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Question Number 108643 by ajfour last updated on 18/Aug/20

Answered by mr W last updated on 18/Aug/20

Commented by mr W last updated on 18/Aug/20

$$\frac{r}{AD}=\frac{BC}{AC}=\frac{a}{\sqrt{a^2+c^2}}$$$$\frac{r}{c-r}=\frac{a}{\sqrt{a^2+c^2}}$$$$\Rightarrow r=\frac{ac}{a+\sqrt{a^2+c^2}}$$$$similarly$$$$\Rightarrow R=\frac{ac}{c+\sqrt{a^2+c^2}}$$$$\cos \mathrm{\angle }DBP=\frac{BP}{2r}$$$$\cos \mathrm{\angle }EBP=\frac{BP}{2R}=\sin \mathrm{\angle }DBP$$$$\frac{{BP}^2}{4}(\frac{1}{r^2}+\frac{1}{R^2})=1$$$${BP}^2[{(a+\sqrt{a^2+c^2})}^2+{(c+\sqrt{a^2+c^2})}^2]=4a^2{c}^2$$$${BP}^2[3a^2+3c^2+2(a+c)\sqrt{a^2+c^2}]=4a^2{c}^2$$$$\Rightarrow BP=\frac{2ac}{\sqrt{3(a^2+c^2)+2(a+c)\sqrt{a^2+c^2}}}$$

Commented by ajfour last updated on 18/Aug/20

$$\mathcal{A}wesome!Greatwork,$$$$thanksalotSir.$$

Answered by ajfour last updated on 19/Aug/20

$$R(1+\frac{\sqrt{a^2+c^2}}{c})=a$$$$r(1+\frac{\sqrt{a^2+c^2}}{a})=c$$$${(x-R)}^2+y^2=R^2...\left(i\right)$$$$x^2+{(y-r)}^2=r^2....\left(ii\right)$$$$\Rightarrow 2Rx-2ry=0$$$$\Rightarrow y=\left(\frac{R}{r}\right)x$$$${BP}^{}=\sqrt{x^2+y^2}=x\left(\frac{\sqrt{R^2+r^2}}{r}\right)$$$$Adding\left(i\right)\mathrm{\&}\left(ii\right)$$$$x^2+y^2=Rx+ry$$$$\Rightarrow x\left(\frac{r^2+R^2}{r^2}\right)=2R$$$$\Rightarrow BP=\frac{2Rr}{\sqrt{R^2+r^2}}=\frac{2}{\sqrt{\frac{1}{r^2}+\frac{1}{R^2}}}$$$$=\frac{2}{\sqrt{{(\frac{1}{c}+\frac{\sqrt{a^2+c^2}}{ac})}^2+{(\frac{1}{a}+\frac{\sqrt{a^2+c^2}}{ac})}^2}}$$$$BP=\frac{2ac}{\sqrt{3(a^2+c^2)+2(a+c)\sqrt{a^2+c^2}}}.$$

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