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Question-52223

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Question Number 52223 by Tawa1 last updated on 04/Jan/19
Commented by tanmay.chaudhury50@gmail.com last updated on 05/Jan/19
Commented by Tawa1 last updated on 05/Jan/19
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Answered by mr W last updated on 05/Jan/19
Commented by Tawa1 last updated on 05/Jan/19
God bless you sir. I appreciate your time.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{time}.\: \\ $$
Commented by mr W last updated on 05/Jan/19
r=2 R=3 AB=a DC=R−r=3−2=1 AB=2r sin θ=a DE=r−r cos θ=r(1−cos θ) EC=DE+DC=r(1−cos θ)+R−r=((√3)/2)a ⇒r(1−cos θ)+R−r=((√3)/2)×2r sin θ ⇒2(1−cos θ)+1=2(√3) sin θ ⇒(√3) sin θ+cos θ=(3/2) ⇒((√3)/2) sin θ+(1/2)cos θ=(3/4) ⇒cos (π/6) sin θ+sin (π/6)cos θ=(3/4) ⇒sin (θ+(π/6))=(3/4) ⇒θ+(π/6)=sin^(−1) (3/4) or π−sin^(−1) (3/4) ⇒θ=sin^(−1) (3/4)−(π/6) (≈18.59°) or ⇒θ=((5π)/6)−sin^(−1) (3/4) (≈101.41°) i.e. two equilateral triangles are possible. a=2r sin θ=4 sin θ sin θ=sin (sin^(−1) (3/4)−(π/6))=(3/4)×((√3)/2)−((√7)/4)×(1/2)=((3(√3)−(√7))/8) sin θ=sin (((5π)/6)−sin^(−1) (3/4))=sin ((π/6)+sin^(−1) (3/4))=(3/4)×((√3)/2)+((√7)/4)×(1/2)=((3(√3)+(√7))/8) ⇒a=((3(√3)−(√7))/2) (≈1.275) or ⇒a=((3(√3)+(√7))/2) (≈3.921) product of possible side lengthes is P=((3(√3)−(√7))/2)×((3(√3)+(√7))/2)=((9×3−7)/4)=5 ⇒the answer is 5.
$${r}=\mathrm{2} \\ $$$${R}=\mathrm{3} \\ $$$${AB}={a} \\ $$$${DC}={R}−{r}=\mathrm{3}−\mathrm{2}=\mathrm{1} \\ $$$${AB}=\mathrm{2}{r}\:\mathrm{sin}\:\theta={a} \\ $$$${DE}={r}−{r}\:\mathrm{cos}\:\theta={r}\left(\mathrm{1}−\mathrm{cos}\:\theta\right) \\ $$$${EC}={DE}+{DC}={r}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)+{R}−{r}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{a}\: \\ $$$$\Rightarrow{r}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)+{R}−{r}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}×\mathrm{2}{r}\:\mathrm{sin}\:\theta \\ $$$$\Rightarrow\mathrm{2}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)+\mathrm{1}=\mathrm{2}\sqrt{\mathrm{3}}\:\mathrm{sin}\:\theta \\ $$$$\Rightarrow\sqrt{\mathrm{3}}\:\mathrm{sin}\:\theta+\mathrm{cos}\:\theta=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\Rightarrow\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\mathrm{sin}\:\theta+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\theta=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\Rightarrow\mathrm{cos}\:\frac{\pi}{\mathrm{6}}\:\mathrm{sin}\:\theta+\mathrm{sin}\:\frac{\pi}{\mathrm{6}}\mathrm{cos}\:\theta=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\Rightarrow\mathrm{sin}\:\left(\theta+\frac{\pi}{\mathrm{6}}\right)=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\Rightarrow\theta+\frac{\pi}{\mathrm{6}}=\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{3}}{\mathrm{4}}\:{or}\:\pi−\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\Rightarrow\theta=\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{3}}{\mathrm{4}}−\frac{\pi}{\mathrm{6}}\:\left(\approx\mathrm{18}.