Question Number 204970 by mr W last updated on 04/Mar/24
Commented by mr W last updated on 04/Mar/24
$${the}\:{vertices}\:{of}\:{triangle}\:{PQR}\:{lie}\:{on} \\ $$$${each}\:{of}\:{the}\:{three}\:{touching}\:{circles} \\ $$$${with}\:{radii}\:{a},\:{b},\:{c}\:{respectively}. \\ $$$${find}\:{the}\:{maximum}\:{area}\:{of}\:{the} \\ $$$${triangle}. \\ $$
Answered by mr W last updated on 02/May/24
Commented by mr W last updated on 02/May/24
![Δ=(√(abc(a+b+c))) R=((abc)/(4Δ))=(1/4)(√((abc)/(a+b+c))) sin ∠A=((2R)/(b+c))=(1/(2(b+c)))(√((abc)/(a+b+c))) sin ∠B=((2R)/(c+a))=(1/(2(c+a)))(√((abc)/(a+b+c))) sin ∠C=((2R)/(a+b))=(1/(2(a+b)))(√((abc)/(a+b+c))) cos ∠A=(((c+a)^2 +(a+b)^2 −(b+c)^2 )/(2(c+a)(a+b)))=((a(a+b+c)−bc)/((c+a)(a+b))) cos ∠B=(((a+b)^2 +(b+c)^2 −(c+a)^2 )/(2(a+b)(b+c)))=((b(a+b+c)−ca)/((a+b)(b+c))) cos ∠C=(((b+c)^2 +(c+a)^2 −(a+b)^2 )/(2(b+c)(c+a)))=((c(a+b+c)−ab)/((b+c)(c+a))) α+β+γ=2π according to Q206922 sin (α−∠A)=sin α cos ∠A−cos α sin ∠A sin (α−∠A)=((a(a+b+c)−bc)/((c+a)(a+b)))×sin α−(1/(2(b+c)))(√((abc)/(a+b+c)))×cos α (c+a)(a+b) sin (α−∠A)=[a(a+b+c)−bc] sin α−(((c+a)(a+b))/(2(b+c)))(√((abc)/(a+b+c)))×cos α cos (α−∠A)=cos α cos ∠A+sin α sin ∠A cos (α−∠A)=((a(a+b+c)−bc)/((c+a)(a+b)))×cos α+(1/(2(b+c)))(√((abc)/(a+b+c)))×sin α (c+a)(a+b)cos (α−∠A)=[a(a+b+c)−bc] cos α+(((c+a)(a+b))/(2(b+c)))(√((abc)/(a+b+c)))×sin α x=(((c+a)(a+b) sin (α−∠A))/( (√((c+a)^2 sin^2 γ+(a+b)^2 sin^2 β+2(c+a)(a+b) sin β sin γ cos (α−∠A))))) ⇒x=(([a(a+b+c)−bc] sin α−(((c+a)(a+b))/(2(b+c)))(√((abc)/(a+b+c)))×cos α)/( (√((c+a)^2 sin^2 γ+(a+b)^2 sin^2 β+2 sin β sin γ {[a(a+b+c)−bc] cos α+(((c+a)(a+b))/(2(b+c)))(√((abc)/(a+b+c)))×sin α})))) y=(((a+b)(b+c) sin (β−∠B))/( (√((a+b)^2 sin^2 α+(b+c)^2 sin^2 γ+2(a+b)(b+c) sin γ sin a cos (β−∠B))))) z=(((b+c)(c+a) sin (γ−∠C))/( (√((b+c)^2 sin^2 β+(c+a)^2 sin^2 α+2(b+c)(c+a) sin α sin β cos (γ−∠C))))) p=(√((y+b)^2 +(z+c)^2 −2(y+b)(z+c) cos α)) q=(√((z+c)^2 +(x+a)^2 −2(z+c)(x+a) cos β)) r=(√((x+a)^2 +(y+b)^2 −2(x+a)(y+b) cos γ)) ((sin ∠Q_2 )/(z+c))=((sin ∠R_1 )/(y+b))=((sin α)/p) ⇒sin ∠Q_2 =(((z+c) sin α)/( (√((y+b)^2 +(z+c)^2 −2(y+b)(z+c) cos α)))) ⇒sin ∠R_1 =(((y+b) sin α)/( (√((y+b)^2 +(z+c)^2 −2(y+b)(z+c) cos