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Question Number 13735 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 22/May/17

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 22/May/17

∡ABC=90,S_(ABC) =1,AB=BC ∡ABD=∡DBE=∡EBC EG⊥AB, DF⊥BC ......................................................... show that:(R_(ABC) /R_(DHE) )=2+(√3) R_(ABC) =radius of circumscribed circle of triangle AB^Δ C and so R_(DHE) for DH^Δ E.

$$\measuredangle{ABC}=\mathrm{90},{S}_{{ABC}} =\mathrm{1},{AB}={BC} \\ $$$$\measuredangle{ABD}=\measuredangle{DBE}=\measuredangle{EBC} \\ $$$${EG}\bot{AB},\:{DF}\bot{BC} \\ $$$$......................................................... \\ $$$${show}\:{that}:\frac{{R}_{{ABC}} }{{R}_{{DHE}} }=\mathrm{2}+\sqrt{\mathrm{3}} \\ $$$${R}_{{ABC}} ={radius}\:{of}\:{circumscribed}\:{circle} \\ $$$${of}\:{triangle}\:{A}\overset{\Delta} {{B}C}\:{and}\:{so}\:{R}_{{DHE}} \:{for}\:{D}\overset{\Delta} {{H}E}. \\ $$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 23/May/17

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 23/May/17

what about inscribed circels? is there any ratio?

$${what}\:{about}\:{inscribed}\:{circels}? \\ $$$${is}\:{there}\:{any}\:{ratio}? \\ $$

Commented by ajfour last updated on 23/May/17

Commented by ajfour last updated on 23/May/17

let AB=BC=a equation of AC: x+y=a ......... of BD: y=(x/(√3)) ......... of BE: y=(√3)x for point D x+(x/(√3))=a ⇒ x_D =((a(√3))/((√3)+1)) for point E x+x(√3)=a ⇒ x_E =(a/((√3)+1)) x_D −x_E =((a((√3)−1))/((√3)+1)) R_(DHE) =((x_D −x_E )/(√2)) =((a((√3)−1))/((√2)((√3)+1))) R_(ABC) =(a/(√2)) so, (R_(ABC) /R_(DHE) )=(((√3)+1)/((√3)−1))=2+(√3) •

$${let}\:{AB}={BC}={a} \\ $$$${equation}\:{of}\:{AC}:\:\:{x}+{y}={a} \\ $$$$\:.........\:\:\:\:\:{of}\:{BD}:\:\:{y}=\frac{{x}}{\sqrt{\mathrm{3}}} \\ $$$$\:.........\:\:\:\:\:{of}\:{BE}:\:\:{y}=\sqrt{\mathrm{3}}{x} \\ $$$${for}\:{point}\:{D} \\ $$$${x}+\frac{{x}}{\sqrt{\mathrm{3}}}={a}\:\:\:\:\Rightarrow\:\:{x}_{{D}} =\frac{{a}\sqrt{\mathrm{3}}}{\sqrt{\mathrm{3}}+\mathrm{1}} \\ $$$${for}\:{point}\:{E} \\ $$$${x}+{x}\sqrt{\mathrm{3}}={a}\:\:\Rightarrow\:\:{x}_{{E}} =\frac{{a}}{\sqrt{\mathrm{3}}+\mathrm{1}} \\ $$$${x}_{{D}} −{x}_{{E}} =\frac{{a}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)}{\sqrt{\mathrm{3}}+\mathrm{1}} \\ $$$${R}_{{DHE}} =\frac{{x}_{{D}} −{x}_{{E}} }{\sqrt{\mathrm{2}}}\:=\frac{{a}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)}{\sqrt{\mathrm{2}}\left(\sqrt{\mathrm{3}}+\mathrm{1}\right)} \\ $$$$\:{R}_{{ABC}} =\frac{{a}}{\sqrt{\mathrm{2}}} \\ $$$${so},\:\:\frac{{R}_{{ABC}} }{{R}_{{DHE}} }=\frac{\sqrt{\mathrm{3}}+\mathrm{1}}{\sqrt{\mathrm{3}}−\mathrm{1}}=\mathrm{2}+\sqrt{\mathrm{3}}\:\:\bullet \\ $$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 25/May/17

thank you mr. ajfour.

$${thank}\:{you}\:{mr}.\:{ajfour}. \\ $$

Answered by mrW1 last updated on 22/May/17

let AB=BC=a ⇒AC=(√2)a ((AD)/(sin 30))=((AB)/(sin (45+30))) ⇒AD=((sin 30)/(sin 75))×a=(a/(2cos 15))=(a/(2(√((1+((√3)/2))/2))))=(a/(√(2+(√3)))) DE=(√2)a−((2a)/(√(2+(√3)))) (R_(DHE) /R_(ABC) )=((DE)/(AC))=1−(√(2/(2+(√3))))

$${let}\:{AB}={BC}={a} \\ $$$$\Rightarrow{AC}=\sqrt{\mathrm{2}}{a} \\ $$$$\frac{{AD}}{\mathrm{sin}\:\mathrm{30}}=\frac{{AB}}{\mathrm{sin}\:\left(\mathrm{45}+\mathrm{30}\right)} \\ $$$$\Rightarrow{AD}=\frac{\mathrm{sin}\:\mathrm{30}}{\mathrm{sin}\:\mathrm{75}}×{a}=\frac{{a}}{\mathrm{2cos}\:\mathrm{15}}=\frac{{a}}{\mathrm{2}\sqrt{\frac{\mathrm{1}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}{\mathrm{2}}}}=\frac{{a}}{\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}} \\ $$$${DE}=\sqrt{\mathrm{2}}{a}−\frac{\mathrm{2}{a}}{\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}} \\ $$$$\frac{{R}_{{DHE}} }{{R}_{{ABC}} }=\frac{{DE}}{{AC}}=\mathrm{1}−\sqrt{\frac{\mathrm{2}}{\mathrm{2}+\sqrt{\mathrm{3}}}} \\ $$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 23/May/17

