Question Number 26205 by ajfour last updated on 22/Dec/17
Commented by ajfour last updated on 22/Dec/17
$${Find}\:{the}\:{area}\:{of}\:{the}\:{uncoloured} \\ $$$${region}\:{inside}\:{triangle}\:{in}\:{terms} \\ $$$${of}\:{radii}\:{a}\:,\:{and}\:{b}\:{of}\:{the}\:{lower} \\ $$$${circles}. \\ $$
Answered by mrW1 last updated on 22/Dec/17
![One side of the triangle is 2(a+b). The other two sides are d and the radius of the big circle is c. d=b+c. (b+c)^2 =(a+b)^2 +(a+c)^2 b^2 +2bc+c^2 =a^2 +b^2 +2ab+a^2 +2ac+c^2 (b−a)c=a(b+a) ⇒c=(((b+a)a)/(b−a))>0 ⇒b>a angle of triangle at center of circle c: tan (θ/2)=((b+a)/(a+c))=((b+a)/(a+(((b+a)a)/(b−a))))=((b^2 −a^2 )/(2ab)) ⇒θ=2tan^(−1) ((b^2 −a^2 )/(2ab))=π−2ϕ angle of triangle at center of circle b: tan ϕ=((a+c)/(b+a))=((2ab)/(b^2 −a^2 )) ⇒ϕ=tan^(−1) ((2ab)/(b^2 −a^2 )) area of triangle: A_1 =((2(a+b)(c+a))/2)=(a+b)(((b+a)/(b−a))+1)a=((2ab(a+b))/(b−a)) area of part of triangle a: A_2 =((πa^2 )/2) area of parts of triangle b: A_3 =((2ϕ)/(2π))×πb^2 =ϕb^2 =b^2 tan^(−1) ((2ab)/(b^2 −a^2 )) area of part of triangle c: A_4 =(θ/(2π))×πc^2 =((π−2ϕ)/2)×c^2 =((π/2)−tan^(−1) ((2ab)/(b^2 −a^2 )))×((a^2 (a+b)^2 )/((b−a)^2 )) area of the rest: A=A_1 −(A_2 +A_3 +A_4 ) =((2ab(a+b))/(b−a))−((πa^2 )/2)−b^2 tan^(−1) ((2ab)/(b^2 −a^2 ))−((π/2)−tan^(−1) ((2ab)/(b^2 −a^2 )))×((a^2 (a+b)^2 )/((b−a)^2 )) =((2ab(a+b))/(b−a))−((πa^2 )/2)[1+(((a+b)^2 )/((b−a)^2 ))]−[b^2 −((a^2 (a+b)^2 )/((b−a)^2 ))] tan^(−1) ((2ab)/(b^2 −a^2 )) =((2ab(a+b))/(b−a))−πa^2 [((a^2 +b^2 )/((b−a)^2 ))]−[(((a^2 +b^2 )(b^2 −2ab−a^2 ))/((b−a)^2 ))] tan^(−1) ((2ab)/(b^2 −a^2 )) ⇒A=((2ab(a+b))/(b−a))−((a^2 +b^2 )/((b−a)^2 ))×[πa^2 +(b^2 −a^2 −2ab) tan^(−1) ((2ab)/(b^2 −a^2 ))]](Q26212.png)
$${One}\:{side}\:{of}\:{the}\:{triangle}\:{is}\:\mathrm{2}\left({a}+{b}\right). \\ $$$${The}\:{other}\:{two}\:{sides}\:{are}\:{d}\:{and}\:{the} \\ $$$${radius}\:{of}\:{the}\:{big}\:{circle}\:{is}\:{c}. \\ $$$${d}={b}+{c}. \\ $$$$\left({b}+{c}\right)^{\mathrm{2}} =\left({a}+{b}\right)^{\mathrm{2}} +\left({a}+{c}\right)^{\mathrm{2}} \\ $$$${b}^{\mathrm{2}} +\mathrm{2}{bc}+{c}^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{ab}+{a}^{\mathrm{2}} +\mathrm{2}{ac}+{c}^{\mathrm{2}} \\ $$$$\left({b}−{a}\right){c}={a}\left({b}+{a}\right) \\ $$$$\Rightarrow{c}=\frac{\left({b}+{a}\right){a}}{{b}−{a}}>\mathrm{0} \\ $$$$\Rightarrow{b}>{a} \\ $$$$ \\ $$$${angle}\:{of}\:{triangle}\:{at}\:{center}\:{of}\:{circle}\:{c}: \\ $$$$\mathrm{tan}\:\frac{\theta}{\mathrm{2}}=\frac{{b}+{a}}{{a}+{c}}=\frac{{b}+{a}}{{a}+\frac{\left({b}+{a}\right){a}}{{b}−{a}}}=\frac{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{ab}} \\ $$$$\Rightarrow\theta=\mathrm{2tan}^{−\mathrm{1}} \frac{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{ab}}=\pi−\mathrm{2}\varphi \\ $$$$ \\ $$$${angle}\:{of}\:{triangle}\:{at}\:{center}\:{of}\:{circle}\:{b}: \\ $$$$\mathrm{tan}\:\varphi=\frac{{a}+{c}}{{b}+{a}}=\frac{\mathrm{2}{ab}}{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} } \\ $$$$\Rightarrow\varphi=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{2}{ab}}{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} } \\ $$$$ \\ $$$${area}\:{of}\:{triangle}: \\ $$$${A}_{\mathrm{1}} =\frac{\mathrm{2}\left({a}+{b}\right)\left({c}+{a}\right)}{\mathrm{2}}=\left({a}+{b}\right)\left(\frac{{b}+{a}}{{b}−{a}}+\mathrm{1}\right){a}=\frac{\mathrm{2}{ab}\left({a}+{b}\right)}{{b}−{a}} \\ $$$${area}\:{of}\:{part}\:{of}\:{triangle}\:{a}: \\ $$$${A}_{\mathrm{2}} =\frac{\pi{a}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${area}\:{of}\:{parts}\:{of}\:{triangle}\:{b}: \\ $$$${A}_{\mathrm{3}} =\frac{\mathrm{2}\varphi}{\mathrm{2}\pi}×\pi{b}^{\mathrm{2}} =\varphi{b}^{\mathrm{2}} ={b}^{\mathrm{2}} \:\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{2}{ab}}{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} } \\ $$$${area}\:{of}\:{part}\:{of}\:{triangle}\:{c}: \\ $$$${A}_{\mathrm{4}} =\frac{\theta}{\mathrm{2}\pi}×\pi{c}^{\mathrm{2}} =\frac{\pi−\mathrm{2}\varphi}{\mathrm{2}}×{c}^{\mathrm{2}} \\ $$$$=\left(\frac{\pi}{\mathrm{2}}−\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{2}{ab}}{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }\right)×\frac{{a}^{\mathrm{2}} \left({a}+{b}\right)^{\mathrm{2}} }{\left({b}−{a}\right)^{\mathrm{2}} } \\ $$$$ \\ $$$${area}\:{of}\:{the}\:{rest}: \\ $$$${A}={A}_{\mathrm{1}} −\left({A}_{\mathrm{2}} +{A}_{\mathrm{3}} +{A}_{\mathrm{4}} \right) \\ $$$$=\frac{\mathrm{2}{ab}\left({a}+{b}\right)}{{b}−{a}}−\frac{\pi{a}^{\mathrm{2}} }{\mathrm{2}}−{b}^{\mathrm{2}} \:\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{2}{ab}}{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }−\left(\frac{\pi}{\mathrm{2}}−\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{2}{ab}}{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }\right)×\frac{{a}^{\mathrm{2}} \left({a}+{b}\right)^{\mathrm{2}} }{\left({b}−{a}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2}{ab}\left({a}+{b}\right)}{{b}−{a}}−\frac{\pi{a}^{\mathrm{2}} }{\mathrm{2}}\left[\mathrm{1}+\frac{\left({a}+{b}\right)^{\mathrm{2}} }{\left({b}−{a}\right)^{\mathrm{2}} }\right]−\left[{b}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} \left({a}+{b}\right)^{\mathrm{2}} }{\left({b}−{a}\right)^{\mathrm{2}} }\right]\:\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{2}{ab}}{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2}{ab}\left({a}+{b}\right)}{{b}−{a}}−\pi{a}^{\mathrm{2}} \left[\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{\left({b}−{a}\right)^{\mathrm{2}} }\right]−\left[\frac{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left({b}^{\mathrm{2}} −\mathrm{2}{ab}−{a}^{\mathrm{2}} \right)}{\left({b}−{a}\right)^{\mathrm{2}} }\right]\:\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{2}{ab}}{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} } \\ $$$$\Rightarrow{A}=\frac{\mathrm{2}{ab}\left({a}+{b}\right)}{{b}−{a}}−\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{\left({b}−{a}\right)^{\mathrm{2}} }×\left[\pi{a}^{\mathrm{2}} +\left({b}^{\mathrm{2}} −{a}^{\mathrm{2}} −\mathrm{2}{ab}\right)\:\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{2}{ab}}{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }\right] \\ $$
Commented by ajfour last updated on 22/Dec/17
$${Great}\:{Sir},\:{Magnificient}\:! \\ $$