\mathrm{59}°\right)\:{or} \\ $$$$\Rightarrow\theta=\frac{\mathrm{5}\pi}{\mathrm{6}}−\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{3}}{\mathrm{4}}\:\left(\approx\mathrm{101}.\mathrm{41}°\right) \\ $$$${i}.{e}.\:{two}\:{equilateral}\:{triangles}\:{are}\:{possible}. \\ $$$$ \\ $$$${a}=\mathrm{2}{r}\:\mathrm{sin}\:\theta=\mathrm{4}\:\mathrm{sin}\:\theta \\ $$$$\mathrm{sin}\:\theta=\mathrm{sin}\:\left(\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{3}}{\mathrm{4}}−\frac{\pi}{\mathrm{6}}\right)=\frac{\mathrm{3}}{\mathrm{4}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−\frac{\sqrt{\mathrm{7}}}{\mathrm{4}}×\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{3}\sqrt{\mathrm{3}}−\sqrt{\mathrm{7}}}{\mathrm{8}} \\ $$$$\mathrm{sin}\:\theta=\mathrm{sin}\:\left(\frac{\mathrm{5}\pi}{\mathrm{6}}−\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{3}}{\mathrm{4}}\right)=\mathrm{sin}\:\left(\frac{\pi}{\mathrm{6}}+\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{3}}{\mathrm{4}}\right)=\frac{\mathrm{3}}{\mathrm{4}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}+\frac{\sqrt{\mathrm{7}}}{\mathrm{4}}×\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{3}\sqrt{\mathrm{3}}+\sqrt{\mathrm{7}}}{\mathrm{8}} \\ $$$$\Rightarrow{a}=\frac{\mathrm{3}\sqrt{\mathrm{3}}−\sqrt{\mathrm{7}}}{\mathrm{2}}\:\left(\approx\mathrm{1}.\mathrm{275}\right)\:{or} \\ $$$$\Rightarrow{a}=\frac{\mathrm{3}\sqrt{\mathrm{3}}+\sqrt{\mathrm{7}}}{\mathrm{2}}\:\left(\approx\mathrm{3}.\mathrm{921}\right) \\ $$$${product}\:{of}\:{possible}\:{side}\:{lengthes}\:{is} \\ $$$${P}=\frac{\mathrm{3}\sqrt{\mathrm{3}}−\sqrt{\mathrm{7}}}{\mathrm{2}}×\frac{\mathrm{3}\sqrt{\mathrm{3}}+\sqrt{\mathrm{7}}}{\mathrm{2}}=\frac{\mathrm{9}×\mathrm{3}−\mathrm{7}}{\mathrm{4}}=\mathrm{5} \\ $$$$ \\ $$$$\Rightarrow{the}\:{answer}\:{is}\:\mathrm{5}. \\ $$
Commented by mr W last updated on 05/Jan/19
alternative way: AB=a OE=(√(r^2 −(a^2 /4))) CE=R−(√(r^2 −(a^2 /4)))=(((√3)a)/2) 3−(√(4−(a^2 /4)))=(((√3)a)/2) (√(16−a^2 ))=6−(√3)a 16−a^2 =36−12(√3)a+3a^2 ⇒a^2 −3(√3)a+5=0 there are two solutions for a. the sum of both solutions is 3(√3) and the product of both solutions is 5. ⇒answer is 5.
$${alternative}\:{way}: \\ $$$${AB}={a} \\ $$$${OE}=\sqrt{{r}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{\mathrm{4}}} \\ $$$${CE}={R}−\sqrt{{r}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{\mathrm{4}}}=\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{2}} \\ $$$$\mathrm{3}−\sqrt{\mathrm{4}−\frac{{a}^{\mathrm{2}} }{\mathrm{4}}}=\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{2}} \\ $$$$\sqrt{\mathrm{16}−{a}^{\mathrm{2}} }=\mathrm{6}−\sqrt{\mathrm{3}}{a} \\ $$$$\mathrm{16}−{a}^{\mathrm{2}} =\mathrm{36}−\mathrm{12}\sqrt{\mathrm{3}}{a}+\mathrm{3}{a}^{\mathrm{2}} \\ $$$$\Rightarrow{a}^{\mathrm{2}} −\mathrm{3}\sqrt{\mathrm{3}}{a}+\mathrm{5}=\mathrm{0} \\ $$$${there}\:{are}\:{two}\:{solutions}\:{for}\:{a}.\:{the} \\ $$$${sum}\:{of}\:{both}\:{solutions}\:{is}\:\mathrm{3}\sqrt{\mathrm{3}}\:{and}\:{the} \\ $$$${product}\:{of}\:{both}\:{solutions}\:{is}\:\mathrm{5}. \\ $$$$ \\ $$$$\Rightarrow{answer}\:{is}\:\mathrm{5}. \\ $$

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