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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

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Answered by mr W last updated on 05/Jan/19

Commented by Tawa1 last updated on 05/Jan/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.\mathrm{I}\mathrm{appreciate}\mathrm{your}\mathrm{time}.$$

Commented by mr W last updated on 05/Jan/19

$$r=2$$$$R=3$$$$AB=a$$$$DC=R-r=3-2=1$$$$AB=2r\sin \theta =a$$$$DE=r-r\cos \theta =r(1-\cos \theta )$$$$EC=DE+DC=r(1-\cos \theta )+R-r=\frac{\sqrt{3}}{2}a$$$$\Rightarrow r(1-\cos \theta )+R-r=\frac{\sqrt{3}}{2}\times 2r\sin \theta$$$$\Rightarrow 2(1-\cos \theta )+1=2\sqrt{3}\sin \theta$$$$\Rightarrow \sqrt{3}\sin \theta +\cos \theta =\frac{3}{2}$$$$\Rightarrow \frac{\sqrt{3}}{2}\sin \theta +\frac{1}{2}\cos \theta =\frac{3}{4}$$$$\Rightarrow \cos \frac{\pi }{6}\sin \theta +\sin \frac{\pi }{6}\cos \theta =\frac{3}{4}$$$$\Rightarrow \sin (\theta +\frac{\pi }{6})=\frac{3}{4}$$$$\Rightarrow \theta +\frac{\pi }{6}={\sin }^{-1}\frac{3}{4}or\pi -{\sin }^{-1}\frac{3}{4}$$$$\Rightarrow \theta ={\sin }^{-1}\frac{3}{4}-\frac{\pi }{6}(\approx 18.59°)or$$$$\Rightarrow \theta =\frac{5\pi }{6}-{\sin }^{-1}\frac{3}{4}(\approx 101.41°)$$$$i.e.twoequilateraltrianglesarepossible.$$$$$$$$a=2r\sin \theta =4\sin \theta$$$$\sin \theta =\sin ({\sin }^{-1}\frac{3}{4}-\frac{\pi }{6})=\frac{3}{4}\times \frac{\sqrt{3}}{2}-\frac{\sqrt{7}}{4}\times \frac{1}{2}=\frac{3\sqrt{3}-\sqrt{7}}{8}$$$$\sin \theta =\sin (\frac{5\pi }{6}-{\sin }^{-1}\frac{3}{4})=\sin (\frac{\pi }{6}+{\sin }^{-1}\frac{3}{4})=\frac{3}{4}\times \frac{\sqrt{3}}{2}+\frac{\sqrt{7}}{4}\times \frac{1}{2}=\frac{3\sqrt{3}+\sqrt{7}}{8}$$$$\Rightarrow a=\frac{3\sqrt{3}-\sqrt{7}}{2}(\approx 1.275)or$$$$\Rightarrow a=\frac{3\sqrt{3}+\sqrt{7}}{2}(\approx 3.921)$$$$productofpossiblesidelengthesis$$$$P=\frac{3\sqrt{3}-\sqrt{7}}{2}\times \frac{3\sqrt{3}+\sqrt{7}}{2}=\frac{9\times 3-7}{4}=5$$$$$$$$\Rightarrow theansweris5.$$

Commented by mr W last updated on 05/Jan/19

$$alternativeway:$$$$AB=a$$$$OE=\sqrt{r^2-\frac{a^2}{4}}$$$$CE=R-\sqrt{r^2-\frac{a^2}{4}}=\frac{\sqrt{3}a}{2}$$$$3-\sqrt{4-\frac{a^2}{4}}=\frac{\sqrt{3}a}{2}$$$$\sqrt{16-a^2}=6-\sqrt{3}a$$$$16-a^2=36-12\sqrt{3}a+3a^2$$$$\Rightarrow a^2-3\sqrt{3}a+5=0$$$$therearetwosolutionsfora.the$$$$sumofbothsolutionsis3\sqrt{3}andthe$$$$productofbothsolutionsis5.$$$$$$$$\Rightarrow answeris5.$$

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