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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(s) in a given triangle with side lengthes a, b, c.
$${Find}\:{the}\:{side}\:{length}\:{s}\:{of}\:{the}\:{inscribed} \\ $$$${equilateral}\:{triangle}\left({s}\right)\:{in}\:{a}\:{given} \\ $$$${triangle}\:{with}\:{side}\:{lengthes}\:{a},\:{b},\:{c}. \\ $$
Commented by ajfour last updated on 18/Feb/20
Sir, i am wondering about the locus of the centroid of the inscribed equilateral △.
$${Sir},\:{i}\:{am}\:{wondering}\:{about}\:{the} \\ $$$${locus}\:{of}\:{the}\:{centroid}\:{of}\:{the} \\ $$$${inscribed}\:{equilateral}\:\bigtriangleup. \\ $$
Commented by mr W last updated on 18/Feb/20
what can be said about the locus?
$${what}\:{can}\:{be}\:{said}\:{about}\:{the}\:{locus}? \\ $$
Answered by mr W last updated on 19/Feb/20
Part I: Find the maximum equilateral s_(max) ∠A=α, ∠B=β, ∠C=γ if α=β=γ, then it′s clear that the maximum equilateral is the triangle itself, i.e. s_(max) =a(=b=c). otherwise we will have at least one angle which is larger than 60°. let′s assume β>γ>α. it′s clear that β>60°. we have two cases: case 1: β>γ>α, γ<60° case 2: β>γ>α, γ≥60°
$${Part}\:{I}:\: \\ $$$${Find}\:{the}\:{maximum}\:{equilateral}\:{s}_{{max}} \\ $$$$\angle{A}=\alpha,\:\angle{B}=\beta,\:\angle{C}=\gamma \\ $$$$ \\ $$$${if}\:\alpha=\beta=\gamma,\:{then}\:{it}'{s}\:{clear}\:{that}\:{the} \\ $$$${maximum}\:{equilateral}\:{is}\:{the}\:{triangle} \\ $$$${itself},\:{i}.{e}.\:{s}_{{max}} ={a}\left(={b}={c}\right).\:{otherwise} \\ $$$${we}\:{will}\:{have}\:{at}\:{least}\:{one}\:{angle}\:{which} \\ $$$${is}\:{larger}\:{than}\:\mathrm{60}°. \\ $$$$ \\ $$$${let}'{s}\:{assume}\:\beta>\gamma>\alpha.\:{it}'{s}\:{clear}\:{that} \\ $$$$\beta>\mathrm{60}°.\:{we}\:{have}\:{two}\:{cases}: \\ $$$${case}\:\mathrm{1}:\:\beta>\gamma>\alpha,\:\gamma<\mathrm{60}° \\ $$$${case}\:\mathrm{2}:\:\beta>\gamma>\alpha,\:\gamma\geqslant\mathrm{60}° \\ $$
Commented by mr W last updated on 19/Feb/20
Commented by mr W last updated on 19/Feb/20
Case 1: it′s clear the red equilateral is the maximum. (s_(max) /(sin γ))=(a/(sin (γ+β−60°))) ⇒s_(max) =((a sin γ)/(sin (β+γ−60°)))
$${Case}\:\mathrm{1}: \\ $$$${it}'{s}\:{clear}\:{the}\:{red}\:{equilateral}\:{is}\:{the} \\ $$$${maximum}. \\ $$$$\frac{{s}_{{max}} }{\mathrm{sin}\:\gamma}=\frac{{a}}{\mathrm{sin}\:\left(\gamma+\beta−\mathrm{60}°\right)} \\ $$$$\Rightarrow{s}_{{max}} =\frac{{a}\:\mathrm{sin}\:\gamma}{\mathrm{sin}\:\left(\beta+\gamma−\mathrm{60}°\right)} \\ $$
