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
$${the}\:{green}\:{one}\:{is}\:{the}\:{smallest}\: \\ $$$${equilateral}.\:{i}\:{can}\:{determine}\:{it}\:{only} \\ $$$${numerically}. \\ $$
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).](https://www.tinkutara.com/question/Q150955.png)
$$\mathrm{sin}\:\alpha=\frac{{a}}{{b}−{a}} \\ $$$${A}\left[\sqrt{\left({b}−{a}\right)^{\mathrm{2}} −{a}^{\mathrm{2}} },\:{a}\right]\equiv\left[{c},{a}\right] \\ $$$${let}\:\:{T}\left({p},{q}\right) \\ $$$${let}\:{s}\:{be}\:{circumradius}\:{of}\:{such}\:{an} \\ $$$${equilateral}\:{triangle}. \\ $$$${P}\left[{p}−{s}\mathrm{cos}\:\theta,\:{q}+{s}\mathrm{sin}\:\theta\right] \\ $$$${Q}\left[{p}+{s}\mathrm{cos}\:\left(\mathrm{60}°−\theta\right),\:{q}+{s}\mathrm{sin}\:\left(\mathrm{60}°−\theta\right)\right] \\ $$$${R}\left[{p}+{s}\mathrm{cos}\:\left(\mathrm{60}°+\theta\right),\:{q}−{s}\mathrm{sin}\:\left(\mathrm{60}°+\theta\right)\right] \\ $$$${now} \\ $$$$\left({p}−{s}\mathrm{cos}\:\theta−{c}\right)^{\mathrm{2}} +\left({q}+{s}\mathrm{sin}\:\theta−{a}\right)^{\mathrm{2}} \\ $$$$\:\:\:=\:{a}^{\mathrm{2}} \:\:\:\:…\left({i}\right) \\ $$$$\left[{p}+{s}\mathrm{cos}\:\left(\mathrm{60}°−\theta\right)\right]^{\mathrm{2}} +\left[{q}+{s}\mathrm{sin}\:\left(\mathrm{60}°−\theta\right)\right]^{\mathrm{2}} \\ $$$$\:=\:{b}^{\mathrm{2}} \:\:\:\:\:\:…\left({ii}\right) \\ $$$${q}={s}\mathrm{sin}\:\left(\mathrm{60}°+\theta\right) \\ $$$$\Rightarrow\:\left({p}−{s}\mathrm{cos}\:\theta−{c}\right)^{\mathrm{2}} \\ $$$$\:\:+\left[\frac{{s}\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{cos}\:\theta+\frac{\mathrm{3}{s}}{\mathrm{2}}\mathrm{sin}\:\theta−{a}\right]^{\mathrm{2}} ={a}^{\mathrm{2}} \\ $$$$\:\:\:……..\left({I}\right) \\ $$$$\left[{p}+\frac{{s}}{\mathrm{2}}\mathrm{cos}\:\theta+\frac{{s}\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{sin}\:\theta\right]^{\mathrm{2}} \\ $$$$\:\:+\left[{s}\sqrt{\mathrm{3}}\mathrm{cos}\:\theta\right]^{\mathrm{2}} ={b}^{\mathrm{2}} \:\:\:\:\:\:….\left({II}\right) \\ $$$$\left({II}\right)−\left({I}\right) \\ $$$$\left\{\frac{\mathrm{3}{s}}{\mathrm{2}}\mathrm{cos}\:\theta+\frac{{s}\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{sin}\:\theta+{c}\right\} \\ $$$$×\left\{\mathrm{2}{p}−\frac{{s}}{\mathrm{2}}\mathrm{cos}\:\theta+\frac{{s}\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{sin}\:\theta−{c}\right\} \\ $$$$+ \\ $$$$\:\:\left\{\frac{{s}\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{cos}\:\theta−\frac{\mathrm{3}{s}}{\mathrm{2}}\mathrm{sin}\:\theta+{a}\right\} \\ $$$$×\left\{\frac{\mathrm{3}\sqrt{\mathrm{3}}{s}}{\mathrm{2}}\mathrm{cos}\:\theta+\frac{\mathrm{3}{s}}{\mathrm{2}}\mathrm{sin}\:\theta−{a}\right\} \\ $$$$\:\:=\:{b}^{\mathrm{2}} −{a}^{\mathrm{2}} \\ $$$${from}\:{here} \\ $$$${p}={f}\left({s},\theta\right) \\ $$$${now}\:{using}\:\left({I}\right)\:{or}\:\left({II}\right) \\ $$$${h}\left({s},\theta\right)=\mathrm{0} \\ $$$${from}\:{here}\:{we}\:{get}\:{s}_{{min}} . \\ $$$${Minimum}\:{side}\:{length}\:{of} \\ $$$${such}\:{an}\:{equilateral}\:\bigtriangleup\:{would} \\ $$$${then}\:{be}\:{s}_{{min}} \sqrt{\mathrm{3}}. \\ $$$$ \\ $$$$ \\ $$
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) .](https://www.tinkutara.com/question/Q151019.png)
$${center}\:{of}\:{semicircle}\:{origin}. \\ $$$${center}\:{of}\:{circle} \\ $$$$\:{A}\left[\sqrt{\left({b}−{a}\right)^{\mathrm{2}} −{a}^{\mathrm{2}} },\:{a}\right]\equiv\left({c},{a}\right) \\ $$$${let}\:{bottom}\:{corner}\:{be}\:{B}\left({p},\mathrm{0}\right) \\ $$$${let}\:{side}\:{of}\:\bigtriangleup\:{be}\:{s}. \\ $$$${x}_{{C}} ^{\mathrm{2}} +{y}_{{C}} ^{\mathrm{2}} ={b}^{\mathrm{2}} \\ $$$$\left({x}_{{C}} −{p}\right)^{\mathrm{2}} +{y}_{{C}} ^{\mathrm{2}} ={s}^{\mathrm{2}} \\ $$$${subtracting} \\ $$$$\mathrm{2}{px}_{{C}} −{p}^{\mathrm{2}} ={b}^{\mathrm{2}} −{s}^{\mathrm{2}} \\ $$$${x}_{{C}} =\frac{{b}^{\mathrm{2}} −{s}^{\mathrm{2}} +{p}^{\mathrm{2}} }{\mathrm{2}{p}} \\ $$$$\frac{{x}_{{c}} +{p}}{\mathrm{2}}−\frac{{s}\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{sin}\:\phi={x}_{{D}} \\ $$$$\frac{{y}_{{C}} }{\mathrm{2}}+\frac{{s}\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{cos}\:\phi={y}_{{D}} \\ $$$$\left({x}_{{D}} −{c}\right)^{\mathrm{2}} +\left({y}_{{D}} −{a}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} \\ $$$$\mathrm{cos}\:\phi=\frac{{x}_{{C}} −{p}}{{s}}\:\:;\:\:\mathrm{sin}\:\phi=\frac{{y}_{{C}} }{{s}} \\ $$$${x}_{{C}} +{p}−{s}\sqrt{\mathrm{3}}\left(\frac{\sqrt{{b}^{\mathrm{2}} −{x}_{{C}} ^{\mathrm{2}} }}{{s}}\right)=\mathrm{2}{x}_{{D}} \\ $$$${y}_{{C}} +{s}\sqrt{\mathrm{3}}\left(\frac{{x}_{{C}} −{p}}{{s}}\right)=\mathrm{2}{y}_{{D}} \\ $$$${Now} \\ $$$$\left\{{x}_{{C}} +{p}−{s}\sqrt{\mathrm{3}}\left(\frac{{y}_{{C}} }{{s}}\right)−\mathrm{2}{c}\right\}^{\mathrm{2}} \\ $$$$+\left\{{y}_{{C}} +{s}\sqrt{\mathrm{3}}\left(\frac{{x}_{{C}} −{p}}{{s}}\right)−{a}\right\}^{\mathrm{2}} ={a}^{\mathrm{2}} \\ $$$${but}\:\:{x}_{{C}} =\frac{{b}^{\mathrm{2}} −{s}^{\mathrm{2}} +{p}^{\mathrm{2}} }{\mathrm{2}{p}}\:\:;\: \\ $$$$\:\:{y}_{{C}} =\sqrt{{b}^{\mathrm{2}} −{x}_{{C}} ^{\mathrm{2}} } \\ $$$$=\sqrt{{b}^{\mathrm{2}} −\left(\frac{{b}^{\mathrm{2}} −{s}^{\mathrm{2}} +{p}^{\mathrm{2}} }{\mathrm{2}{p}}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow \\ $$$$\left\{\frac{{b}^{\mathrm{2}} −{s}^{\mathrm{2}} +{p}^{\mathrm{2}} }{\mathrm{2}{p}}+{p}−\sqrt{\mathrm{3}}\sqrt{{b}^{\mathrm{2}} −\left(\frac{{b}^{\mathrm{2}} −{s}^{\mathrm{2}} +{p}^{\mathrm{2}} }{\mathrm{2}{p}}\right)^{\mathrm{2}} }−\mathrm{2}{c}\right\}^{\mathrm{2}} \\ $$$$+\left\{\sqrt{{b}^{\mathrm{2}} −\left(\frac{{b}^{\mathrm{2}} −{s}^{\mathrm{2}} +{p}^{\mathrm{2}} }{\mathrm{2}{p}}\right)^{\mathrm{2}} }+\sqrt{\mathrm{3}}\left(\frac{{b}^{\mathrm{2}} −{s}^{\mathrm{2}} +{p}^{\mathrm{2}} }{\mathrm{2}{p}}\right)−\sqrt{\mathrm{3}}{p}−{a}\right\}^{\mathrm{2}} ={a}^{\mathrm{2}} \\ $$$${eq}.\:{is}\:{implicit}\:{in}\:\:{s}\:{and}\:{p}. \\ $$$${we}\:{can}\:{graph}\:{it}\:{and}\:{can}\:{find} \\ $$$${the}\:{s}_{{min}} . \\ $$$$ \\ $$$$ \\ $$
Commented by mr W last updated on 17/Aug/21
$${i}\:{also}\:{can}\:{only}\:{solve}\:{in}\:{this}\:{way}\:{to}\:{find} \\ $$$${the}\:{s}_{{min}} . \\ $$
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?](https://www.tinkutara.com/question/Q151024.png)
$${O}\left({center}\:{of}\:{semicircle}\right) \\ $$$${z}_{{A}} =\sqrt{\left({b}−{a}\right)^{\mathrm{2}} −{a}^{\mathrm{2}} }+{ia}={c}+{ia} \\ $$$${z}_{{B}} ={p} \\ $$$${z}_{{C}} ={p}+{s}\left(\mathrm{cos}\:\theta+{i}\mathrm{sin}\:\theta\right) \\ $$$${z}_{{D}} ={p}+{s}\left[−\mathrm{cos}\:\left(\frac{\mathrm{2}\pi}{\mathrm{3}}−\theta\right)+{i}\mathrm{sin}\:\left(\frac{\mathrm{2}\pi}{\mathrm{3}}−\theta\right)\right] \\ $$$$\mid{z}_{{D}} −{z}_{{A}} \mid={a} \\ $$$$\mid{z}_{{C}} \mid={b} \\ $$$$\left({p}+{s}\mathrm{cos}\:\theta\right)^{\mathrm{2}} +{s}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta={b}^{\mathrm{2}} \\ $$$$\Rightarrow \\ $$$${p}^{\mathrm{2}} +\mathrm{2}{sp}\mathrm{cos}\:\theta+{s}^{\mathrm{2}} ={b}^{\mathrm{2}} \:\:\:\:\:….\left({i}\right) \\ $$$$\left\{{p}+{s}\mathrm{cos}\:\left(\frac{\mathrm{2}\pi}{\mathrm{3}}−\theta\right)−{c}\right\}^{\mathrm{2}} \\ $$$$\:\:+\left\{{s}\mathrm{sin}\:\left(\frac{\mathrm{2}\pi}{\mathrm{3}}−\theta\right)−{a}\right\}^{\mathrm{2}} ={a}^{\mathrm{2}} \:\:\:..\left({ii}\right) \\ $$$${say}\:\:\frac{\mathrm{2}\pi}{\mathrm{3}}−\theta=\phi\:,\:{then} \\ $$$$\Rightarrow\:\:{p}^{\mathrm{2}} +{s}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}{sp}\mathrm{cos}\:\phi \\ $$$$\:\:\:\:\:=\mathrm{2}{cp}+\mathrm{2}{cs}\mathrm{cos}\:\phi+\mathrm{2}{as}\mathrm{sin}\:\phi \\ $$$$\Rightarrow\:\:{b}^{\mathrm{2}} +\mathrm{2}{sp}\left(\mathrm{cos}\:\phi−\mathrm{cos}\:\theta\right)+{c}^{\mathrm{2}} \\ $$$$\:\:=\:\mathrm{2}{cp}+\mathrm{2}{s}\left({c}\mathrm{cos}\:\phi+{a}\mathrm{sin}\:\phi\right) \\ $$$${p}=\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}{s}\left({c}\mathrm{cos}\:\phi+{a}\mathrm{sin}\:\phi\right)}{\mathrm{2}{c}−\mathrm{2}{s}\left(\mathrm{cos}\:\phi−\mathrm{cos}\:\theta\right)} \\ $$$${using}\:{this}\:{in} \\ $$$${p}^{\mathrm{2}} +\mathrm{2}{sp}\mathrm{cos}\:\theta+{s}^{\mathrm{2}} ={b}^{\mathrm{2}} \:\:\:\:\:….