Question Number 190694 by cortano12 last updated on 09/Apr/23
Commented by nikif99 last updated on 10/Apr/23
Answered by nikif99 last updated on 10/Apr/23
![(b/(CD))=tan 60 ⇒CD=((b(√3))/3), d=((x(√3))/2) (1) A_1 =bx=b(2−2CD)=2b(((3−b(√3))/3))(2), x=((2(3−b(√3)))/3) (3) △BFG: ((d−R)/R)=tan 60 ⇒d=R(1+(√3))⇒^((1)) ((x(√3))/2)=R(1+(√3))⇒^((3)) R=((2(3−b(√3))(√3))/(2×3(1+(√3)))) ⇒R=(((3−b(√3))(√3))/(3(1+(√3)))) (4) A_2 =((πR^2 )/2)=(π/2)[(((3−b(√3))(√3))/(3(1+(√3))))]^2 ⇒... ⇒ A_2 =((π[(2−(√3))b^2 +2(3−2(√3))b+3(2−(√3))])/4) (5) a=((2R(d−R))/(2R+(d+R)))^((∗) see comment) =((2R×R(√3))/(R+d))=((2R^2 (√3))/(R+R(1+(√3)))) ⇒^((4)) a=((2[(((3−b(√3))(√3))/(3(1+(√3))))](√3))/(2+(√3))) ⇒... ⇒a=((2(3−b(√3)))/(5+3(√3))) (6) A_3 =a^2 =((4(3−b(√3))^2 )/((5+3(√3))^2 )) ⇒... ⇒A_3 =((6(b^2 −2b(√3)+3))/(15(√3)+26)) (7) A_1 +A_2 +A_3 =f(b)=...=((6π−8(√3)−3π(√3))/(12))b^2 + +((10(√3)−2π(√3)+3π−14)/2)b+((3(12(√3)−20+2π−π(√3))/4) f(x)=kx^2 +mx+n has maximum at −(m/(2k)) f(b) has maximum at 0.493=b ⇒^((3)) x=1.431](Q190758.png)
$$ \\ $$$$\frac{{b}}{{CD}}=\mathrm{tan}\:\mathrm{60}\:\Rightarrow{CD}=\frac{{b}\sqrt{\mathrm{3}}}{\mathrm{3}},\:{d}=\frac{{x}\sqrt{\mathrm{3}}}{\mathrm{2}}\:\left(\mathrm{1}\right) \\ $$$${A}_{\mathrm{1}} ={bx}={b}\left(\mathrm{2}−\mathrm{2}{CD}\right)=\mathrm{2}{b}\left(\frac{\mathrm{3}−{b}\sqrt{\mathrm{3}}}{\mathrm{3}}\right)\left(\mathrm{2}\right),\:{x}=\frac{\mathrm{2}\left(\mathrm{3}−{b}\sqrt{\mathrm{3}}\right)}{\mathrm{3}}\:\left(\mathrm{3}\right) \\ $$$$\bigtriangleup{BFG}:\:\frac{{d}−{R}}{{R}}=\mathrm{tan}\:\mathrm{60}\:\Rightarrow{d}={R}\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)\overset{\left(\mathrm{1}\right)} {\Rightarrow}\frac{{x}\sqrt{\mathrm{3}}}{\mathrm{2}}={R}\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)\overset{\left(\mathrm{3}\right)} {\Rightarrow} \\ $$$${R}=\frac{\mathrm{2}\left(\mathrm{3}−{b}\sqrt{\mathrm{3}}\right)\sqrt{\mathrm{3}}}{\mathrm{2}×\mathrm{3}\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)}\:\Rightarrow{R}=\frac{\left(\mathrm{3}−{b}\sqrt{\mathrm{3}}\right)\sqrt{\mathrm{3}}}{\mathrm{3}\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)}\:\left(\mathrm{4}\right) \\ $$$${A}_{\mathrm{2}} =\frac{\pi{R}^{\mathrm{2}} }{\mathrm{2}}=\frac{\pi}{\mathrm{2}}\left[\frac{\left(\mathrm{3}−{b}\sqrt{\mathrm{3}}\right)\sqrt{\mathrm{3}}}{\mathrm{3}\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)}\right]^{\mathrm{2}} \Rightarrow...