Commented by mr W last updated on 19/Feb/20
Commented by mr W last updated on 20/Feb/20
Case 2: blue equilateral with s_(blue) (s_(blue) /(sin γ))=(a/(sin (γ+β−60°))) ⇒s_(blue) =((a sin γ)/(sin (γ+β−60°))) similarly ⇒s_(red) =((a sin β)/(sin (β+γ−60°))) since β>γ, s_(red) >s_(blue) . ⇒s_(max) =((a sin β)/(sin (β+γ−60°))) to be exact and for all cases: if ∣90°−β∣<∣90°−γ∣, then ⇒s_(max) =((a sin β)/(sin (β+γ−60°)))=((a sin β)/(sin (120°−α))) if ∣90°−β∣>∣90°−γ∣, then ⇒s_(max) =((a sin γ)/(sin (β+γ−60°)))=((a sin γ)/(sin (120°−α))) i.e. s_(max) =((a max(sin β, sin γ))/(cos (30°−α)))
$${Case}\:\mathrm{2}: \\ $$$${blue}\:{equilateral}\:{with}\:{s}_{{blue}} \\ $$$$\frac{{s}_{{blue}} }{\mathrm{sin}\:\gamma}=\frac{{a}}{\mathrm{sin}\:\left(\gamma+\beta−\mathrm{60}°\right)} \\ $$$$\Rightarrow{s}_{{blue}} =\frac{{a}\:\mathrm{sin}\:\gamma}{\mathrm{sin}\:\left(\gamma+\beta−\mathrm{60}°\right)} \\ $$$${similarly} \\ $$$$\Rightarrow{s}_{{red}} =\frac{{a}\:\mathrm{sin}\:\beta}{\mathrm{sin}\:\left(\beta+\gamma−\mathrm{60}°\right)} \\ $$$${since}\:\beta>\gamma,\:{s}_{{red}} >{s}_{{blue}} . \\ $$$$\Rightarrow{s}_{{max}} =\frac{{a}\:\mathrm{sin}\:\beta}{\mathrm{sin}\:\left(\beta+\gamma−\mathrm{60}°\right)} \\ $$$$ \\ $$$${to}\:{be}\:{exact}\:{and}\:{for}\:{all}\:{cases}: \\ $$$${if}\:\mid\mathrm{90}°−\beta\mid<\mid\mathrm{90}°−\gamma\mid,\:{then} \\ $$$$\Rightarrow{s}_{{max}} =\frac{{a}\:\mathrm{sin}\:\beta}{\mathrm{sin}\:\left(\beta+\gamma−\mathrm{60}°\right)}=\frac{{a}\:\mathrm{sin}\:\beta}{\mathrm{sin}\:\left(\mathrm{120}°−\alpha\right)} \\ $$$${if}\:\mid\mathrm{90}°−\beta\mid>\mid\mathrm{90}°−\gamma\mid,\:{then} \\ $$$$\Rightarrow{s}_{{max}} =\frac{{a}\:\mathrm{sin}\:\gamma}{\mathrm{sin}\:\left(\beta+\gamma−\mathrm{60}°\right)}=\frac{{a}\:\mathrm{sin}\:\gamma}{\mathrm{sin}\:\left(\mathrm{120}°−\alpha\right)} \\ $$$${i}.{e}. \\ $$$${s}_{{max}} =\frac{{a}\:{max}\left(\mathrm{sin}\:\beta,\:\mathrm{sin}\:\gamma\right)}{\mathrm{cos}\:\left(\mathrm{30}°−\alpha\right)} \\ $$
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 α,β,γ 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 α,β,γ. 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=1 we construct circumtriangles with inner angles α,β,γ and find the largest one. if the largest one has the side lengthes a′,b′ and c′, then we have (s_(min) /1)=(a/(a′))=(b/(b′))=(c/(c′)). through this way we get the side length s_(min) of the minimum inscribed equilateral for the given triangle ABC.
