Question-164018

Question Number 164018 by cortano1 last updated on 13/Jan/22

Answered by mr W last updated on 13/Jan/22

Commented by cortano1 last updated on 13/Jan/22

s u p e r b s o l u t i o n

Commented by mr W last updated on 13/Jan/22

b = 2 R

a = \sqrt{{(R + r)}^{2} - {(R - r)}^{2}} + \sqrt{{(R + r)}^{2} - {(2 R - r)}^{2}}

a = 2 \sqrt{R r} + \sqrt{3 R (2 r - R)}

B C = 2 \sqrt{{(R + r)}^{2} - r^{2}} = 2 \sqrt{R (R + 2 r)}

a = \sqrt{{B C}^{2} - R^{2}} = \sqrt{4 R (R + 2 r) - R^{2}} = \sqrt{R (3 R + 8 r)}

2 \sqrt{R r} + \sqrt{3 R (2 r - R)} = \sqrt{R (3 R + 8 r)}

l e t λ = \frac{R}{r}

2 + \sqrt{3 (2 - λ)} = \sqrt{3 λ + 8}

3 (2 - λ) = 3 λ + 8 + 4 - 4 \sqrt{3 λ + 8}

3 (λ + 1) = 2 \sqrt{3 λ + 8}

9 (λ^{2} + 2 λ + 1) = 4 (3 λ + 8)

9 λ^{2} + 6 λ - 23 = 0

λ = \frac{- 1 + 2 \sqrt{6}}{3}

\Rightarrow \frac{R}{r} = \frac{2 \sqrt{6} - 1}{3} ✓

Commented by Tawa11 last updated on 13/Jan/22

Great sir

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