldsymbol{{m}}_{\mathrm{2}} }\right) \\ $$$$\boldsymbol{{T}}{he}\:\Delta{ABC}\:{is}\:{now}\:{in}\:{place}\:{and} \\ $$$$\:{equilateral}\:. \\ $$$$\: \\ $$
Commented by ajfour last updated on 20/Jun/17
Is it correct Sir ? I checked it only for α=π/6 and β=0 . It answers well; equal to a/2 (that is correct i think). if α=π/4 , β=π/4 even then it answers well. s= a((√3)−1) =((2a)/( (√3)+1)) = a (√(4−2(√3))) And θ=π/6 .(kindly refer to the diagram of my solution when checking these) .
$${Is}\:{it}\:{correct}\:{Sir}\:? \\ $$$${I}\:{checked}\:{it}\:{only}\:{for}\:\alpha=\pi/\mathrm{6} \\ $$$$\:{and}\:\beta=\mathrm{0}\:.\:{It}\:{answers}\:{well};\: \\ $$$${equal}\:{to}\:{a}/\mathrm{2}\:\left({that}\:{is}\:{correct}\:{i}\:{think}\right). \\ $$$${if}\:\alpha=\pi/\mathrm{4}\:,\:\beta=\pi/\mathrm{4}\:{even}\:{then} \\ $$$${it}\:{answers}\:{well}.\:\:{s}=\:{a}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right) \\ $$$$\:\:\:\:\:=\frac{\mathrm{2}{a}}{\:\sqrt{\mathrm{3}}+\mathrm{1}}\:=\:{a}\:\sqrt{\mathrm{4}−\mathrm{2}\sqrt{\mathrm{3}}}\:\: \\ $$$${And}\:\theta=\pi/\mathrm{6}\:.\left({kindly}\:{refer}\:{to}\:{the}\right. \\ $$$$\:{diagram}\:{of}\:{my}\:{solution}\:{when} \\ $$$$\left.{checking}\:{these}\right)\:. \\ $$
Commented by mrW1 last updated on 20/Jun/17
I have no own calculation, so I can not compare. Your calculation seems to be correct. I′ll check the result later. Generally there are two such triangles, one to left side and one to right side of point A.
$$\mathrm{I}\:\mathrm{have}\:\mathrm{no}\:\mathrm{own}\:\mathrm{calculation},\:\mathrm{so}\:\mathrm{I}\:\mathrm{can} \\ $$$$\mathrm{not}\:\mathrm{compare}. \\ $$$$\mathrm{Your}\:\mathrm{calculation}\:\:\mathrm{seems}\:\mathrm{to}\:\mathrm{be}\:\mathrm{correct}. \\ $$$$\mathrm{I}'\mathrm{ll}\:\mathrm{check}\:\mathrm{the}\:\mathrm{result}\:\mathrm{later}. \\ $$$$\mathrm{Generally}\:\mathrm{there}\:\mathrm{are}\:\mathrm{two}\:\mathrm{such}\:\mathrm{triangles}, \\ $$$$\mathrm{one}\:\mathrm{to}\:\mathrm{left}\:\mathrm{side}\:\mathrm{and}\:\mathrm{one}\:\mathrm{to}\:\mathrm{right}\:\mathrm{side} \\ $$$$\mathrm{of}\:\mathrm{point}\:\mathrm{A}. \\ $$
Commented by mrW1 last updated on 21/Jun/17
Following picture shows how to draw the requested triangles only with ruler and compass.
$$\mathrm{Following}\:\mathrm{picture}\:\mathrm{shows}\:\mathrm{how}\:\mathrm{to} \\ $$$$\mathrm{draw}\:\mathrm{the}\:\mathrm{requested}\:\mathrm{triangles}\:\mathrm{only} \\ $$$$\mathrm{with}\:\mathrm{ruler}\:\mathrm{and}\:\mathrm{compass}. \\ $$
Commented by mrW1 last updated on 20/Jun/17
Commented by ajfour last updated on 20/Jun/17
tbank you Sir.
$${tbank}\:{you}\:{Sir}. \\ $$
Commented by mrW1 last updated on 21/Jun/17
How does your formula change for the triangle on the richt side of point A?
$$\mathrm{How}\:\mathrm{does}\:\mathrm{your}\:\mathrm{formula}\:\mathrm{change}\:\mathrm{for} \\ $$$$\mathrm{the}\:\mathrm{triangle}\:\mathrm{on}\:\mathrm{the}\:\mathrm{richt}\:\mathrm{side}\:\mathrm{of}\:\mathrm{point} \\ $$$$\mathrm{A}? \\ $$
Commented by ajfour last updated on 21/Jun/17
s=((2a(√(m_1 ^2 m_2 ^2 +m_1 ^2 +m_2 ^2 +m_1 m_2 +(√3)m_1 m_2 (m_1 +m_2 ))))/( (√3)(1−m_1 m_2 )−(m_1 +m_2 ))) this is it Sir, when △ is on the right side of point A . i understand, then α→ π−α and β → π−β , for the other case.
$${s}=\frac{\mathrm{2}{a}\sqrt{{m}_{\mathrm{1}} ^{\mathrm{2}} {m}_{\mathrm{2}} ^{\mathrm{2}} +{m}_{\mathrm{1}} ^{\mathrm{2}} +{m}_{\mathrm{2}} ^{\mathrm{2}} +{m}_{\mathrm{1}} {m}_{\mathrm{2}} +\sqrt{\mathrm{3}}{m}_{\mathrm{1}} {m}_{\mathrm{2}} \left({m}_{\mathrm{1}} +{m}_{\mathrm{2}} \right)}}{\:\sqrt{\mathrm{3}}\left(\mathrm{1}−{m}_{\mathrm{1}} {m}_{\mathrm{2}} \right)−\left({m}_{\mathrm{1}} +{m}_{\mathrm{2}} \right)} \\ $$$$\:{this}\:{is}\:{it}\:{Sir},\:{when}\:\bigtriangleup\:{is}\:{on}\:{the} \\ $$$${right}\:{side}\:{of}\:{point}\:{A}\:. \\ $$$${i}\:{understand},\:{then}\:\alpha\rightarrow\:\pi−\alpha \\ $$$$\:{and}\:\beta\:\rightarrow\:\pi−\beta\:,\:{for}\:{the}\:{other}\:{case}. \\ $$
Commented by mrW1 last updated on 21/Jun/17
absolutely smart work!
$$\mathrm{absolutely}\:\mathrm{smart}\:\mathrm{work}! \\ $$

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