$${Part}\:{II} \\ $$$${Find}\:{the}\:{minimum}\:{equilateral}\:{s}_{{min}} . \\ $$$${let}'{s}\:{say}\:{for}\:{the}\:{given}\:{triangle}\:{ABC} \\ $$$${with}\:{side}\:{lengthes}\:{a},{b}\:{c}\:{and}\:{inner} \\ $$$${angles}\:\alpha,\beta,\gamma\:{the}\:{minimum}\:{inscribed} \\ $$$${equilateral}\:{triangle}\:{is}\:{that}\:{one}\:{with} \\ $$$${the}\:{side}\:{length}\:{s}. \\ $$$${now}\:{we}\:{image}\:{that}\:{we}\:{have}\:{an} \\ $$$${equilateral}\:{triangle}\:{with}\:{side}\:{length}\:{s}. \\ $$$${we}\:{can}\:{construct}\:{infinite}\:{circum}− \\ $$$${triangles}\:{with}\:{the}\:{inner}\:{angles} \\ $$$$\alpha,\beta,\gamma.\:{all}\:{of}\:{these}\:{circumtriangles} \\ $$$${are}\:{similar}\:{to}\:{each}\:{other}\:{and}\:{similar} \\ $$$${to}\:{the}\:{given}\:{triangle}\:{ABC}.\:{it}\:{is}\:{clear} \\ $$$${that}\:{the}\:{largest}\:{one}\:{of}\:{these}\:{circum}−\: \\ $$$${triangles}\:{has}\:{exactly}\:{the}\:{same}\:{size}\:{as}\: \\ $$$${the}\:{given}\:{triangle}.\:{this}\:{consideration} \\ $$$${gives}\:{us}\:{following}\:{way}\:{to}\:{solve}\:{our} \\ $$$${problem}: \\ $$$${for}\:{an}\:{equilateral}\:{triangle}\:{with}\:{side} \\ $$$${length}\:{s}=\mathrm{1}\:{we}\:{construct}\:{circumtriangles} \\ $$$${with}\:{inner}\:{angles}\:\alpha,\beta,\gamma\:{and}\:{find} \\ $$$${the}\:{largest}\:{one}.\:{if}\:{the}\:{largest}\:{one} \\ $$$${has}\:{the}\:{side}\:{lengthes}\:{a}',{b}'\:{and}\:{c}',\:{then} \\ $$$${we}\:{have} \\ $$$$\frac{{s}_{{min}} }{\mathrm{1}}=\frac{{a}}{{a}'}=\frac{{b}}{{b}'}=\frac{{c}}{{c}'}. \\ $$$${through}\:{this}\:{way}\:{we}\:{get}\:{the}\:{side} \\ $$$${length}\:{s}_{{min}} \:{of}\:{the}\:{minimum}\:{inscribed} \\ $$$${equilateral}\:{for}\:{the}\:{given}\:{triangle}\:{ABC}. \\ $$
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=1. the circumtriangle must have the same angles as the given triangle ABC, namely α,β,γ.
$${now}\:{we}\:{just}\:{need}\:{to}\:{find}\:{the}\:{largest} \\ $$$${circumtriangle}\:{for}\:{an}\:{equilateral} \\ $$$${triangle}\:{with}\:{side}\:{length}\:{s}=\mathrm{1}.\:{the} \\ $$$${circumtriangle}\:{must}\:{have}\:{the}\:{same} \\ $$$${angles}\:{as}\:{the}\:{given}\:{triangle}\:{ABC}, \\ $$$${namely}\:\alpha,\beta,\gamma. \\ $$
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.
