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Question-150944




Question Number 150944 by ajfour last updated on 16/Aug/21
Commented by mr W last updated on 17/Aug/21
Commented by ajfour last updated on 17/Aug/21
okay find minimum side length, sir.
Commented by mr W last updated on 17/Aug/21
the green one is the smallest   equilateral. i can determine it only  numerically.
$${the}\:{green}\:{one}\:{is}\:{the}\:{smallest}\: \\ $$$${equilateral}.\:{i}\:{can}\:{determine}\:{it}\:{only} \\ $$$${numerically}. \\ $$
Answered by ajfour last updated on 17/Aug/21
sin α=(a/(b−a))  A[(√((b−a)^2 −a^2 )), a]≡[c,a]  let  T(p,q)  let s be circumradius of such an  equilateral triangle.  P[p−scos θ, q+ssin θ]  Q[p+scos (60°−θ), q+ssin (60°−θ)]  R[p+scos (60°+θ), q−ssin (60°+θ)]  now  (p−scos θ−c)^2 +(q+ssin θ−a)^2      = a^2     ...(i)  [p+scos (60°−θ)]^2 +[q+ssin (60°−θ)]^2    = b^2       ...(ii)  q=ssin (60°+θ)  ⇒ (p−scos θ−c)^2     +[((s(√3))/2)cos θ+((3s)/2)sin θ−a]^2 =a^2      ........(I)  [p+(s/2)cos θ+((s(√3))/2)sin θ]^2     +[s(√3)cos θ]^2 =b^2       ....(II)  (II)−(I)  {((3s)/2)cos θ+((s(√3))/2)sin θ+c}  ×{2p−(s/2)cos θ+((s(√3))/2)sin θ−c}  +    {((s(√3))/2)cos θ−((3s)/2)sin θ+a}  ×{((3(√3)s)/2)cos θ+((3s)/2)sin θ−a}    = b^2 −a^2   from here  p=f(s,θ)  now using (I) or (II)  h(s,θ)=0  from here we get s_(min) .  Minimum side length of  such an equilateral △ would  then be s_(min) (√3).
$$\mathrm{sin}\:\alpha=\frac{{a}}{{b}−{a}} \\ $$$${A}\left[\sqrt{\left({b}−{a}\right)^{\mathrm{2}} −{a}^{\mathrm{2}} },\:{a}\right]\equiv\left[{c},{a}\right] \\ $$$${let}\:\:{T}\left({p},{q}\right) \\ $$$${let}\:{s}\:{be}\:{circumradius}\:{of}\:{such}\:{an} \\ $$$${equilateral}\:{triangle}. \\ $$$${P}\left[{p}−{s}\mathrm{cos}\:\theta,\:{q}+{s}\mathrm{sin}\:\theta\right] \\ $$$${Q}\left[{p}+{s}\mathrm{cos}\:\left(\mathrm{60}°−\theta\right),\:{q}+{s}\mathrm{sin}\:\left(\mathrm{60}°−\theta\right)\right] \\ $$$${R}\left[{p}+{s}\mathrm{cos}\:\left(\mathrm{60}°+\theta\right),\:{q}−{s}\mathrm{sin}\:\left(\mathrm{60}°+\theta\right)\right] \\ $$$${now} \\ $$$$\left({p}−{s}\mathrm{cos}\:\theta−{c}\right)^{\mathrm{2}} +\left({q}+{s}\mathrm{sin}\:\theta−{a}\right)^{\mathrm{2}} \\ $$$$\:\:\:=\:{a}^{\mathrm{2}} \:\:\:\:…\left({i}\right) \\ $$$$\left[{p}+{s}\mathrm{cos}\:\left(\mathrm{60}°−\theta\right)\right]^{\mathrm{2}} +\left[{q}+{s}\mathrm{sin}\:\left(\mathrm{60}°−\theta\right)\right]^{\mathrm{2}} \\ $$$$\:=\:{b}^{\mathrm{2}} \:\:\:\:\:\:…\left({ii}\right) \\ $$$${q}={s}\mathrm{sin}\:\left(\mathrm{60}°+\theta\right) \\ $$$$\Rightarrow\:\left({p}−{s}\mathrm{cos}\:\theta−{c}\right)^{\mathrm{2}} \\ $$$$\:\:+\left[\frac{{s}\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{cos}\:\theta+\frac{\mathrm{3}{s}}{\mathrm{2}}\mathrm{sin}\:\theta−{a}\right]^{\mathrm{2}} ={a}^{\mathrm{2}} \\ $$$$\:\:\:……..