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Question Number 213859 by ajfour last updated on 19/Nov/24

Commented by mr W last updated on 19/Nov/24

$$0<AB<1$$$$nomaximumorminimumexists.$$

Commented by ajfour last updated on 19/Nov/24

Commented by ajfour last updated on 19/Nov/24

nice, thank you.

Commented by A5T last updated on 19/Nov/24

$$\left(Fortheseconddiagram\right)$$$$LetcirclewithcentreA,Bhaveradiusa,bresp.$$$$\sqrt{{(a+b)}^2-b^2}+b+a=1\Rightarrow \sqrt{a(a+2b)}=1-a-b$$$$\Rightarrow a^2+2ab=1+a^2+b^2-2a-2b+2ab$$$$\Rightarrow 1+b^2-2a-2b=0\Rightarrow a+b=\frac{1+b^2}{2}$$$${(1-b)}^2=b^2+b^2\Rightarrow b^2+2b-1=0\Rightarrow b=-1+\sqrt{2}$$$$\Rightarrow AB=a+b=2-\sqrt{2}$$

Commented by ajfour last updated on 19/Nov/24

$$FindmaximumAB.$$

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