$${r}_{{A}} =\frac{{s}}{\mathrm{2}\:\mathrm{sin}\:\alpha}=\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{sin}\:\alpha} \\ $$$${r}_{{B}} =\frac{{s}}{\mathrm{2}\:\mathrm{sin}\:\beta}=\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{sin}\:\beta} \\ $$$${r}_{{C}} =\frac{{s}}{\mathrm{2}\:\mathrm{sin}\:\gamma}=\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{sin}\:\gamma} \\ $$$$ \\ $$$${when}\:{its}\:{vertexes}\:{lie}\:{on}\:{the}\:{three} \\ $$$${circles}\:{with}\:{radii}\:{r}_{{A}} ,{r}_{{B}} ,{r}_{{C}} \:{and}\:{centers} \\ $$$${at}\:{A}',{B}',{C}',\:{then}\:{the}\:{circumtriangle} \\ $$$${always}\:{has}\:{the}\:{inner}\:{angles}\:\alpha,\beta,\gamma. \\ $$$$ \\ $$$${the}\:{maximum}\:{circumtriangle}\:{is} \\ $$$${A}''{B}''{C}''.\:{we}\:{know}\:{this}\:{is}\:{the}\:{case} \\ $$$${when}\:{A}''{D},\:{B}''{D}\:{and}\:{C}''{D}\:{are}\:{the} \\ $$$${diameters}\:{of}\:{the}\:{circles}. \\ $$$$ \\ $$$${B}'{C}'^{\mathrm{2}} ={r}_{{B}} ^{\mathrm{2}} +{r}_{{C}} ^{\mathrm{2}} −\mathrm{2}{r}_{{B}} {r}_{{C}} \mathrm{cos}\:\left(\mathrm{60}°+\mathrm{90}°−\beta+\mathrm{90}°−\gamma\right) \\ $$$${B}'{C}'^{\mathrm{2}} ={r}_{{B}} ^{\mathrm{2}} +{r}_{{C}} ^{\mathrm{2}} −\mathrm{2}{r}_{{B}} {r}_{{C}} \mathrm{cos}\:\left(\mathrm{60}°+\alpha°\right) \\ $$$${B}'{C}'^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}\left[\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \:\beta}+\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \:\gamma}−\frac{\mathrm{2}\:\mathrm{cos}\:\left(\mathrm{60}°+\alpha\right)}{\mathrm{sin}\:\beta\:\mathrm{sin}\:\gamma}\right] \\ $$$${B}'{C}'=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \:\beta}+\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \:\gamma}−\frac{\mathrm{2}\:\mathrm{cos}\:\left(\mathrm{60}°+\alpha\right)}{\mathrm{sin}\:\beta\:\mathrm{sin}\:\gamma}} \\ $$$${a}'={B}''{C}''=\mathrm{2}{B}'{C}' \\ $$$$\Rightarrow{a}'=\sqrt{\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \:\beta}+\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \:\gamma}−\frac{\mathrm{2}\:\mathrm{cos}\:\left(\mathrm{60}°+\alpha\right)}{\mathrm{sin}\:\beta\:\mathrm{sin}\:\gamma}} \\ $$$${similarly} \\ $$$$\Rightarrow{b}'=\sqrt{\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \:\gamma}+\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \:\alpha}−\frac{\mathrm{2}\:\mathrm{cos}\:\left(\mathrm{60}°+\beta\right)}{\mathrm{sin}\:\gamma\:\mathrm{sin}\:\alpha}} \\ $$$$\Rightarrow{c}'=\sqrt{\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \:\alpha}+\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \:\beta}−\frac{\mathrm{2}\:\mathrm{cos}\:\left(\mathrm{60}°+\gamma\right)}{\mathrm{sin}\:\alpha\:\mathrm{sin}\:\beta}} \\ $$$$ \\ $$$${as}\:{described}\:{above}, \\ $$$${s}_{{min}} =\frac{{a}}{{a}'} \\ $$$${s}_{{min}} =\frac{{a}}{\:\sqrt{\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \:\beta}+\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \:\gamma}−\frac{\mathrm{2}\:\mathrm{cos}\:\left(\mathrm{60}°+\alpha\right)}{\mathrm{sin}\:\beta\:\mathrm{sin}\:\gamma}}} \\ $$$${or} \\ $$$${s}_{{min}} =\frac{{b}}{\:\sqrt{\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \:\gamma}+\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \:\alpha}−\frac{\mathrm{2}\:\mathrm{cos}\:\left(\mathrm{60}°+\beta\right)}{\mathrm{sin}\:\gamma\:\mathrm{sin}\:{a}}}} \\ $$$${or} \\ $$$${s}_{{min}} =\frac{{c}}{\:\sqrt{\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \:\alpha}+\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \:\beta}−\frac{\mathrm{2}\:\mathrm{cos}\:\left(\mathrm{60}°+\gamma\right)}{\mathrm{sin}\:\alpha\:\mathrm{sin}\:\beta}}} \\ $$$$ \\ $$$$\mathrm{sin}\:\alpha=\frac{{a}}{\mathrm{2}{R}},\:\mathrm{sin}\:\beta=\frac{{b}}{\mathrm{2}{R}},\:\mathrm{sin}\:\gamma=\frac{{c}}{\mathrm{2}{R}} \\ $$$$\mathrm{cos}\:\left(\mathrm{60}°+\gamma\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{cos}\:\gamma−\sqrt{\mathrm{3}}\:\mathrm{sin}\:\gamma\right) \\ $$$$\mathrm{cos}\:\left(\mathrm{60}°+\gamma\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{ab}}−\frac{\sqrt{\mathrm{3}}{c}}{\mathrm{2}{R}}\right) \\ $$$${s}_{{min}} =\frac{{c}}{\:\sqrt{\frac{\mathrm{4}{R}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{\mathrm{4}{R}^{\mathrm{2}} }{{b}^{\mathrm{2}} }−\frac{\mathrm{4}{R}^{\mathrm{2}} }{{ab}}\left(\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{ab}}−\frac{\sqrt{\mathrm{3}}{c}}{\mathrm{2}{R}}\right)}} \\ $$$${s}_{{min}} =\frac{{c}}{\:\sqrt{\mathrm{4}{R}^{\mathrm{2}} \left(\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{2}{a}^{\mathrm{2}} {b}^{\mathrm{2}} }\right)+\frac{\mathrm{2}{R}\sqrt{\mathrm{3}}{c}}{{ab}}}} \\ $$$${s}_{{min}} =\frac{\mathrm{1}}{\:\sqrt{\frac{\mathrm{2}{R}^{\mathrm{2}} }{{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} }\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)+\frac{\mathrm{2}{R}\sqrt{\mathrm{3}}}{{abc}}}} \\ $$$${with}\:{R}=\frac{{abc}}{\mathrm{4}\Delta} \\ $$$${s}_{{min}} =\frac{\mathrm{1}}{\:\sqrt{\frac{\mathrm{1}}{\mathrm{8}\Delta^{\mathrm{2}} }\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}\Delta}}} \\ $$$$\Rightarrow{s}_{{min}} =\frac{\mathrm{2}\sqrt{\mathrm{2}}\Delta}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{4}\sqrt{\mathrm{3}}\Delta}} \\ $$$${with}\:\Delta={area}\:{of}\:\Delta{ABC}. \\ $$
Commented by mr W last updated on 19/Feb/20
example 1: a=b=c=4 sin α=sin β=sin γ=((√3)/2) s_(min) =(4/( (√(2((2/( (√3))))^2 +((2/( (√3))))^2 ))))=2 example 2: a=3, b=4, c=5 sin α=(3/5), sin β=(4/5), sin γ=1 s_(min) =(3/( (√(((25)/(16))+1−2((1/2)×(4/5)−((√3)/2)×(3/5))×(5/4)×1))))=((12)/( (√(25+12(√3))))) s_(min) =(4/( (√(((25)/9)+1−2((1/2)×(3/5)−((√3)/2)×(4/5))×(5/3)×1))))=((12)/( (√(25+12(√3))))) s_(min) =(5/( (√(((25)/(16))+((25)/9)−2((1/2)×0−((√3)/2)×1)×(5/3)×(5/4)))))=((12)/( (√(25+12(√3)))))
