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Question Number 58675 by ajfour last updated on 27/Apr/19

Commented by ajfour last updated on 27/Apr/19

ABCD is a square; while, PQM is a sector. Find in what proportion are radii b, r, g (blue, red, green respectively).

$$\mathrm{ABCD}\:\mathrm{is}\:\mathrm{a}\:\mathrm{square};\:\mathrm{while},\:\mathrm{PQM}\:\mathrm{is} \\ $$$$\mathrm{a}\:\mathrm{sector}.\:\mathrm{Find}\:\mathrm{in}\:\mathrm{what}\:\mathrm{proportion} \\ $$$$\mathrm{are}\:\mathrm{radii}\:\mathrm{b},\:\mathrm{r},\:\mathrm{g}\:\left(\mathrm{blue},\:\mathrm{red},\:\mathrm{green}\right. \\ $$$$\left.\mathrm{respectively}\right). \\ $$

Commented by ajfour last updated on 27/Apr/19

Commented by ajfour last updated on 27/Apr/19

let square side be a. ⇒ (r/(sin α))+r+2g=a (g+b)cos β+b=g (r/(sin α))+r+g =(g/(cos α)) ........ ........

$$\mathrm{let}\:\mathrm{square}\:\mathrm{side}\:\mathrm{be}\:\mathrm{a}. \\ $$$$\Rightarrow\:\frac{\mathrm{r}}{\mathrm{sin}\:\alpha}+\mathrm{r}+\mathrm{2g}=\mathrm{a} \\ $$$$\:\:\:\:\:\left(\mathrm{g}+\mathrm{b}\right)\mathrm{cos}\:\beta+\mathrm{b}=\mathrm{g} \\ $$$$\:\:\:\:\frac{\mathrm{r}}{\mathrm{sin}\:\alpha}+\mathrm{r}+\mathrm{g}\:=\frac{\mathrm{g}}{\mathrm{cos}\:\alpha} \\ $$$$\:\:........ \\ $$$$\:\:........ \\ $$

Commented by mr W last updated on 01/May/19

MP=MQ=MN=PQ=a ⇒ΔMPQ=equilateral ⇒α=60° EM=a−g EH=g EM=2EH a−g=2g ⇒g=(a/3) FG=r FM=a−2g−r=2FG=2r ⇒r=((a−2g)/3) ⇒r=(a/9) ML=a−b MJ=(√(ML^2 −JL^2 ))=(√((a−b)^2 −b^2 ))=(√(a(a−2b))) MH=(√3)EH=(√3)g=(((√3)a)/3) HJ=MJ−MH=(√(a(a−2b)))−(((√3)a)/3) HJ^2 =(g+b)^2 −(g−b)^2 =4gb=((4ab)/3) ⇒a(a−2b)+(a^2 /3)−((2a)/3)(√(3a(a−2b)))=((4ab)/3) ⇒((4a^2 )/3)−((2a)/3)(√(3a(a−2b)))=((10ab)/3) ⇒2a−(√(3a(a−2b)))=5b ⇒2a−5b=(√(3a(a−2b))) ⇒4a^2 +25b^2 −20ab=3a^2 −6ab ⇒25b^2 −14ab+a^2 =0 ⇒b=((7−2(√6))/(25))a ≈(a/(11.9))=0.084a<r ⇒g/r/b=3/1/0.756

$${MP}={MQ}={MN}={PQ}={a} \\ $$$$\Rightarrow\Delta{MPQ}={equilateral} \\ $$$$\Rightarrow\alpha=\mathrm{60}° \\ $$$${EM}={a}−{g} \\ $$$${EH}={g} \\ $$$${EM}=\mathrm{2}{EH} \\ $$$${a}−{g}=\mathrm{2}{g} \\ $$$$\Rightarrow{g}=\frac{{a}}{\mathrm{3}} \\ $$$${FG}={r} \\ $$$${FM}={a}−\mathrm{2}{g}−{r}=\mathrm{2}{FG}=\mathrm{2}{r} \\ $$$$\Rightarrow{r}=\frac{{a}−\mathrm{2}{g}}{\mathrm{3}} \\ $$$$\Rightarrow{r}=\frac{{a}}{\mathrm{9}} \\ $$$${ML}={a}−{b} \\ $$$${MJ}=\sqrt{{ML}^{\mathrm{2}} −{JL}^{\mathrm{2}} }=\sqrt{\left({a}−{b}\right)^{\mathrm{2}} −{b}^{\mathrm{2}} }=\sqrt{{a}\left({a}−\mathrm{2}{b}\right)} \\ $$$${MH}=\sqrt{\mathrm{3}}{EH}=\sqrt{\mathrm{3}}{g}=\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{3}} \\ $$$${HJ}={MJ}−{MH}=\sqrt{{a}\left({a}−\mathrm{2}{b}\right)}−\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{3}} \\ $$$${HJ}^{\mathrm{2}} =\left({g}+{b}\right)^{\mathrm{2}} −\left({g}−{b}\right)^{\mathrm{2}} =\mathrm{4}{gb}=\frac{\mathrm{4}{ab}}{\mathrm{3}} \\ $$$$\Rightarrow{a}\left({a}−\mathrm{2}{b}\right)+\frac{{a}^{\mathrm{2}} }{\mathrm{3}}−\frac{\mathrm{2}{a}}{\mathrm{3}}\sqrt{\mathrm{3}{a}\left({a}−\mathrm{2}{b}\right)}=\frac{\mathrm{4}{ab}}{\mathrm{3}} \\ $$$$\Rightarrow\frac{\mathrm{4}{a}^{\mathrm{2}} }{\mathrm{3}}−\frac{\mathrm{2}{a}}{\mathrm{3}}\sqrt{\mathrm{3}{a}\left({a}−\mathrm{2}{b}\right)}=\frac{\mathrm{10}{ab}}{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{2}{a}−\sqrt{\mathrm{3}{a}\left({a}−\mathrm{2}{b}\right)}=\mathrm{5}{b} \\ $$$$\Rightarrow\mathrm{2}{a}−\mathrm{5}{b}=\sqrt{\mathrm{3}{a}\left({a}−\mathrm{2}{b}\right)} \\ $$$$\Rightarrow\mathrm{4}{a}^{\mathrm{2}} +\mathrm{25}{b}^{\mathrm{2}} −\mathrm{20}{ab}=\mathrm{3}{a}^{\mathrm{2}} −\mathrm{6}{ab} \\ $$$$\Rightarrow\mathrm{25}{b}^{\mathrm{2}} −\mathrm{14}{ab}+{a}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{b}=\frac{\mathrm{7}−\mathrm{2}\sqrt{\mathrm{6}}}{\mathrm{25}}{a}\:\approx\frac{{a}}{\mathrm{11}.\mathrm{9}}=\mathrm{0}.\mathrm{084}{a}<{r} \\ $$$$ \\ $$$$\Rightarrow{g}/{r}/{b}=\mathrm{3}/\mathrm{1}/\mathrm{0}.\mathrm{756} \\ $$

Commented by ajfour last updated on 01/May/19

VERY WISE n NICE Sir. Thank you very much!

$$\mathcal{VERY}\:\mathcal{WISE}\:\mathrm{n}\:\mathcal{NICE}\:\:\mathcal{S}{ir}. \\ $$$${Thank}\:{you}\:{very}\:{much}! \\ $$

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