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




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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