Question Number 82131 by mr W last updated on 18/Feb/20
Commented by ajfour last updated on 18/Feb/20
Commented by mr W last updated on 18/Feb/20
$${Find}\:{the}\:{side}\:{length}\:{s}\:{of}\:{the}\:{inscribed} \\ $$$${equilateral}\:{triangle}\left({s}\right)\:{in}\:{a}\:{given} \\ $$$${triangle}\:{with}\:{side}\:{lengthes}\:{a},\:{b},\:{c}. \\ $$
Commented by ajfour last updated on 18/Feb/20
$${Sir},\:{i}\:{am}\:{wondering}\:{about}\:{the} \\ $$$${locus}\:{of}\:{the}\:{centroid}\:{of}\:{the} \\ $$$${inscribed}\:{equilateral}\:\bigtriangleup. \\ $$
Commented by mr W last updated on 18/Feb/20
$${what}\:{can}\:{be}\:{said}\:{about}\:{the}\:{locus}? \\ $$
Answered by mr W last updated on 19/Feb/20
$${Part}\:{I}:\: \\ $$$${Find}\:{the}\:{maximum}\:{equilateral}\:{s}_{{max}} \\ $$$$\angle{A}=\alpha,\:\angle{B}=\beta,\:\angle{C}=\gamma \\ $$$$ \\ $$$${if}\:\alpha=\beta=\gamma,\:{then}\:{it}'{s}\:{clear}\:{that}\:{the} \\ $$$${maximum}\:{equilateral}\:{is}\:{the}\:{triangle} \\ $$$${itself},\:{i}.{e}.\:{s}_{{max}} ={a}\left(={b}={c}\right).\:{otherwise} \\ $$$${we}\:{will}\:{have}\:{at}\:{least}\:{one}\:{angle}\:{which} \\ $$$${is}\:{larger}\:{than}\:\mathrm{60}°. \\ $$$$ \\ $$$${let}'{s}\:{assume}\:\beta>\gamma>\alpha.\:{it}'{s}\:{clear}\:{that} \\ $$$$\beta>\mathrm{60}°.\:{we}\:{have}\:{two}\:{cases}: \\ $$$${case}\:\mathrm{1}:\:\beta>\gamma>\alpha,\:\gamma<\mathrm{60}° \\ $$$${case}\:\mathrm{2}:\:\beta>\gamma>\alpha,\:\gamma\geqslant\mathrm{60}° \\ $$
Commented by mr W last updated on 19/Feb/20
Commented by mr W last updated on 19/Feb/20
$${Case}\:\mathrm{1}: \\ $$$${it}'{s}\:{clear}\:{the}\:{red}\:{equilateral}\:{is}\:{the} \\ $$$${maximum}. \\ $$$$\frac{{s}_{{max}} }{\mathrm{sin}\:\gamma}=\frac{{a}}{\mathrm{sin}\:\left(\gamma+\beta−\mathrm{60}°\right)} \\ $$$$\Rightarrow{s}_{{max}} =\frac{{a}\:\mathrm{sin}\:\gamma}{\mathrm{sin}\:\left(\beta+\gamma−\mathrm{60}°\right)} \\ $$
Commented by mr W last updated on 19/Feb/20
Commented by mr W last updated on 20/Feb/20
$${Case}\:\mathrm{2}: \\ $$$${blue}\:{equilateral}\:{with}\:{s}_{{blue}} \\ $$$$\frac{{s}_{{blue}} }{\mathrm{sin}\:\gamma}=\frac{{a}}{\mathrm{sin}\:\left(\gamma+\beta−\mathrm{60}°\right)} \\ $$$$\Rightarrow{s}_{{blue}} =\frac{{a}\:\mathrm{sin}\:\gamma}{\mathrm{sin}\:\left(\gamma+\beta−\mathrm{60}°\right)} \\ $$$${similarly} \\ $$$$\Rightarrow{s}_{{red}} =\frac{{a}\:\mathrm{sin}\:\beta}{\mathrm{sin}\:\left(\beta+\gamma−\mathrm{60}°\right)} \\ $$$${since}\:\beta>\gamma,\:{s}_{{red}} >{s}_{{blue}} . \\ $$$$\Rightarrow{s}_{{max}} =\frac{{a}\:\mathrm{sin}\:\beta}{\mathrm{sin}\:\left(\beta+\gamma−\mathrm{60}°\right)} \\ $$$$ \\ $$$${to}\:{be}\:{exact}\:{and}\:{for}\:{all}\:{cases}: \\ $$$${if}\:\mid\mathrm{90}°−\beta\mid<\mid\mathrm{90}°−\gamma\mid,\:{then} \\ $$$$\Rightarrow{s}_{{max}} =\frac{{a}\:\mathrm{sin}\:\beta}{\mathrm{sin}\:\left(\beta+\gamma−\mathrm{60}°\right)}=\frac{{a}\:\mathrm{sin}\:\beta}{\mathrm{sin}\:\left(\mathrm{120}°−\alpha\right)} \\ $$$${if}\:\mid\mathrm{90}°−\beta\mid>\mid\mathrm{90}°−\gamma\mid,\:{then} \\ $$$$\Rightarrow{s}_{{max}} =\frac{{a}\:\mathrm{sin}\:\gamma}{\mathrm{sin}\:\left(\beta+\gamma−\mathrm{60}°\right)}=\frac{{a}\:\mathrm{sin}\:\gamma}{\mathrm{sin}\:\left(\mathrm{120}°−\alpha\right)} \\ $$$${i}.{e}. \\ $$$${s}_{{max}} =\frac{{a}\:{max}\left(\mathrm{sin}\:\beta,\:\mathrm{sin}\:\gamma\right)}{\mathrm{cos}\:\left(\mathrm{30}°−\alpha\right)} \\ $$
Answered by mr W last updated on 19/Feb/20
$${Part}\:{II} \\ $$$${Find}\:{the}\:{minimum}\:{equilateral}\:{s}_{{min}} . \\ $$$${let}'{s}\:{say}\:{for}\:{the}\:{given}\:{triangle}\:{ABC} \\ $$$${with}\:{side}\:{lengthes}\:{a},{b}\:{c}\:{and}\:{inner} \\ $$$${angles}\:\alpha,\beta,\gamma\:{the}\:{minimum}\:{inscribed} \\ $$$${equilateral}\:{triangle}\:{is}\:{that}\:{one}\:{with} \\ $$$${the}\:{side}\:{length}\:{s}. \\ $$$${now}\:{we}\:{image}\:{that}\:{we}\:{have}\:{an} \\ $$$${equilateral}\:{triangle}\:{with}\:{side}\:{length}\:{s}. \\ $$$${we}\:{can}\:{construct}\:{infinite}\:{circum}− \\ $$$${triangles}\:{with}\:{the}\:{inner}\:{angles} \\ $$$$\alpha,\beta,\gamma.\:{all}\:{of}\:{these}\:{circumtriangles} \\ $$$${are}\:{similar}\:{to}\:{each}\:{other}\:{and}\:{similar} \\ $$$${to}\:{the}\:{given}\:{triangle}\:{ABC}.\:{it}\:{is}\:{clear} \\ $$$${that}\:{the}\:{largest}\:{one}\:{of}\:{these}\:{circum}−\: \\ $$$${triangles}\:{has}\:{exactly}\:{the}\:{same}\:{size}\:{as}\: \\ $$$${the}\:{given}\:{triangle}.\:{this}\:{consideration} \\ $$$${gives}\:{us}\:{following}\:{way}\:{to}\:{solve}\:{our} \\ $$$${problem}: \\ $$$${for}\:{an}\:{equilateral}\:{triangle}\:{with}\:{side} \\ $$$${length}\:{s}=\mathrm{1}\:{we}\:{construct}\:{circumtriangles} \\ $$$${with}\:{inner}\:{angles}\:\alpha,\beta,\gamma\:{and}\:{find} \\ $$$${the}\:{largest}\:{one}.\:{if}\:{the}\:{largest}\:{one} \\ $$$${has}\:{the}\:{side}\:{lengthes}\:{a}',{b}'\:{and}\:{c}',\:{then} \\ $$$${we}\:{have} \\ $$$$\frac{{s}_{{min}} }{\mathrm{1}}=\frac{{a}}{{a}'}=\frac{{b}}{{b}'}=\frac{{c}}{{c}'}. \\ $$$${through}\:{this}\:{way}\:{we}\:{get}\:{the}\:{side} \\ $$$${length}\:{s}_{{min}} \:{of}\:{the}\:{minimum}\:{inscribed} \\ $$$${equilateral}\:{for}\:{the}\:{given}\:{triangle}\:{ABC}. \\ $$
