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Question Number 104411 by ajfour last updated on 21/Jul/20

Commented by ajfour last updated on 21/Jul/20

$$Relate\mathit{a},\mathit{b},\mathit{c}\left(whentheyrreal\right).$$

Commented by mr W last updated on 21/Jul/20

$$a+b=\frac{b+c}{2}$$$$c+a=\frac{\sqrt{3}(b+c)}{2}$$$$with\alpha =\frac{a}{c},\beta =\frac{b}{c}$$$$\Rightarrow 2\alpha +\beta =1$$$$\Rightarrow -2\alpha +\sqrt{3}\beta =2-\sqrt{3}$$$$\Rightarrow \alpha =2-\sqrt{3}\approx 0.268$$$$\Rightarrow \beta =2\sqrt{3}-3\approx 0.464$$

Commented by ajfour last updated on 21/Jul/20

$$It{isn}^{\prime }tlikethatSir...$$$$pleaseviewQ.104445$$

Commented by mr W last updated on 21/Jul/20

$$youarerightsir!whatishowedis$$$$onlyaspecialcase.$$$$generallywehave:$$$${(a+c)}^2={(a+b)}^2+{(b+c)}^2-(a+b)(b+c)$$$$\Rightarrow b^2+(a+c)b-3ac=0$$$$or$$$$\Rightarrow c=\frac{(a+b)b}{3a-b}$$

Commented by ajfour last updated on 21/Jul/20

$$Excellentandverypragmaticnotice$$$$Sir,ifollowitwell,thanksalot.$$

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