\left({I}\right) \\ $$$$\left[{p}+\frac{{s}}{\mathrm{2}}\mathrm{cos}\:\theta+\frac{{s}\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{sin}\:\theta\right]^{\mathrm{2}} \\ $$$$\:\:+\left[{s}\sqrt{\mathrm{3}}\mathrm{cos}\:\theta\right]^{\mathrm{2}} ={b}^{\mathrm{2}} \:\:\:\:\:\:….\left({II}\right) \\ $$$$\left({II}\right)−\left({I}\right) \\ $$$$\left\{\frac{\mathrm{3}{s}}{\mathrm{2}}\mathrm{cos}\:\theta+\frac{{s}\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{sin}\:\theta+{c}\right\} \\ $$$$×\left\{\mathrm{2}{p}−\frac{{s}}{\mathrm{2}}\mathrm{cos}\:\theta+\frac{{s}\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{sin}\:\theta−{c}\right\} \\ $$$$+ \\ $$$$\:\:\left\{\frac{{s}\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{cos}\:\theta−\frac{\mathrm{3}{s}}{\mathrm{2}}\mathrm{sin}\:\theta+{a}\right\} \\ $$$$×\left\{\frac{\mathrm{3}\sqrt{\mathrm{3}}{s}}{\mathrm{2}}\mathrm{cos}\:\theta+\frac{\mathrm{3}{s}}{\mathrm{2}}\mathrm{sin}\:\theta−{a}\right\} \\ $$$$\:\:=\:{b}^{\mathrm{2}} −{a}^{\mathrm{2}} \\ $$$${from}\:{here} \\ $$$${p}={f}\left({s},\theta\right) \\ $$$${now}\:{using}\:\left({I}\right)\:{or}\:\left({II}\right) \\ $$$${h}\left({s},\theta\right)=\mathrm{0} \\ $$$${from}\:{here}\:{we}\:{get}\:{s}_{{min}} . \\ $$$${Minimum}\:{side}\:{length}\:{of} \\ $$$${such}\:{an}\:{equilateral}\:\bigtriangleup\:{would} \\ $$$${then}\:{be}\:{s}_{{min}} \sqrt{\mathrm{3}}. \\ $$$$ \\ $$$$ \\ $$
Answered by ajfour last updated on 17/Aug/21
center of semicircle origin.  center of circle   A[(√((b−a)^2 −a^2 )), a]≡(c,a)  let bottom corner be B(p,0)  let side of △ be s.  x_C ^2 +y_C ^2 =b^2   (x_C −p)^2 +y_C ^2 =s^2   subtracting  2px_C −p^2 =b^2 −s^2   x_C =((b^2 −s^2 +p^2 )/(2p))  ((x_c +p)/2)−((s(√3))/2)sin φ=x_D   (y_C /2)+((s(√3))/2)cos φ=y_D   (x_D −c)^2 +(y_D −a)^2 =a^2   cos φ=((x_C −p)/s)  ;  sin φ=(y_C /s)  x_C +p−s(√3)(((√(b^2 −x_C ^2 ))/s))=2x_D   y_C +s(√3)(((x_C −p)/s))=2y_D   Now  {x_C +p−s(√3)((y_C /s))−2c}^2   +{y_C +s(√3)(((x_C −p)/s))−a}^2 =a^2   but  x_C =((b^2 −s^2 +p^2 )/(2p))  ;     y_C =(√(b^2 −x_C ^2 ))  =(√(b^2 −(((b^2 −s^2 +p^2 )/(2p)))^2 ))  ⇒  {((b^2 −s^2 +p^2 )/(2p))+p−(√3)(√(b^2 −(((b^2 −s^2 +p^2 )/(2p)))^2 ))−2c}^2   +{(√(b^2 −(((b^2 −s^2 +p^2 )/(2p)))^2 ))+(√3)(((b^2 −s^2 +p^2 )/(2p)))−(√3)p−a}^2 =a^2   eq. is implicit in  s and p.  we can graph it and can find  the s_(min) .