Commented by mr W last updated on 19/Feb/20
Commented by mr W last updated on 19/Feb/20
$${now}\:{we}\:{just}\:{need}\:{to}\:{find}\:{the}\:{largest} \\ $$$${circumtriangle}\:{for}\:{an}\:{equilateral} \\ $$$${triangle}\:{with}\:{side}\:{length}\:{s}=\mathrm{1}.\:{the} \\ $$$${circumtriangle}\:{must}\:{have}\:{the}\:{same} \\ $$$${angles}\:{as}\:{the}\:{given}\:{triangle}\:{ABC}, \\ $$$${namely}\:\alpha,\beta,\gamma. \\ $$
Commented by mr W last updated on 19/Feb/20
Commented by mr W last updated on 19/Feb/20
Commented by mr W last updated on 20/Feb/20
![r_A =(s/(2 sin α))=(1/(2 sin α)) r_B =(s/(2 sin β))=(1/(2 sin β)) r_C =(s/(2 sin γ))=(1/(2 sin γ)) when its vertexes lie on the three circles with radii r_A ,r_B ,r_C and centers at A′,B′,C′, then the circumtriangle always has the inner angles α,β,γ. the maximum circumtriangle is A′′B′′C′′. we know this is the case when A′′D, B′′D and C′′D are the diameters of the circles. B′C′^2 =r_B ^2 +r_C ^2 −2r_B r_C cos (60°+90°−β+90°−γ) B′C′^2 =r_B ^2 +r_C ^2 −2r_B r_C cos (60°+α°) B′C′^2 =(1/4)[(1/(sin^2 β))+(1/(sin^2 γ))−((2 cos (60°+α))/(sin β sin γ))] B′C′=(1/2)(√((1/(sin^2 β))+(1/(sin^2 γ))−((2 cos (60°+α))/(sin β sin γ)))) a′=B′′C′′=2B′C′ ⇒a′=(√((1/(sin^2 β))+(1/(sin^2 γ))−((2 cos (60°+α))/(sin β sin γ)))) similarly ⇒b′=(√((1/(sin^2 γ))+(1/(sin^2 α))−((2 cos (60°+β))/(sin γ sin α)))) ⇒c′=(√((1/(sin^2 α))+(1/(sin^2 β))−((2 cos (60°+γ))/(sin α sin β)))) as described above, s_(min) =(a/(a′)) s_(min) =(a/(√((1/(sin^2 β))+(1/(sin^2 γ))−((2 cos (60°+α))/(sin β sin γ))))) or s_(min) =(b/(√((1/(sin^2 γ))+(1/(sin^2 α))−((2 cos (60°+β))/(sin γ sin a))))) or s_(min) =(c/(√((1/(sin^2 α))+(1/(sin^2 β))−((2 cos (60°+γ))/(sin α sin β))))) sin α=(a/(2R)), sin β=(b/(2R)), sin γ=(c/(2R)) cos (60°+γ)=(1/2)(cos γ−(√3) sin γ) cos (60°+γ)=(1/2)(((a^2 +b^2 −c^2 )/(2ab))−(((√3)c)/(2R))) s_(min) =(c/(√(((4R^2 )/a^2 )+((4R^2 )/b^2 )−((4R^2 )/(ab))(((a^2 +b^2 −c^2 )/(2ab))−(((√3)c)/(2R)))))) s_(min) =(c/(√(4R^2 (((a^2 +b^2 +c^2 )/(2a^2 b^2 )))+((2R(√3)c)/(ab))))) s_(min) =(1/(√(((2R^2 )/(a^2 b^2 c^2 ))(a^2 +b^2 +c^2 )+((2R(√3))/(abc))))) with R=((abc)/(4Δ)) s_(min) =(1/(√((1/(8Δ^2 ))(a^2 +b^2 +c^2 )+((√3)/(2Δ))))) ⇒s_(min) =((2(√2)Δ)/(√(a^2 +b^2 +c^2 +4(√3)Δ))) with Δ=area of ΔABC.](Q82263.png)