α)))) similarly sin ∠R_2 =(((x+a) sin β)/( (√((z+c)^2 +(x+a)^2 −2(z+c)(x+a) cos β)))) sin ∠P_1 =(((z+c) sin β)/( (√((z+c)^2 +(x+a)^2 −2(z+c)(x+a) cos β)))) sin ∠P_2 =(((y+b) sin γ)/( (√((x+a)^2 +(y+b)^2 −2(x+a)(y+b) cos γ)))) sin ∠Q_1 =(((x+a) sin γ)/( (√((x+a)^2 +(y+b)^2 −2(x+a)(y+b) cos γ)))) ∠P_1 =∠Q_2 ⇒(((z+c) sin β)/( (√((z+c)^2 +(x+a)^2 −2(z+c)(x+a) cos β))))=(((z+c) sin α)/( (√((y+b)^2 +(z+c)^2 −2(y+b)(z+c) cos α)))) ⇒((sin α)/(sin β))=((√((y+b)^2 +(z+c)^2 −2(y+b)(z+c) cos α))/( (√((z+c)^2 +(x+a)^2 −2(z+c)(x+a) cos β)))) ...(i) ∠Q_1 =∠R_2 ⇒(((x+a) sin γ)/( (√((x+a)^2 +(y+b)^2 −2(x+a)(y+b) cos γ))))=(((x+a) sin β)/( (√((z+c)^2 +(x+a)^2 −2(z+c)(x+a) cos β)))) ⇒((sin β)/(sin γ))=((√((z+c)^2 +(x+a)^2 −2(z+c)(x+a) cos β))/( (√((x+a)^2 +(y+b)^2 −2(x+a)(y+b) cos γ)))) ...(ii) ∠R_1 =∠P_2 ⇒(((y+b) sin α)/( (√((y+b)^2 +(z+c)^2 −2(y+b)(z+c) cos α))))=(((y+b) sin γ)/( (√((x+a)^2 +(y+b)^2 −2(x+a)(y+b) cos γ)))) ⇒((sin γ)/(sin α))=((√((x+a)^2 +(y+b)^2 −2(x+a)(y+b) cos γ))/( (√((y+b)^2 +(z+c)^2 −2(y+b)(z+c) cos α)))) ...(iii) two of these three equations can be solved to get α, β and γ.](https://www.tinkutara.com/question/Q206936.png)
$$\Delta=\sqrt{{abc}\left({a}+{b}+{c}\right)} \\ $$$${R}=\frac{{abc}}{\mathrm{4}\Delta}=\frac{\mathrm{1}}{\mathrm{4}}\sqrt{\frac{{abc}}{{a}+{b}+{c}}} \\ $$$$\mathrm{sin}\:\angle{A}=\frac{\mathrm{2}{R}}{{b}+{c}}=\frac{\mathrm{1}}{\mathrm{2}\left({b}+{c}\right)}\sqrt{\frac{{abc}}{{a}+{b}+{c}}} \\ $$$$\mathrm{sin}\:\angle{B}=\frac{\mathrm{2}{R}}{{c}+{a}}=\frac{\mathrm{1}}{\mathrm{2}\left({c}+{a}\right)}\sqrt{\frac{{abc}}{{a}+{b}+{c}}} \\ $$$$\mathrm{sin}\:\angle{C}=\frac{\mathrm{2}{R}}{{a}+{b}}=\frac{\mathrm{1}}{\mathrm{2}\left({a}+{b}\right)}\sqrt{\frac{{abc}}{{a}+{b}+{c}}} \\ $$$$\mathrm{cos}\:\angle{A}=\frac{\left({c}+{a}\right)^{\mathrm{2}} +\left({a}+{b}\right)^{\mathrm{2}} −\left({b}+{c}\right)^{\mathrm{2}} }{\mathrm{2}\left({c}+{a}\right)\left({a}+{b}\right)}=\frac{{a}\left({a}+{b}+{c}\right)−{bc}}{\left({c}+{a}\right)\left({a}+{b}\right)} \\ $$$$\mathrm{cos}\:\angle{B}=\frac{\left({a}+{b}\right)^{\mathrm{2}} +\left({b}+{c}\right)^{\mathrm{2}} −\left({c}+{a}\right)^{\mathrm{2}} }{\mathrm{2}\left({a}+{b}\right)\left({b}+{c}\right)}=\frac{{b}\left({a}+{b}+{c}\right)−{ca}}{\left({a}+{b}\right)\left({b}+{c}\right)} \\ $$$$\mathrm{cos}\:\angle{C}=\frac{\left({b}+{c}\right)^{\mathrm{2}} +\left({c}+{a}\right)^{\mathrm{2}} −\left({a}+{b}\right)^{\mathrm{2}} }{\mathrm{2}\left({b}+{c}\right)\left({c}+{a}\right)}=\frac{{c}\left({a}+{b}+{c}\right)−{ab}}{\left({b}+{c}\right)\left({c}+{a}\right)} \\ $$$$\alpha+\beta+\gamma=\mathrm{2}\pi \\ $$$${according}\:{to}\:{Q}\mathrm{206922} \\ $$$$\mathrm{sin}\:\left(\alpha−\angle{A}\right)=\mathrm{sin}\:\alpha\:\mathrm{cos}\:\angle{A}−\mathrm{cos}\:\alpha\:\mathrm{sin}\:\angle{A} \\ $$$$\mathrm{sin}\:\left(\alpha−\angle{A}\right)=\frac{{a}\left({a}+{b}+{c}\right)−{bc}}{\left({c}+{a}\right)\left({a}+{b}\right)}×\mathrm{sin}\:\alpha−\frac{\mathrm{1}}{\mathrm{2}\left({b}+{c}\right)}\sqrt{\frac{{abc}}{{a}+{b}+{c}}}×\mathrm{cos}\:\alpha \\ $$$$\left({c}+{a}\right)\left({a}+{b}\right)\:\mathrm{sin}\:\left(\alpha−\angle{A}\right)=\left[{a}\left({a}+{b}+{c}\right)−{bc}\right]\:\mathrm{sin}\:\alpha−\frac{\left({c}+{a}\right)\left({a}+{b}\right)}{\mathrm{2}\left({b}+{c}\right)}\sqrt{\frac{{abc}}{{a}+{b}+{c}}}×\mathrm{cos}\:\alpha \\ $$$$\mathrm{cos}\:\left(\alpha−\angle{A}\right)=\mathrm{cos}\:\alpha\:\mathrm{cos}\:\angle{A}+\mathrm{sin}\:\alpha\:\mathrm{sin}\:\angle{A} \\ $$$$\mathrm{cos}\:\left(\alpha−\angle{A}\right)=\frac{{a}\left({a}+{b}+{c}\right)−{bc}}{\left({c}+{a}\right)\left({a}+{b}\right)}×\mathrm{cos}\:\alpha+\frac{\mathrm{1}}{\mathrm{2}\left({b}+{c}\right)}\sqrt{\frac{{abc}}{{a}+{b}+{c}}}×\mathrm{sin}\:\alpha \\ $$$$\left({c}+{a}\right)\left({a}+{b}\right)\mathrm{cos}\:\:\left(\alpha−\angle{A}\right)=\left[{a}\left({a}+{b}+{c}\right)−{bc}\right]\:\mathrm{cos}\:\alpha+\frac{\left({c}+{a}\right)\left({a}+{b}\right)}{\mathrm{2}\left({b}+{c}\right)}\sqrt{\frac{{abc}}{{a}+{b}+{c}}}×\mathrm{sin}\:\alpha \\ $$$${x}=\frac{\left({c}+{a}\right)\left({a}+{b}\right)\:\mathrm{sin}\:\left(\alpha−\angle{A}\right)}{\:\sqrt{\left({c}+{a}\right)^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\gamma+\left({a}+{b}\right)^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\beta+\mathrm{2}\left({c}+{a}\right)\left({a}+{b}\right)\:\mathrm{sin}\:\beta\:\mathrm{sin}\:\gamma\:\mathrm{cos}\:\left(\alpha−\angle{A}\right)}} \\ $$$$\Rightarrow{x}=\frac{\left[{a}\left({a}+{b}+{c}\right)−{bc}\right]\:\mathrm{sin}\:\alpha−\frac{\left({c}+{a}\right)\left({a}+{b}\right)}{\mathrm{2}\left({b}+{c}\right)}\sqrt{\frac{{abc}}{{a}+{b}+{c}}}×\mathrm{cos}\:\alpha}{\:\sqrt{\left({c}+{a}\right)^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\gamma+\left({a}+{b}\right)^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\beta+\mathrm{2}\:\mathrm{sin}\:\beta\:\mathrm{sin}\:\gamma\:\left\{\left[{a}\left({a}+{b}+{c}\right)−{bc}\right]\:\mathrm{cos}\:\alpha+\frac{\left({c}+{a}\right)\left({a}+{b}\right)}{\mathrm{2}\left({b}+{c}\right)}\sqrt{\frac{{abc}}{{a}+{b}+{c}}}×\mathrm{sin}\:\alpha\right\}}} \\ $$$${y}=\frac{\left({a}+{b}\right)\left({b}+{c}\right)\:\mathrm{sin}\:\left(\beta−\angle{B}\right)}{\:\sqrt{\left({a}+{b}\right)^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\alpha+\left({b}+{c}\right)^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\gamma+\mathrm{2}\left({a}+{b}\right)\left({b}+{c}\right)\:\mathrm{sin}\:\gamma\:\mathrm{sin}\:{a}\:\mathrm{cos}\:\left(\beta−\angle{B}\right)}} \\ $$$${z}=\frac{\left({b}+{c}\right)\left({c}+{a}\right)\:\mathrm{sin}\:\left(\gamma−\angle{C}\right)}{\:\sqrt{\left({b}+{c}\right)^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\beta+\left({c}+{a}\right)^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\alpha+\mathrm{2}\left({b}+{c}\right)\left({c}+{a}\right)\:\mathrm{sin}\:\alpha\:\mathrm{sin}\:\beta\:\mathrm{cos}\:\left(\gamma−\angle{C}\right)}} \\ $$$${p}=\sqrt{\left({y}+{b}\right)^{\mathrm{2}} +\left({z}+{c}\right)^{\mathrm{2}} −\mathrm{2}\left({y}+{b}\right)\left({z}+{c}\right)\:\mathrm{cos}\:\alpha} \\ $$$${q}=\sqrt{\left({z}+{c}\right)^{\mathrm{2}} +\left({x}+{a}\right)^{\mathrm{2}} −\mathrm{2}\left({z}+{c}\right)\left({x}+{a}\right)\:\mathrm{cos}\:\beta} \\ $$$${r}=\sqrt{\left({x}+{a}\right)^{\mathrm{2}} +\left({y}+{b}\right)^{\mathrm{2}} −\mathrm{2}\left({x}+{a}\right)\left({y}+{b}\right)\:\mathrm{cos}\:\gamma} \\ $$$$\frac{\mathrm{sin}\:\angle{Q}_{\mathrm{2}} }{{z}+{c}}=\frac{\mathrm{sin}\:\angle{R}_{\mathrm{1}} }{{y}+{b}}=\frac{\mathrm{sin}\:\alpha}{{p}} \\ $$$$\Rightarrow\mathrm{sin}\:\angle{Q}_{\mathrm{2}} =\frac{\left({z}+{c}\right)\:\mathrm{sin}\:\alpha}{\:\sqrt{\left({y}+{b}\right)^{\mathrm{2}} +\left({z}+{c}\right)^{\mathrm{2}} −\mathrm{2}\left({y}+{b}\right)\left({z}+{c}\right)\:\mathrm{cos}\:\alpha}} \\ $$$$\Rightarrow\mathrm{sin}\:\angle{R}_{\mathrm{1}} =\frac{\left({y}+{b}\right)\:\mathrm{sin}\:\alpha}{\:\sqrt{\left({y}+{b}\right)^{\mathrm{2}} +\left({z}+{c}\right)^{\mathrm{2}} −\mathrm{2}\left({y}+{b}\right)\left({z}+{c}\right)\:\mathrm{cos}\:\alpha}} \\ $$$${similarly} \\ $$$$\mathrm{sin}\:\angle{R}_{\mathrm{2}} =\frac{\left({x}+{a}\right)\:\mathrm{sin}\:\beta}{\:\sqrt{\left({z}+{c}\right)^{\mathrm{2}} +\left({x}+{a}\right)^{\mathrm{2}} −\mathrm{2}\left({z}+{c}\right)\left({x}+{a}\right)\:\mathrm{cos}\:\beta}} \\ $$$$\mathrm{sin}\:\angle{P}_{\mathrm{1}} =\frac{\left({z}+{c}\right)\:\mathrm{sin}\:\beta}{\:\sqrt{\left({z}+{c}\right)^{\mathrm{2}} +\left({x}+{a}\right)^{\mathrm{2}} −\mathrm{2}\left({z}+{c}\right)\left({x}+{a}\right)\:\mathrm{cos}\:\beta}} \\ $$$$\mathrm{sin}\:\angle{P}_{\mathrm{2}} =\frac{\left({y}+{b}\right)\:\mathrm{sin}\:\gamma}{\:\sqrt{\left({x}+{a}\right)^{\mathrm{2}} +\left({y}+{b}\right)^{\mathrm{2}} −\mathrm{2}\left({x}+{a}\right)\left({y}+{b}\right)\:\mathrm{cos}\:\gamma}} \\ $$$$\mathrm{sin}\:\angle{Q}_{\mathrm{1}} =\frac{\left({x}+{a}\right)\:\mathrm{sin}\:\gamma}{\:\sqrt{\left({x}+{a}\right)^{\mathrm{2}} +\left({y}+{b}\right)^{\mathrm{2}} −\mathrm{2}\left({x}+{a}\right)\left({y}+{b}\right)\:\mathrm{cos}\:\gamma}} \\ $$$$\angle{P}_{\mathrm{1}} =\angle{Q}_{\mathrm{2}} \\ $$$$\Rightarrow\frac{\left({z}+{c}\right)\:\mathrm{sin}\:\beta}{\:\sqrt{\left({z}+{c}\right)^{\mathrm{2}} +\left({x}+{a}\right)^{\mathrm{2}} −\mathrm{2}\left({z}+{c}\right)\left({x}+{a}\right)\:\mathrm{cos}\:\beta}}=\frac{\left({z}+{c}\right)\:\mathrm{sin}\:\alpha}{\:\sqrt{\left({y}+{b}\right)^{\mathrm{2}} +\left({z}+{c}\right)^{\mathrm{2}} −\mathrm{2}\left({y}+{b}\right)\left({z}+{c}\right)\:\mathrm{cos}\:\alpha}} \\ $$$$\Rightarrow\frac{\mathrm{sin}\:\alpha}{\mathrm{sin}\:\beta}=\frac{\sqrt{\left({y}+{b}\right)^{\mathrm{2}} +\left({z}+{c}\right)^{\mathrm{2}} −\mathrm{2}\left({y}+{b}\right)\left({z}+{c}\right)\:\mathrm{cos}\:\alpha}}{\:\sqrt{\left({z}+{c}\right)^{\mathrm{2}} +\left({x}+{a}\right)^{\mathrm{2}} −\mathrm{2}\left({z}+{c}\right)\left({x}+{a}\right)\:\mathrm{cos}\:\beta}}\:\:\:…\left({i}\right) \\ $$$$\angle{Q}_{\mathrm{1}} =\angle{R}_{\mathrm{2}} \\ $$$$\Rightarrow\frac{\left({x}+{a}\right)\:\mathrm{sin}\:\gamma}{\:\sqrt{\left({x}+{a}\right)^{\mathrm{2}} +\left({y}+{b}\right)^{\mathrm{2}} −\mathrm{2}\left({x}+{a}\right)\left({y}+{b}\right)\:\mathrm{cos}\:\gamma}}=\frac{\left({x}+{a}\right)\:\mathrm{sin}\:\beta}{\:\sqrt{\left({z}+{c}\right)^{\mathrm{2}} +\left({x}+{a}\right)^{\mathrm{2}} −\mathrm{2}\left({z}+{c}\right)\left({x}+{a}\right)\:\mathrm{cos}\:\beta}} \\ $$$$\Rightarrow\frac{\mathrm{sin}\:\beta}{\mathrm{sin}\:\gamma}=\frac{\sqrt{\left({z}+{c}\right)^{\mathrm{2}} +\left({x}+{a}\right)^{\mathrm{2}} −\mathrm{2}\left({z}+{c}\right)\left({x}+{a}\right)\:\mathrm{cos}\:\beta}}{\:\sqrt{\left({x}+{a}\right)^{\mathrm{2}} +\left({y}+{b}\right)^{\mathrm{2}} −\mathrm{2}\left({x}+{a}\right)\left({y}+{b}\right)\:\mathrm{cos}\:\gamma}}\:\:\:…\left({ii}\right) \\ $$$$\angle{R}_{\mathrm{1}} =\angle{P}_{\mathrm{2}} \\ $$$$\Rightarrow\frac{\left({y}+{b}\right)\:\mathrm{sin}\:\alpha}{\:\sqrt{\left({y}+{b}\right)^{\mathrm{2}} +\left({z}+{c}\right)^{\mathrm{2}} −\mathrm{2}\left({y}+{b}\right)\left({z}+{c}\right)\:\mathrm{cos}\:\alpha}}=\frac{\left({y}+{b}\right)\:\mathrm{sin}\:\gamma}{\:\sqrt{\left({x}+{a}\right)^{\mathrm{2}} +\left({y}+{b}\right)^{\mathrm{2}} −\mathrm{2}\left({x}+{a}\right)\left({y}+{b}\right)\:\mathrm{cos}\:\gamma}} \\ $$$$\Rightarrow\frac{\mathrm{sin}\:\gamma}{\mathrm{sin}\:\alpha}=\frac{\sqrt{\left({x}+{a}\right)^{\mathrm{2}} +\left({y}+{b}\right)^{\mathrm{2}} −\mathrm{2}\left({x}+{a}\right)\left({y}+{b}\right)\:\mathrm{cos}\:\gamma}}{\:\sqrt{\left({y}+{b}\right)^{\mathrm{2}} +\left({z}+{c}\right)^{\mathrm{2}} −\mathrm{2}\left({y}+{b}\right)\left({z}+{c}\right)\:\mathrm{cos}\:\alpha}}\:\:\:…\left({iii}\right) \\ $$$${two}\:{of}\:{these}\:{three}\:{equations}\:{can}\:{be} \\ $$$${solved}\:{to}\:{get}\:\alpha,\:\beta\:{and}\:\gamma. \\ $$