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Question Number 189861 by Rupesh123 last updated on 23/Mar/23

Answered by mr W last updated on 23/Mar/23

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

$$A=2$$$$B=4$$$$C=5$$$$D=3$$$$A=\frac{pa}{2}\Rightarrow a=\frac{2A}{p}$$$$B=\frac{qb}{2}\Rightarrow b=\frac{2B}{q}$$$$C=\frac{(a+b)c}{2}\Rightarrow c=\frac{2C}{a+b}$$$$D=\frac{(a+b)d}{2}\Rightarrow d=\frac{2D}{a+b}$$$$\frac{c+d-p}{a}=\frac{q-(c+d)}{b}$$$$(c+d)(\frac{1}{a}+\frac{1}{b})=\frac{p}{a}+\frac{q}{b}$$$$\left(\frac{2C+2D}{a+b}\right)\left(\frac{a+b}{ab}\right)=\frac{aq+pb}{ab}$$$$2(C+D)=\frac{2Aq}{p}+\frac{2Bp}{q}$$$$C+D=\frac{Aq}{p}+\frac{Bp}{q}$$$$let\lambda =\frac{q}{p}$$$$A\lambda +\frac{B}{\lambda }-(C+D)=0$$$$A{\lambda }^2-(C+D)\lambda +B=0$$$$\lambda =\frac{C+D\pm \sqrt{{(C+D)}^2-4AB}}{2A}$$$$\lambda =\frac{8\pm \sqrt{8^2-4\times 8}}{2\times 2}=2\pm \sqrt{2}$$$$sincegivenAB>CD,\Rightarrow \lambda =2+\sqrt{2}✓$$

Commented by Rupesh123 last updated on 23/Mar/23

Perfect ��

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