$${r}_{{A}} =\frac{{s}}{\mathrm{2}\:\mathrm{sin}\:\alpha}=\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{sin}\:\alpha} \\ $$$${r}_{{B}} =\frac{{s}}{\mathrm{2}\:\mathrm{sin}\:\beta}=\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{sin}\:\beta} \\ $$$${r}_{{C}} =\frac{{s}}{\mathrm{2}\:\mathrm{sin}\:\gamma}=\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{sin}\:\gamma} \\ $$$$ \\ $$$${when}\:{its}\:{vertexes}\:{lie}\:{on}\:{the}\:{three} \\ $$$${circles}\:{with}\:{radii}\:{r}_{{A}} ,{r}_{{B}} ,{r}_{{C}} \:{and}\:{centers} \\ $$$${at}\:{A}',{B}',{C}',\:{then}\:{the}\:{circumtriangle} \\ $$$${always}\:{has}\:{the}\:{inner}\:{angles}\:\alpha,\beta,\gamma. \\ $$$$ \\ $$$${the}\:{maximum}\:{circumtriangle}\:{is} \\ $$$${A}''{B}''{C}''.\:{we}\:{know}\:{this}\:{is}\:{the}\:{case} \\ $$$${when}\:{A}''{D},\:{B}''{D}\:{and}\:{C}''{D}\:{are}\:{the} \\ $$$${diameters}\:{of}\:{the}\:{circles}. \\ $$$$ \\ $$$${B}'{C}'^{\mathrm{2}} ={r}_{{B}} ^{\mathrm{2}} +{r}_{{C}} ^{\mathrm{2}} −\mathrm{2}{r}_{{B}} {r}_{{C}} \mathrm{cos}\:\left(\mathrm{60}°+\mathrm{90}°−\beta+\mathrm{90}°−\gamma\right) \\ $$$${B}'{C}'^{\mathrm{2}} ={r}_{{B}} ^{\mathrm{2}} +{r}_{{C}} ^{\mathrm{2}} −\mathrm{2}{r}_{{B}} {r}_{{C}} \mathrm{cos}\:\left(\mathrm{60}°+\alpha°\right) \\ $$$${B}'{C}'^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}\left[\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \:\beta}+\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \:\gamma}−\frac{\mathrm{2}\:\mathrm{cos}\:\left(\mathrm{60}°+\alpha\right)}{\mathrm{sin}\:\beta\:\mathrm{sin}\:\gamma}\right] \\ $$$${B}'{C}'=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \:\beta}+\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \:\gamma}−\frac{\mathrm{2}\:\mathrm{cos}\:\left(\mathrm{60}°+\alpha\right)}{\mathrm{sin}\:\beta\:\mathrm{sin}\:\gamma}} \\ $$$${a}'={B}''{C}''=\mathrm{2}{B}'{C}' \\ $$$$\Rightarrow{a}'=\sqrt{\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \:\beta}+\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \:\gamma}−\frac{\mathrm{2}\:\mathrm{cos}\:\left(\mathrm{60}°+\alpha\right)}{\mathrm{sin}\:\beta\:\mathrm{sin}\:\gamma}} \\ $$$${similarly} \\ $$$$\Rightarrow{b}'=\sqrt{\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \:\gamma}+\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \:\alpha}−\frac{\mathrm{2}\:\mathrm{cos}\:\left(\mathrm{60}°+\beta\right)}{\mathrm{sin}\:\gamma\:\mathrm{sin}\:\alpha}} \\ $$$$\Rightarrow{c}'=\sqrt{\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \:\alpha}+\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \:\beta}−\frac{\mathrm{2}\:\mathrm{cos}\:\left(\mathrm{60}°+\gamma\right)}{\mathrm{sin}\:\alpha\:\mathrm{sin}\:\beta}} \\ $$$$ \\ $$$${as}\:{described}\:{above}, \\ $$$${s}_{{min}} =\frac{{a}}{{a}'} \\ $$$${s}_{{min}} =\frac{{a}}{\sqrt{\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \:\beta}+\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \:\gamma}−\frac{\mathrm{2}\:\mathrm{cos}\:\left(\mathrm{60}°+\alpha\right)}{\mathrm{sin}\:\beta\:\mathrm{sin}\:\gamma}}} \\ $$$${or} \\ $$$${s}_{{min}} =\frac{{b}}{\sqrt{\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \:\gamma}+\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \:\alpha}−\frac{\mathrm{2}\:\mathrm{cos}\:\left(\mathrm{60}°+\beta\right)}{\mathrm{sin}\:\gamma\:\mathrm{sin}\:{a}}}} \\ $$$${or} \\ $$$${s}_{{min}} =\frac{{c}}{\sqrt{\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \:\alpha}+\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \:\beta}−\frac{\mathrm{2}\:\mathrm{cos}\:\left(\mathrm{60}°+\gamma\right)}{\mathrm{sin}\:\alpha\:\mathrm{sin}\:\beta}}} \\ $$$$ \\ $$$$\mathrm{sin}\:\alpha=\frac{{a}}{\mathrm{2}{R}},\:\mathrm{sin}\:\beta=\frac{{b}}{\mathrm{2}{R}},\:\mathrm{sin}\:\gamma=\frac{{c}}{\mathrm{2}{R}} \\ $$$$\mathrm{cos}\:\left(\mathrm{60}°+\gamma\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{cos}\:\gamma−\sqrt{\mathrm{3}}\:\mathrm{sin}\:\gamma\right) \\ $$$$\mathrm{cos}\:\left(\mathrm{60}°+\gamma\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{ab}}−\frac{\sqrt{\mathrm{3}}{c}}{\mathrm{2}{R}}\right) \\ $$$${s}_{{min}} =\frac{{c}}{\sqrt{\frac{\mathrm{4}{R}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{\mathrm{4}{R}^{\mathrm{2}} }{{b}^{\mathrm{2}} }−\frac{\mathrm{4}{R}^{\mathrm{2}} }{{ab}}\left(\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{ab}}−\frac{\sqrt{\mathrm{3}}{c}}{\mathrm{2}{R}}\right)}} \\ $$$${s}_{{min}} =\frac{{c}}{\sqrt{\mathrm{4}{R}^{\mathrm{2}} \left(\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{2}{a}^{\mathrm{2}} {b}^{\mathrm{2}} }\right)+\frac{\mathrm{2}{R}\sqrt{\mathrm{3}}{c}}{{ab}}}} \\ $$$${s}_{{min}} =\frac{\mathrm{1}}{\sqrt{\frac{\mathrm{2}{R}^{\mathrm{2}} }{{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} }\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)+\frac{\mathrm{2}{R}\sqrt{\mathrm{3}}}{{abc}}}} \\ $$$${with}\:{R}=\frac{{abc}}{\mathrm{4}\Delta} \\ $$$${s}_{{min}} =\frac{\mathrm{1}}{\sqrt{\frac{\mathrm{1}}{\mathrm{8}\Delta^{\mathrm{2}} }\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}\Delta}}} \\ $$$$\Rightarrow{s}_{{min}} =\frac{\mathrm{2}\sqrt{\mathrm{2}}\Delta}{\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{4}\sqrt{\mathrm{3}}\Delta}} \\ $$$${with}\:\Delta={area}\:{of}\:\Delta{ABC}. \\ $$
Commented by mr W last updated on 19/Feb/20