\left({i}\right) \\ $$$$\Rightarrow \\ $$$$\left\{\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}{s}\left({c}\mathrm{cos}\:\phi+{a}\mathrm{sin}\:\phi\right)}{\mathrm{2}{c}−\mathrm{2}{s}\left(\mathrm{cos}\:\phi−\mathrm{cos}\:\theta\right)}+{s}\mathrm{cos}\:\theta\right\}^{\mathrm{2}} \\ $$$$\:\:\:+{s}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta={b}^{\mathrm{2}} \\ $$$$\Rightarrow \\ $$$$\left\{\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}{s}\left[{c}+{a}\mathrm{tan}\:\left(\frac{\mathrm{2}\pi}{\mathrm{3}}−\theta\right)\right]}{\mathrm{2}{c}−\mathrm{2}{s}\left[\mathrm{1}−\frac{\mathrm{cos}\:\theta}{\mathrm{cos}\:\left(\frac{\mathrm{2}\pi}{\mathrm{3}}−\theta\right)}\right]}+{s}\mathrm{cos}\:\theta\right\}^{\mathrm{2}} \\ $$$$\:\:\:+{s}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta={b}^{\mathrm{2}} \\ $$$${for}\:{a}=\mathrm{2},\:{b}=\mathrm{5} \\ $$$${c}=\sqrt{\mathrm{5}} \\ $$$$\left\{\frac{\mathrm{30}−\mathrm{2}{s}\left[\sqrt{\mathrm{5}}+\mathrm{2tan}\:\left(\frac{\mathrm{2}\pi}{\mathrm{3}}−\theta\right)\right]}{\mathrm{2}\sqrt{\mathrm{5}}−\mathrm{2}{s}\left[\mathrm{1}−\frac{\mathrm{cos}\:\theta}{\mathrm{cos}\:\left(\frac{\mathrm{2}\pi}{\mathrm{3}}−\theta\right)}\right]}+{s}\mathrm{cos}\:\theta\right\}^{\mathrm{2}} \\ $$$$\:\:+{s}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta=\mathrm{25} \\ $$$$\left\{\frac{\mathrm{30}−\mathrm{2}{s}\sqrt{\mathrm{5}}−\mathrm{2}{s}\left(\frac{\sqrt{\mathrm{3}}\mathrm{cos}\:\theta−\mathrm{sin}\:\theta}{\mathrm{cos}\:\theta−\sqrt{\mathrm{3}}\mathrm{sin}\:\theta}\right)}{\mathrm{2}\sqrt{\mathrm{5}}−\mathrm{2}{s}+\mathrm{4}{s}\left(\frac{\mathrm{cos}\:\theta}{\:\sqrt{\mathrm{3}}\mathrm{sin}\:\theta+\mathrm{cos}\:\theta}\right)}+{s}\mathrm{cos}\:\theta\right\}^{\mathrm{2}} \\ $$$$\:+{s}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta=\mathrm{25} \\ $$$${graph}\:{of}\:{this}\:{f}\left({s},\theta\right)=\mathrm{25} \\ $$$${should}\:{give}\:{s}_{{min}} . \\ $$$${sir}\:{can}\:{u}\:{help}? \\ $$$$ \\ $$$$ \\ $$
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°)](https://www.tinkutara.com/question/Q151040.png)