$${example}\:\mathrm{1}:\:{a}={b}={c}=\mathrm{4} \\ $$$$\mathrm{sin}\:\alpha=\mathrm{sin}\:\beta=\mathrm{sin}\:\gamma=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${s}_{{min}} =\frac{\mathrm{4}}{\:\sqrt{\mathrm{2}\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} }}=\mathrm{2} \\ $$$$ \\ $$$${example}\:\mathrm{2}:\:{a}=\mathrm{3},\:{b}=\mathrm{4},\:{c}=\mathrm{5} \\ $$$$\mathrm{sin}\:\alpha=\frac{\mathrm{3}}{\mathrm{5}},\:\mathrm{sin}\:\beta=\frac{\mathrm{4}}{\mathrm{5}},\:\mathrm{sin}\:\gamma=\mathrm{1} \\ $$$${s}_{{min}} =\frac{\mathrm{3}}{\:\sqrt{\frac{\mathrm{25}}{\mathrm{16}}+\mathrm{1}−\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{4}}{\mathrm{5}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}×\frac{\mathrm{3}}{\mathrm{5}}\right)×\frac{\mathrm{5}}{\mathrm{4}}×\mathrm{1}}}=\frac{\mathrm{12}}{\:\sqrt{\mathrm{25}+\mathrm{12}\sqrt{\mathrm{3}}}} \\ $$$${s}_{{min}} =\frac{\mathrm{4}}{\:\sqrt{\frac{\mathrm{25}}{\mathrm{9}}+\mathrm{1}−\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{3}}{\mathrm{5}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}×\frac{\mathrm{4}}{\mathrm{5}}\right)×\frac{\mathrm{5}}{\mathrm{3}}×\mathrm{1}}}=\frac{\mathrm{12}}{\:\sqrt{\mathrm{25}+\mathrm{12}\sqrt{\mathrm{3}}}} \\ $$$${s}_{{min}} =\frac{\mathrm{5}}{\:\sqrt{\frac{\mathrm{25}}{\mathrm{16}}+\frac{\mathrm{25}}{\mathrm{9}}−\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{0}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}×\mathrm{1}\right)×\frac{\mathrm{5}}{\mathrm{3}}×\frac{\mathrm{5}}{\mathrm{4}}}}=\frac{\mathrm{12}}{\:\sqrt{\mathrm{25}+\mathrm{12}\sqrt{\mathrm{3}}}} \\ $$
Commented by mr W last updated on 19/Feb/20
or s_(min) =((2(√2)Δ)/( (√(a^2 +b^2 +c^2 +4(√3)Δ))))=((12(√2))/( (√(50+24(√3)))))=((12)/( (√(25+12(√3)))))
$${or} \\ $$$${s}_{{min}} =\frac{\mathrm{2}\sqrt{\mathrm{2}}\Delta}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{4}\sqrt{\mathrm{3}}\Delta}}=\frac{\mathrm{12}\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{50}+\mathrm{24}\sqrt{\mathrm{3}}}}=\frac{\mathrm{12}}{\:\sqrt{\mathrm{25}+\mathrm{12}\sqrt{\mathrm{3}}}} \\ $$
Commented by ajfour last updated on 22/Feb/20
Sir, would you please let me know the Q.# where you had proved the concept used here for largest cirumtriangle construction.. rest of your solution, i have followed entirely, very clear and clever Sir..
$$\mathrm{Sir},\:\mathrm{would}\:\mathrm{you}\:\mathrm{please}\:\mathrm{let}\:\mathrm{me}\:\mathrm{know} \\ $$$$\mathrm{the}\:\mathrm{Q}.#\:\mathrm{where}\:\mathrm{you}\:\mathrm{had}\:\mathrm{proved} \\ $$$$\mathrm{the}\:\mathrm{concept}\:\mathrm{used}\:\mathrm{here}\:\mathrm{for}\:\mathrm{largest} \\ $$$$\mathrm{cirumtriangle}\:\mathrm{construction}.. \\ $$$$\mathrm{rest}\:\mathrm{of}\:\mathrm{your}\:\mathrm{solution},\:\mathrm{i}\:\mathrm{have}\: \\ $$$$\mathrm{followed}\:\mathrm{entirely},\:\mathrm{very}\:\mathrm{clear} \\ $$$$\mathrm{and}\:\mathrm{clever}\:\mathrm{Sir}.. \\ $$
Commented by mr W last updated on 23/Feb/20
thanks for reviewing sir! see Q60313.
$${thanks}\:{for}\:{reviewing}\:{sir}! \\ $$$${see}\:{Q}\mathrm{60313}. \\ $$

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