$${example}\:\mathrm{1}:\:{a}={b}={c}=\mathrm{4} \\ $$$$\mathrm{sin}\:\alpha=\mathrm{sin}\:\beta=\mathrm{sin}\:\gamma=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${s}_{{min}} =\frac{\mathrm{4}}{\sqrt{\mathrm{2}\left(\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} }}=\mathrm{2} \\ $$$$ \\ $$$${example}\:\mathrm{2}:\:{a}=\mathrm{3},\:{b}=\mathrm{4},\:{c}=\mathrm{5} \\ $$$$\mathrm{sin}\:\alpha=\frac{\mathrm{3}}{\mathrm{5}},\:\mathrm{sin}\:\beta=\frac{\mathrm{4}}{\mathrm{5}},\:\mathrm{sin}\:\gamma=\mathrm{1} \\ $$$${s}_{{min}} =\frac{\mathrm{3}}{\sqrt{\frac{\mathrm{25}}{\mathrm{16}}+\mathrm{1}−\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{4}}{\mathrm{5}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}×\frac{\mathrm{3}}{\mathrm{5}}\right)×\frac{\mathrm{5}}{\mathrm{4}}×\mathrm{1}}}=\frac{\mathrm{12}}{\sqrt{\mathrm{25}+\mathrm{12}\sqrt{\mathrm{3}}}} \\ $$$${s}_{{min}} =\frac{\mathrm{4}}{\sqrt{\frac{\mathrm{25}}{\mathrm{9}}+\mathrm{1}−\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{3}}{\mathrm{5}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}×\frac{\mathrm{4}}{\mathrm{5}}\right)×\frac{\mathrm{5}}{\mathrm{3}}×\mathrm{1}}}=\frac{\mathrm{12}}{\sqrt{\mathrm{25}+\mathrm{12}\sqrt{\mathrm{3}}}} \\ $$$${s}_{{min}} =\frac{\mathrm{5}}{\sqrt{\frac{\mathrm{25}}{\mathrm{16}}+\frac{\mathrm{25}}{\mathrm{9}}−\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{0}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}×\mathrm{1}\right)×\frac{\mathrm{5}}{\mathrm{3}}×\frac{\mathrm{5}}{\mathrm{4}}}}=\frac{\mathrm{12}}{\sqrt{\mathrm{25}+\mathrm{12}\sqrt{\mathrm{3}}}} \\ $$
Commented by mr W last updated on 19/Feb/20
$${or} \\ $$$${s}_{{min}} =\frac{\mathrm{2}\sqrt{\mathrm{2}}\Delta}{\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{4}\sqrt{\mathrm{3}}\Delta}}=\frac{\mathrm{12}\sqrt{\mathrm{2}}}{\sqrt{\mathrm{50}+\mathrm{24}\sqrt{\mathrm{3}}}}=\frac{\mathrm{12}}{\sqrt{\mathrm{25}+\mathrm{12}\sqrt{\mathrm{3}}}} \\ $$
Commented by ajfour last updated on 22/Feb/20
$$\mathrm{Sir},\:\mathrm{would}\:\mathrm{you}\:\mathrm{please}\:\mathrm{let}\:\mathrm{me}\:\mathrm{know} \\ $$$$\mathrm{the}\:\mathrm{Q}.#\:\mathrm{where}\:\mathrm{you}\:\mathrm{had}\:\mathrm{proved} \\ $$$$\mathrm{the}\:\mathrm{concept}\:\mathrm{used}\:\mathrm{here}\:\mathrm{for}\:\mathrm{largest} \\ $$$$\mathrm{cirumtriangle}\:\mathrm{construction}.. \\ $$$$\mathrm{rest}\:\mathrm{of}\:\mathrm{your}\:\mathrm{solution},\:\mathrm{i}\:\mathrm{have}\: \\ $$$$\mathrm{followed}\:\mathrm{entirely},\:\mathrm{very}\:\mathrm{clear} \\ $$$$\mathrm{and}\:\mathrm{clever}\:\mathrm{Sir}.. \\ $$
Commented by mr W last updated on 23/Feb/20
$${thanks}\:{for}\:{reviewing}\:{sir}! \\ $$$${see}\:{Q}\mathrm{60313}. \\ $$