\:\Rightarrow \\ $$$${A}_{\mathrm{2}} =\frac{\pi\left[\left(\mathrm{2}−\sqrt{\mathrm{3}}\right){b}^{\mathrm{2}} +\mathrm{2}\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{3}}\right){b}+\mathrm{3}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)\right]}{\mathrm{4}}\:\left(\mathrm{5}\right) \\ $$$${a}=\frac{\mathrm{2}{R}\left({d}−{R}\right)}{\mathrm{2}{R}+\left({d}+{R}\right)}\:^{\left(\ast\right)\:{see}\:{comment}} =\frac{\mathrm{2}{R}×{R}\sqrt{\mathrm{3}}}{{R}+{d}}=\frac{\mathrm{2}{R}^{\mathrm{2}} \sqrt{\mathrm{3}}}{{R}+{R}\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)}\:\overset{\left(\mathrm{4}\right)} {\Rightarrow} \\ $$$${a}=\frac{\mathrm{2}\left[\frac{\left(\mathrm{3}−{b}\sqrt{\mathrm{3}}\right)\sqrt{\mathrm{3}}}{\mathrm{3}\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)}\right]\sqrt{\mathrm{3}}}{\mathrm{2}+\sqrt{\mathrm{3}}}\:\Rightarrow...\:\Rightarrow{a}=\frac{\mathrm{2}\left(\mathrm{3}−{b}\sqrt{\mathrm{3}}\right)}{\mathrm{5}+\mathrm{3}\sqrt{\mathrm{3}}}\:\left(\mathrm{6}\right) \\ $$$${A}_{\mathrm{3}} ={a}^{\mathrm{2}} =\frac{\mathrm{4}\left(\mathrm{3}−{b}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }{\left(\mathrm{5}+\mathrm{3}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }\:\Rightarrow...\:\Rightarrow{A}_{\mathrm{3}} =\frac{\mathrm{6}\left({b}^{\mathrm{2}} −\mathrm{2}{b}\sqrt{\mathrm{3}}+\mathrm{3}\right)}{\mathrm{15}\sqrt{\mathrm{3}}+\mathrm{26}}\:\left(\mathrm{7}\right) \\ $$$${A}_{\mathrm{1}} +{A}_{\mathrm{2}} +{A}_{\mathrm{3}} ={f}\left({b}\right)=...=\frac{\mathrm{6}\pi−\mathrm{8}\sqrt{\mathrm{3}}−\mathrm{3}\pi\sqrt{\mathrm{3}}}{\mathrm{12}}{b}^{\mathrm{2}} + \\ $$$$+\frac{\mathrm{10}\sqrt{\mathrm{3}}−\mathrm{2}\pi\sqrt{\mathrm{3}}+\mathrm{3}\pi−\mathrm{14}}{\mathrm{2}}{b}+\frac{\mathrm{3}\left(\mathrm{12}\sqrt{\mathrm{3}}−\mathrm{20}+\mathrm{2}\pi−\pi\sqrt{\mathrm{3}}\right.}{\mathrm{4}} \\ $$$${f}\left({x}\right)={kx}^{\mathrm{2}} +{mx}+{n}\:{has}\:{maximum}\:{at}\:−\frac{{m}}{\mathrm{2}{k}} \\ $$$${f}\left({b}\right)\:{has}\:{maximum}\:{at}\:\mathrm{0}.\mathrm{493}={b}\:\overset{\left(\mathrm{3}\right)} {\Rightarrow}{x}=\mathrm{1}.\mathrm{431} \\ $$
Commented by nikif99 last updated on 10/Apr/23
$${A}\:{square}\:{inscribed}\:{in}\:{a}\:{triangle}\:{of} \\ $$$${base}\:{b}\:{and}\:{height}\:{h}\:{has}\:{side}\:{a}=\frac{{b}×{h}}{{b}+{h}} \\ $$