$${s}={side}\:{length}\:{of}\:{triangle} \\ $$$${OC}={c}=\sqrt{\left({b}−{a}\right)^{\mathrm{2}} −{a}^{\mathrm{2}} }=\sqrt{{b}\left({b}−\mathrm{2}{a}\right)} \\ $$$${say}\:{A}\left({u},\mathrm{0}\right) \\ $$$${u}={c}−{s}\:\mathrm{cos}\:\theta+\sqrt{{b}^{\mathrm{2}} −{s}^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\theta} \\ $$$$ \\ $$$${eqn}.\:{of}\:{small}\:{circle}: \\ $$$${x}^{\mathrm{2}} +\left({y}−{a}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} \\ $$$${x}_{{D}} ={u}+{s}\:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{3}}+\theta\right) \\ $$$${y}_{{D}} ={s}\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}+\theta\right) \\ $$$$\left({u}+{s}\:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{3}}+\theta\right)\right)^{\mathrm{2}} +\left({s}\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}+\theta\right)−{a}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} \\ $$$${u}^{\mathrm{2}} +\mathrm{2}{su}\:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{3}}+\theta\right)+{s}^{\mathrm{2}} −\mathrm{2}{as}\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}+\theta\right)=\mathrm{0} \\ $$$${u}=−{s}\:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{3}}+\theta\right)+\sqrt{{s}\left(\mathrm{2}{a}−{s}\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}+\theta\right)\right)\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}+\theta\right)} \\ $$$$ \\ $$$${c}−{s}\:\mathrm{cos}\:\theta+\sqrt{{b}^{\mathrm{2}} −{s}^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\theta}=−{s}\:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{3}}+\theta\right)+\sqrt{{s}\left(\mathrm{2}{a}−{s}\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}+\theta\right)\right)\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}+\theta\right)} \\ $$$$\Rightarrow{s}\left[\mathrm{cos}\:\theta−\mathrm{cos}\:\left(\frac{\pi}{\mathrm{3}}+\theta\right)\right]−\sqrt{{b}^{\mathrm{2}} −{s}^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\theta}+\sqrt{{s}\left[\mathrm{2}{a}−{s}\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}+\theta\right)\right]\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}+\theta\right)}={c} \\ $$$$\lambda\left[\mathrm{cos}\:\theta−\mathrm{cos}\:\left(\frac{\pi}{\mathrm{3}}+\theta\right)\right]−\sqrt{\mathrm{1}−\lambda^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\theta}+\sqrt{\lambda\left[\mu−\lambda\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}+\theta\right)\right]\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}+\theta\right)}=\sqrt{\mathrm{1}−\mu} \\ $$$${with}\:\mu=\frac{\mathrm{2}{a}}{{b}}\leqslant\mathrm{1},\:\lambda=\frac{{s}}{{b}} \\ $$$${from}\:{this}\:{relationship}\:{between}\:{s}\:{and} \\ $$$$\theta\:{we}\:{can}\:{find}\:{the}\:{s}_{{min}} \:{graphically}. \\ $$$${example}:\:{a}=\mathrm{2},\:{b}=\mathrm{5} \\ $$$${s}_{{min}} =\mathrm{3}.\mathrm{8157}\:{at}\:\theta=\mathrm{1}.\mathrm{4075}\:\left(\mathrm{80}.\mathrm{64}°\right) \\ $$
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