$${center}\:{of}\:{semicircle}\:{origin}. \\ $$$${center}\:{of}\:{circle} \\ $$$$\:{A}\left[\sqrt{\left({b}−{a}\right)^{\mathrm{2}} −{a}^{\mathrm{2}} },\:{a}\right]\equiv\left({c},{a}\right) \\ $$$${let}\:{bottom}\:{corner}\:{be}\:{B}\left({p},\mathrm{0}\right) \\ $$$${let}\:{side}\:{of}\:\bigtriangleup\:{be}\:{s}. \\ $$$${x}_{{C}} ^{\mathrm{2}} +{y}_{{C}} ^{\mathrm{2}} ={b}^{\mathrm{2}} \\ $$$$\left({x}_{{C}} −{p}\right)^{\mathrm{2}} +{y}_{{C}} ^{\mathrm{2}} ={s}^{\mathrm{2}} \\ $$$${subtracting} \\ $$$$\mathrm{2}{px}_{{C}} −{p}^{\mathrm{2}} ={b}^{\mathrm{2}} −{s}^{\mathrm{2}} \\ $$$${x}_{{C}} =\frac{{b}^{\mathrm{2}} −{s}^{\mathrm{2}} +{p}^{\mathrm{2}} }{\mathrm{2}{p}} \\ $$$$\frac{{x}_{{c}} +{p}}{\mathrm{2}}−\frac{{s}\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{sin}\:\phi={x}_{{D}} \\ $$$$\frac{{y}_{{C}} }{\mathrm{2}}+\frac{{s}\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{cos}\:\phi={y}_{{D}} \\ $$$$\left({x}_{{D}} −{c}\right)^{\mathrm{2}} +\left({y}_{{D}} −{a}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} \\ $$$$\mathrm{cos}\:\phi=\frac{{x}_{{C}} −{p}}{{s}}\:\:;\:\:\mathrm{sin}\:\phi=\frac{{y}_{{C}} }{{s}} \\ $$$${x}_{{C}} +{p}−{s}\sqrt{\mathrm{3}}\left(\frac{\sqrt{{b}^{\mathrm{2}} −{x}_{{C}} ^{\mathrm{2}} }}{{s}}\right)=\mathrm{2}{x}_{{D}} \\ $$$${y}_{{C}} +{s}\sqrt{\mathrm{3}}\left(\frac{{x}_{{C}} −{p}}{{s}}\right)=\mathrm{2}{y}_{{D}} \\ $$$${Now} \\ $$$$\left\{{x}_{{C}} +{p}−{s}\sqrt{\mathrm{3}}\left(\frac{{y}_{{C}} }{{s}}\right)−\mathrm{2}{c}\right\}^{\mathrm{2}} \\ $$$$+\left\{{y}_{{C}} +{s}\sqrt{\mathrm{3}}\left(\frac{{x}_{{C}} −{p}}{{s}}\right)−{a}\right\}^{\mathrm{2}} ={a}^{\mathrm{2}} \\ $$$${but}\:\:{x}_{{C}} =\frac{{b}^{\mathrm{2}} −{s}^{\mathrm{2}} +{p}^{\mathrm{2}} }{\mathrm{2}{p}}\:\:;\: \\ $$$$\:\:{y}_{{C}} =\sqrt{{b}^{\mathrm{2}} −{x}_{{C}} ^{\mathrm{2}} } \\ $$$$=\sqrt{{b}^{\mathrm{2}} −\left(\frac{{b}^{\mathrm{2}} −{s}^{\mathrm{2}} +{p}^{\mathrm{2}} }{\mathrm{2}{p}}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow \\ $$$$\left\{\frac{{b}^{\mathrm{2}} −{s}^{\mathrm{2}} +{p}^{\mathrm{2}} }{\mathrm{2}{p}}+{p}−\sqrt{\mathrm{3}}\sqrt{{b}^{\mathrm{2}} −\left(\frac{{b}^{\mathrm{2}} −{s}^{\mathrm{2}} +{p}^{\mathrm{2}} }{\mathrm{2}{p}}\right)^{\mathrm{2}} }−\mathrm{2}{c}\right\}^{\mathrm{2}} \\ $$$$+\left\{\sqrt{{b}^{\mathrm{2}} −\left(\frac{{b}^{\mathrm{2}} −{s}^{\mathrm{2}} +{p}^{\mathrm{2}} }{\mathrm{2}{p}}\right)^{\mathrm{2}} }+\sqrt{\mathrm{3}}\left(\frac{{b}^{\mathrm{2}} −{s}^{\mathrm{2}} +{p}^{\mathrm{2}} }{\mathrm{2}{p}}\right)−\sqrt{\mathrm{3}}{p}−{a}\right\}^{\mathrm{2}} ={a}^{\mathrm{2}} \\ $$$${eq}.\:{is}\:{implicit}\:{in}\:\:{s}\:{and}\:{p}. \\ $$$${we}\:{can}\:{graph}\:{it}\:{and}\:{can}\:{find} \\ $$$${the}\:{s}_{{min}} . \\ $$$$ \\ $$$$ \\ $$
Commented by mr W last updated on 17/Aug/21
i also can only solve in this way to find  the s_(min) .
$${i}\:{also}\:{can}\:{only}\:{solve}\:{in}\:{this}\:{way}\:{to}\:{find} \\ $$$${the}\:{s}_{{min}} . \\ $$
Answered by ajfour last updated on 17/Aug/21
O(center of semicircle)  z_A =(√((b−a)^2 −a^2 ))+ia=c+ia  z_B =p  z_C =p+s(cos θ+isin θ)  z_D =p+s[−cos (((2π)/3)−θ)+isin (((2π)/3)−θ)]  ∣z_D −z_A ∣=a  ∣z_C ∣=b  (p+scos θ)^2 +s^2 sin^2 θ=b^2   ⇒  p^2 +2spcos θ+s^2 =b^2      ....(i)  {p+scos (((2π)/3)−θ)−c}^2     +{ssin (((2π)/3)−θ)−a}^2 =a^2    ..(ii)  say  ((2π)/3)−θ=φ , then  ⇒  p^2 +s^2 +c^2 +2spcos φ       =2cp+2cscos φ+2assin φ  ⇒  b^2 +2sp(cos φ−cos θ)+c^2     = 2cp+2s(ccos φ+asin φ)  p=((b^2 +c^2 −2s(ccos φ+asin φ))/(2c−2s(cos φ−cos θ)))  using this in  p^2 +2spcos θ+s^2 =b^2      ....(i)  ⇒  {((b^2 +c^2 −2s(ccos φ+asin φ))/(2c−2s(cos φ−cos θ)))+scos θ}^2      +s^2 sin^2 θ=b^2   ⇒  {((b^2 +c^2 −2s[c+atan (((2π)/3)−θ)])/(2c−2s[1−((cos θ)/(cos (((2π)/3)−θ)))]))+scos θ}^2      +s^2 sin^2 θ=b^2   for a=2, b=5  c=(√5)  {((30−2s[(√5)+2tan (((2π)/3)−θ)])/(2(√5)−2s[1−((cos θ)/(cos (((2π)/3)−θ)))]))+scos θ}^2     +s^2 sin^2 θ=25  {((30−2s(√5)−2s((((√3)cos θ−sin θ)/(cos θ−(√3)sin θ))))/(2(√5)−2s+4s(((cos θ)/( (√3)sin θ+cos θ)))))+scos θ}^2    +s^2 sin^2 θ=25  graph of this f(s,θ)=25  should give s_(min) .  sir can u help?