1−(√(2/(2+(√3))))=1−(√((2(2−(√(3))))/1))=1−(√(4−2(√3)))= 1−(√(((√3)−1)^2 ))=1−((√3)−1)=2−(√3)=(1/(2+(√3))).

$$\mathrm{1}−\sqrt{\frac{\mathrm{2}}{\mathrm{2}+\sqrt{\mathrm{3}}}}=\mathrm{1}−\sqrt{\frac{\mathrm{2}\left(\mathrm{2}−\sqrt{\left.\mathrm{3}\right)}\right.}{\mathrm{1}}}=\mathrm{1}−\sqrt{\mathrm{4}−\mathrm{2}\sqrt{\mathrm{3}}}= \\ $$$$\mathrm{1}−\sqrt{\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{1}−\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)=\mathrm{2}−\sqrt{\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{2}+\sqrt{\mathrm{3}}}. \\ $$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 25/May/17

thank you mrW1.

$${thank}\:{you}\:{mrW}\mathrm{1}. \\ $$

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 23/May/17

S=(1/2)ac=2⇒ac=2 ⇒^(a=c) a=c=(√2),b=2 R_(ABC) =((abc)/(4S))=(((√2).(√2).2)/(2×4))=(4/4)=1 ((AD)/(sin30))=((AB)/(sin(180−(30+45))))⇒((AD)/(1/2))=((√2)/(sin75))⇒ AD=(1/2)(√2)((((√6)−(√2))/1))=(√3)−1 DE=2−2AD=2−2((√3)−1)=4−2(√3) DH=HE=((DE)/(√2))=((((√3)−1)^2 )/(√2)) S_(DHE) =(1/2)DH.HE=(1/2).((((√3)−1)^2 )/(√2)).((((√3)−1)^2 )/(√2))= =((2(2−(√3)).2×(2−(√3)))/4)=(2−(√3))^2 R_(DHE) =((a^′ b^′ c^′ )/(4S^′ ))=((((((√3)−1)^2 )/(√2)).((((√3)−1)^2 )/(√2)).((√3)−1)^2 )/(4(2−(√3))^2 ))= =((4(2−(√3))^3 )/(4(2−(√3))^2 ))=2−(√3) ⇒(R_(ABC) /R_(DHE) )=(1/(2−(√3)))=2+(√3) .■

$${S}=\frac{\mathrm{1}}{\mathrm{2}}{ac}=\mathrm{2}\Rightarrow{ac}=\mathrm{2}\:\overset{{a}={c}} {\Rightarrow}\:{a}={c}=\sqrt{\mathrm{2}},{b}=\mathrm{2} \\ $$$${R}_{{ABC}} =\frac{{abc}}{\mathrm{4}{S}}=\frac{\sqrt{\mathrm{2}}.\sqrt{\mathrm{2}}.\mathrm{2}}{\mathrm{2}×\mathrm{4}}=\frac{\mathrm{4}}{\mathrm{4}}=\mathrm{1} \\ $$$$\frac{{AD}}{{sin}\mathrm{30}}=\frac{{AB}}{{sin}\left(\mathrm{180}−\left(\mathrm{30}+\mathrm{45}\right)\right)}\Rightarrow\frac{{AD}}{\frac{\mathrm{1}}{\mathrm{2}}}=\frac{\sqrt{\mathrm{2}}}{{sin}\mathrm{75}}\Rightarrow \\ $$$${AD}=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{2}}\left(\frac{\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}}{\mathrm{1}}\right)=\sqrt{\mathrm{3}}−\mathrm{1} \\ $$$${DE}=\mathrm{2}−\mathrm{2}{AD}=\mathrm{2}−\mathrm{2}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)=\mathrm{4}−\mathrm{2}\sqrt{\mathrm{3}} \\ $$$${DH}={HE}=\frac{{DE}}{\sqrt{\mathrm{2}}}=\frac{\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)^{\mathrm{2}} }{\sqrt{\mathrm{2}}} \\ $$$${S}_{{DHE}} =\frac{\mathrm{1}}{\mathrm{2}}{DH}.{HE}=\frac{\mathrm{1}}{\mathrm{2}}.\frac{\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)^{\mathrm{2}} }{\sqrt{\mathrm{2}}}.\frac{\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)^{\mathrm{2}} }{\sqrt{\mathrm{2}}}= \\ $$$$=\frac{\mathrm{2}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right).\mathrm{2}×\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)}{\mathrm{4}}=\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \\ $$$${R}_{{DHE}} =\frac{{a}^{'} {b}^{'} {c}^{'} }{\mathrm{4}{S}^{'} }=\frac{\frac{\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)^{\mathrm{2}} }{\sqrt{\mathrm{2}}}.\frac{\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)^{\mathrm{2}} }{\sqrt{\mathrm{2}}}.\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }= \\ $$$$=\frac{\mathrm{4}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\mathrm{3}} }{\mathrm{4}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }=\mathrm{2}−\sqrt{\mathrm{3}} \\ $$$$\Rightarrow\frac{{R}_{{ABC}} }{{R}_{{DHE}} }=\frac{\mathrm{1}}{\mathrm{2}−\sqrt{\mathrm{3}}}=\mathrm{2}+\sqrt{\mathrm{3}}\:\:.\blacksquare \\ $$

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