Question Number 58675 by ajfour last updated on 27/Apr/19
Commented by ajfour last updated on 27/Apr/19
$$\mathrm{ABCD}\:\mathrm{is}\:\mathrm{a}\:\mathrm{square};\:\mathrm{while},\:\mathrm{PQM}\:\mathrm{is} \\ $$$$\mathrm{a}\:\mathrm{sector}.\:\mathrm{Find}\:\mathrm{in}\:\mathrm{what}\:\mathrm{proportion} \\ $$$$\mathrm{are}\:\mathrm{radii}\:\mathrm{b},\:\mathrm{r},\:\mathrm{g}\:\left(\mathrm{blue},\:\mathrm{red},\:\mathrm{green}\right. \\ $$$$\left.\mathrm{respectively}\right). \\ $$
Commented by ajfour last updated on 27/Apr/19
Commented by ajfour last updated on 27/Apr/19
$$\mathrm{let}\:\mathrm{square}\:\mathrm{side}\:\mathrm{be}\:\mathrm{a}. \\ $$$$\Rightarrow\:\frac{\mathrm{r}}{\mathrm{sin}\:\alpha}+\mathrm{r}+\mathrm{2g}=\mathrm{a} \\ $$$$\:\:\:\:\:\left(\mathrm{g}+\mathrm{b}\right)\mathrm{cos}\:\beta+\mathrm{b}=\mathrm{g} \\ $$$$\:\:\:\:\frac{\mathrm{r}}{\mathrm{sin}\:\alpha}+\mathrm{r}+\mathrm{g}\:=\frac{\mathrm{g}}{\mathrm{cos}\:\alpha} \\ $$$$\:\:…….. \\ $$$$\:\:…….. \\ $$
Commented by mr W last updated on 01/May/19
$${MP}={MQ}={MN}={PQ}={a} \\ $$$$\Rightarrow\Delta{MPQ}={equilateral} \\ $$$$\Rightarrow\alpha=\mathrm{60}° \\ $$$${EM}={a}−{g} \\ $$$${EH}={g} \\ $$$${EM}=\mathrm{2}{EH} \\ $$$${a}−{g}=\mathrm{2}{g} \\ $$$$\Rightarrow{g}=\frac{{a}}{\mathrm{3}} \\ $$$${FG}={r} \\ $$$${FM}={a}−\mathrm{2}{g}−{r}=\mathrm{2}{FG}=\mathrm{2}{r} \\ $$$$\Rightarrow{r}=\frac{{a}−\mathrm{2}{g}}{\mathrm{3}} \\ $$$$\Rightarrow{r}=\frac{{a}}{\mathrm{9}} \\ $$$${ML}={a}−{b} \\ $$$${MJ}=\sqrt{{ML}^{\mathrm{2}} −{JL}^{\mathrm{2}} }=\sqrt{\left({a}−{b}\right)^{\mathrm{2}} −{b}^{\mathrm{2}} }=\sqrt{{a}\left({a}−\mathrm{2}{b}\right)} \\ $$$${MH}=\sqrt{\mathrm{3}}{EH}=\sqrt{\mathrm{3}}{g}=\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{3}} \\ $$$${HJ}={MJ}−{MH}=\sqrt{{a}\left({a}−\mathrm{2}{b}\right)}−\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{3}} \\ $$$${HJ}^{\mathrm{2}} =\left({g}+{b}\right)^{\mathrm{2}} −\left({g}−{b}\right)^{\mathrm{2}} =\mathrm{4}{gb}=\frac{\mathrm{4}{ab}}{\mathrm{3}} \\ $$$$\Rightarrow{a}\left({a}−\mathrm{2}{b}\right)+\frac{{a}^{\mathrm{2}} }{\mathrm{3}}−\frac{\mathrm{2}{a}}{\mathrm{3}}\sqrt{\mathrm{3}{a}\left({a}−\mathrm{2}{b}\right)}=\frac{\mathrm{4}{ab}}{\mathrm{3}} \\ $$$$\Rightarrow\frac{\mathrm{4}{a}^{\mathrm{2}} }{\mathrm{3}}−\frac{\mathrm{2}{a}}{\mathrm{3}}\sqrt{\mathrm{3}{a}\left({a}−\mathrm{2}{b}\right)}=\frac{\mathrm{10}{ab}}{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{2}{a}−\sqrt{\mathrm{3}{a}\left({a}−\mathrm{2}{b}\right)}=\mathrm{5}{b} \\ $$$$\Rightarrow\mathrm{2}{a}−\mathrm{5}{b}=\sqrt{\mathrm{3}{a}\left({a}−\mathrm{2}{b}\right)} \\ $$$$\Rightarrow\mathrm{4}{a}^{\mathrm{2}} +\mathrm{25}{b}^{\mathrm{2}} −\mathrm{20}{ab}=\mathrm{3}{a}^{\mathrm{2}} −\mathrm{6}{ab} \\ $$$$\Rightarrow\mathrm{25}{b}^{\mathrm{2}} −\mathrm{14}{ab}+{a}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{b}=\frac{\mathrm{7}−\mathrm{2}\sqrt{\mathrm{6}}}{\mathrm{25}}{a}\:\approx\frac{{a}}{\mathrm{11}.\mathrm{9}}=\mathrm{0}.\mathrm{084}{a}<{r} \\ $$$$ \\ $$$$\Rightarrow{g}/{r}/{b}=\mathrm{3}/\mathrm{1}/\mathrm{0}.\mathrm{756} \\ $$
Commented by ajfour last updated on 01/May/19
$$\mathcal{VERY}\:\mathcal{WISE}\:\mathrm{n}\:\mathcal{NICE}\:\:\mathcal{S}{ir}. \\ $$$${Thank}\:{you}\:{very}\:{much}! \\ $$