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Question Number 185495 by Mingma last updated on 22/Jan/23

Answered by mr W last updated on 22/Jan/23

Commented by mr W last updated on 23/Jan/23

$${(-\frac{1}{R}+\frac{2}{c}+\frac{1}{r})}^2=2(\frac{1}{R^2}+\frac{2}{c^2}+\frac{1}{r^2})$$$$\frac{1}{r^2}-(\frac{4}{c}-\frac{2}{R})\frac{1}{r}+\frac{1}{R}(\frac{1}{R}+\frac{4}{c})=0$$$$\frac{1}{a}+\frac{1}{b}=\frac{4}{c}-\frac{2}{R}$$$$\frac{1}{ab}=\frac{1}{R}(\frac{1}{R}+\frac{4}{c})=\frac{1}{R}(\frac{3}{R}+\frac{1}{a}+\frac{1}{b})$$$$\Rightarrow (a+b)R+3ab=R^2$$$$example:$$$$R=95$$$$b=28.5$$$$\Rightarrow a=\frac{(R-b)R}{R+3b}=\frac{(95-28.5)\times 95}{95+3\times 28.5}=35$$$$?=2a=70✓$$

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