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Question Number 54170 by ajfour last updated on 30/Jan/19

Commented by ajfour last updated on 30/Jan/19

$$Determinebintermsofa,p,R.$$

Answered by ajfour last updated on 30/Jan/19

$$DF=\sqrt{{(a+b)}^2-{(b-a)}^2}=2\sqrt{ab}$$$$DE=\sqrt{{(R-a)}^2-{(a-p)}^2}=\sqrt{(R-p)(R+p-2a)}$$$$EF=\sqrt{{(R-b)}^2-{(b-p)}^2}=\sqrt{(R-p)(R+p-2b)}$$$$DE+EF=DF,\Rightarrow$$$$\sqrt{(R-p)(R+p-2a)}+\sqrt{(R-p)(R+p-2b)}$$$$=2\sqrt{ab}$$$$let\frac{2a}{R-p}=\alpha ,\frac{2b}{R-p}=\beta ,\frac{R+p}{R-p}=c.$$$$\Rightarrow \sqrt{c-\alpha }+\sqrt{c-\beta }=\sqrt{\alpha \beta }$$$$squaring$$$$2c-(\alpha +\beta )+2\sqrt{c^2-(\alpha +\beta )c+\alpha \beta }=\alpha \beta$$$$\Rightarrow 4(c^2-(\alpha +\beta )c+\alpha \beta )={(\alpha \beta -2c+\alpha +\beta )}^2$$$$\Rightarrow 4c^2-4c(\alpha +\beta )+4\alpha \beta$$$$={\alpha }^2{\beta }^2+4c^2+{(\alpha +\beta )}^2-4c\alpha \beta$$$$-2\alpha \beta (\alpha +\beta )-4c(\alpha +\beta )$$$$\Rightarrow {(\alpha -\beta )}^2+{\alpha }^2{\beta }^2=2\alpha \beta (\alpha +\beta +2c)$$$$\Rightarrow$$$${\beta }^2({\alpha }^2-2\alpha +1)-2\alpha (1+\alpha +2c)+{\alpha }^2=0$$$$\beta =\frac{\alpha (1+\alpha +2c)+\sqrt{{(1+\alpha +2c)}^2-{\alpha }^2{(\alpha -1)}^2}}{{(\alpha -1)}^2}$$$$..$$

Commented by mr W last updated on 30/Jan/19

$$goodsolutionsir!$$

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