$${O}\left({center}\:{of}\:{semicircle}\right) \\ $$$${z}_{{A}} =\sqrt{\left({b}−{a}\right)^{\mathrm{2}} −{a}^{\mathrm{2}} }+{ia}={c}+{ia} \\ $$$${z}_{{B}} ={p} \\ $$$${z}_{{C}} ={p}+{s}\left(\mathrm{cos}\:\theta+{i}\mathrm{sin}\:\theta\right) \\ $$$${z}_{{D}} ={p}+{s}\left[−\mathrm{cos}\:\left(\frac{\mathrm{2}\pi}{\mathrm{3}}−\theta\right)+{i}\mathrm{sin}\:\left(\frac{\mathrm{2}\pi}{\mathrm{3}}−\theta\right)\right] \\ $$$$\mid{z}_{{D}} −{z}_{{A}} \mid={a} \\ $$$$\mid{z}_{{C}} \mid={b} \\ $$$$\left({p}+{s}\mathrm{cos}\:\theta\right)^{\mathrm{2}} +{s}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta={b}^{\mathrm{2}} \\ $$$$\Rightarrow \\ $$$${p}^{\mathrm{2}} +\mathrm{2}{sp}\mathrm{cos}\:\theta+{s}^{\mathrm{2}} ={b}^{\mathrm{2}} \:\:\:\:\:….\left({i}\right) \\ $$$$\left\{{p}+{s}\mathrm{cos}\:\left(\frac{\mathrm{2}\pi}{\mathrm{3}}−\theta\right)−{c}\right\}^{\mathrm{2}} \\ $$$$\:\:+\left\{{s}\mathrm{sin}\:\left(\frac{\mathrm{2}\pi}{\mathrm{3}}−\theta\right)−{a}\right\}^{\mathrm{2}} ={a}^{\mathrm{2}} \:\:\:..\left({ii}\right) \\ $$$${say}\:\:\frac{\mathrm{2}\pi}{\mathrm{3}}−\theta=\phi\:,\:{then} \\ $$$$\Rightarrow\:\:{p}^{\mathrm{2}} +{s}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}{sp}\mathrm{cos}\:\phi \\ $$$$\:\:\:\:\:=\mathrm{2}{cp}+\mathrm{2}{cs}\mathrm{cos}\:\phi+\mathrm{2}{as}\mathrm{sin}\:\phi \\ $$$$\Rightarrow\:\:{b}^{\mathrm{2}} +\mathrm{2}{sp}\left(\mathrm{cos}\:\phi−\mathrm{cos}\:\theta\right)+{c}^{\mathrm{2}} \\ $$$$\:\:=\:\mathrm{2}{cp}+\mathrm{2}{s}\left({c}\mathrm{cos}\:\phi+{a}\mathrm{sin}\:\phi\right) \\ $$$${p}=\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}{s}\left({c}\mathrm{cos}\:\phi+{a}\mathrm{sin}\:\phi\right)}{\mathrm{2}{c}−\mathrm{2}{s}\left(\mathrm{cos}\:\phi−\mathrm{cos}\:\theta\right)} \\ $$$${using}\:{this}\:{in} \\ $$$${p}^{\mathrm{2}} +\mathrm{2}{sp}\mathrm{cos}\:\theta+{s}^{\mathrm{2}} ={b}^{\mathrm{2}} \:\:\:\:\:….\left({i}\right) \\ $$$$\Rightarrow \\ $$$$\left\{\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}{s}\left({c}\mathrm{cos}\:\phi+{a}\mathrm{sin}\:\phi\right)}{\mathrm{2}{c}−\mathrm{2}{s}\left(\mathrm{cos}\:\phi−\mathrm{cos}\:\theta\right)}+{s}\mathrm{cos}\:\theta\right\}^{\mathrm{2}} \\ $$$$\:\:\:+{s}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta={b}^{\mathrm{2}} \\ $$$$\Rightarrow \\ $$$$\left\{\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}{s}\left[{c}+{a}\mathrm{tan}\:\left(\frac{\mathrm{2}\pi}{\mathrm{3}}−\theta\right)\right]}{\mathrm{2}{c}−\mathrm{2}{s}\left[\mathrm{1}−\frac{\mathrm{cos}\:\theta}{\mathrm{cos}\:\left(\frac{\mathrm{2}\pi}{\mathrm{3}}−\theta\right)}\right]}+{s}\mathrm{cos}\:\theta\right\}^{\mathrm{2}} \\ $$$$\:\:\:+{s}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta={b}^{\mathrm{2}} \\ $$$${for}\:{a}=\mathrm{2},\:{b}=\mathrm{5} \\ $$$${c}=\sqrt{\mathrm{5}} \\ $$$$\left\{\frac{\mathrm{30}−\mathrm{2}{s}\left[\sqrt{\mathrm{5}}+\mathrm{2tan}\:\left(\frac{\mathrm{2}\pi}{\mathrm{3}}−\theta\right)\right]}{\mathrm{2}\sqrt{\mathrm{5}}−\mathrm{2}{s}\left[\mathrm{1}−\frac{\mathrm{cos}\:\theta}{\mathrm{cos}\:\left(\frac{\mathrm{2}\pi}{\mathrm{3}}−\theta\right)}\right]}+{s}\mathrm{cos}\:\theta\right\}^{\mathrm{2}} \\ $$$$\:\:+{s}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta=\mathrm{25} \\ $$$$\left\{\frac{\mathrm{30}−\mathrm{2}{s}\sqrt{\mathrm{5}}−\mathrm{2}{s}\left(\frac{\sqrt{\mathrm{3}}\mathrm{cos}\:\theta−\mathrm{sin}\:\theta}{\mathrm{cos}\:\theta−\sqrt{\mathrm{3}}\mathrm{sin}\:\theta}\right)}{\mathrm{2}\sqrt{\mathrm{5}}−\mathrm{2}{s}+\mathrm{4}{s}\left(\frac{\mathrm{cos}\:\theta}{\:\sqrt{\mathrm{3}}\mathrm{sin}\:\theta+\mathrm{cos}\:\theta}\right)}+{s}\mathrm{cos}\:\theta\right\}^{\mathrm{2}} \\ $$$$\:+{s}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta=\mathrm{25} \\ $$$${graph}\:{of}\:{this}\:{f}\left({s},\theta\right)=\mathrm{25} \\ $$$${should}\:{give}\:{s}_{{min}} . \\ $$$${sir}\:{can}\:{u}\:{help}? \\ $$$$ \\ $$$$ \\ $$
Answered by mr W last updated on 17/Aug/21
Commented by mr W last updated on 18/Aug/21
s=side length of triangle  OC=c=(√((b−a)^2 −a^2 ))=(√(b(b−2a)))  say A(u,0)  u=c−s cos θ+(√(b^2 −s^2  sin^2  θ))    eqn. of small circle:  x^2 +(y−a)^2 =a^2   x_D =u+s cos ((π/3)+θ)  y_D =s sin ((π/3)+θ)  (u+s cos ((π/3)+θ))^2 +(s sin ((π/3)+θ)−a)^2 =a^2   u^2 +2su cos ((π/3)+θ)+s^2 −2as sin ((π/3)+θ)=0  u=−s cos ((π/3)+θ)+(√(s(2a−s sin ((π/3)+θ))sin ((π/3)+θ)))    c−s cos θ+(√(b^2 −s^2  sin^2  θ))=−s cos ((π/3)+θ)+(√(s(2a−s sin ((π/3)+θ))sin ((π/3)+θ)))  ⇒s[cos θ−cos ((π/3)+θ)]−(√(b^2 −s^2  sin^2  θ))+(√(s[2a−s sin ((π/3)+θ)]sin ((π/3)+θ)))=c  λ[cos θ−cos ((π/3)+θ)]−(√(1−λ^2  sin^2  θ))+(√(λ[μ−λ sin ((π/3)+θ)]sin ((π/3)+θ)))=(√(1−μ))  with μ=((2a)/b)≤1, λ=(s/b)  from this relationship between s and  θ we can find the s_(min)  graphically.  example: a=2, b=5  s_(min) =3.8157 at θ=1.4075 (80.64°)
$${s}={side}\:{length}\:{of}\:{triangle} \\ $$$${OC}={c}=\sqrt{\left({b}−{a}\right)^{\mathrm{2}} −{a}^{\mathrm{2}} }=\sqrt{{b}\left({b}−\mathrm{2}{a}\right)} \\ $$$${say}\:{A}\left({u},\mathrm{0}\right) \\ $$$${u}={c}−{s}\:\mathrm{cos}\:\theta+\sqrt{{b}^{\mathrm{2}} −{s}^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\theta} \\ $$$$ \\ $$$${eqn}.\:{of}\:{small}\:{circle}: \\ $$$${x}^{\mathrm{2}} +\left({y}−{a}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} \\ $$$${x}_{{D}} ={u}+{s}\:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{3}}+\theta\right) \\ $$$${y}_{{D}} ={s}\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}+\theta\right) \\ $$$$\left({u}+{s}\:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{3}}+\theta\right)\right)^{\mathrm{2}} +\left({s}\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}+\theta\right)−{a}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} \\ $$$${u}^{\mathrm{2}} +\mathrm{2}{su}\:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{3}}+\theta\right)+{s}^{\mathrm{2}} −\mathrm{2}{as}\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}+\theta\right)=\mathrm{0} \\ $$$${u}=−{s}\:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{3}}+\theta\right)+\sqrt{{s}\left(\mathrm{2}{a}−{s}\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}+\theta\right)\right)\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}+\theta\right)} \\ $$$$ \\ $$$${c}−{s}\:\mathrm{cos}\:\theta+\sqrt{{b}^{\mathrm{2}} −{s}^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\theta}=−{s}\:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{3}}+\theta\right)+\sqrt{{s}\left(\mathrm{2}{a}−{s}\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}+\theta\right)\right)\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}+\theta\right)} \\ $$$$\Rightarrow{s}\left[\mathrm{cos}\:\theta−\mathrm{cos}\:\left(\frac{\pi}{\mathrm{3}}+\theta\right)\right]−\sqrt{{b}^{\mathrm{2}} −{s}^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\theta}+\sqrt{{s}\left[\mathrm{2}{a}−{s}\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}+\theta\right)\right]\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}+\theta\right)}={c} \\ $$$$\lambda\left[\mathrm{cos}\:\theta−\mathrm{cos}\:\left(\frac{\pi}{\mathrm{3}}+\theta\right)\right]−\sqrt{\mathrm{1}−\lambda^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\theta}+\sqrt{\lambda\left[\mu−\lambda\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}+\theta\right)\right]\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}+\theta\right)}=\sqrt{\mathrm{1}−\mu} \\ $$$${with}\:\mu=\frac{\mathrm{2}{a}}{{b}}\leqslant\mathrm{1},\:\lambda=\frac{{s}}{{b}} \\ $$$${from}\:{this}\:{relationship}\:{between}\:{s}\:{and} \\ $$$$\theta\:{we}\:{can}\:{find}\:{the}\:{s}_{{min}} \:{graphically}. \\ $$$${example}:\:{a}=\mathrm{2},\:{b}=\mathrm{5} \\ $$$${s}_{{min}} =\mathrm{3}.\mathrm{8157}\:{at}\:\theta=\mathrm{1}.\mathrm{4075}\:\left(\mathrm{80}.\mathrm{64}°\right) \\ $$
Commented by mr W last updated on 17/Aug/21
Commented by mr W last updated on 17/Aug/21
Commented by mr W last updated on 17/Aug/21
Commented by mr W last updated on 18/Aug/21
Commented by mr W last updated on 18/Aug/21

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