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Question Number 65235 by ajfour last updated on 26/Jul/19

Commented by ajfour last updated on 26/Jul/19

$$Findp,q,rintermsofa,b,c.$$

Commented by ajfour last updated on 27/Jul/19

$$doexplainabitsir!thanks.$$

Answered by mr W last updated on 27/Jul/19

Commented by ajfour last updated on 27/Jul/19

$$\sin \beta =\cos A=\frac{b^2+c^2-a^2}{2bc}is$$$$straightawaytrueSir!$$$$as\beta +A=\pi /2.$$

Commented by mr W last updated on 27/Jul/19

$$R=radiusofcircumcircle$$$$\sin \beta =\frac{{AB}^{\prime }}{AB}with{BB}^{\prime }\mathrm{\perp }AC$$$${AB}^{\prime }=\frac{b^2+c^2-a^2}{2b}(\ast seebelow)$$$$\Rightarrow \sin \beta =\frac{b^2+c^2-a^2}{2bc}$$$$\sin \gamma =\frac{{AC}^{\prime }}{AC}with{CC}^{\prime }\mathrm{\perp }AB$$$${AC}^{\prime }=\frac{c^2+b^2-a^2}{2c}$$$$\Rightarrow \sin \gamma =\frac{b^2+c^2-a^2}{2bc}=\sin \beta$$$$\Rightarrow \beta =\gamma$$$$i.e.Aismidpointof\stackrel{⌢}{QR}andOA\mathrm{\perp }QR.$$$$$$$$\mathrm{\angle }AOQ=2\beta$$$$p=2R\sin 2\beta =4R\sin \beta \cos \beta$$$$\sin \beta =\frac{b^2+c^2-a^2}{2bc}$$$$\cos \beta =\frac{\sqrt{{\left(2bc\right)}^2-{(b^2+c^2-a^2)}^2}}{2bc}$$$$=\frac{\sqrt{(2bc+b^2+c^2-a^2)(2bc-b^2-c^2+a^2)}}{2bc}$$$$=\frac{\sqrt{[{(b+c)}^2-a^2][a^2-{(b-c)}^2]}}{2bc}$$$$=\frac{\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}{2bc}$$$$=\frac{a}{2}\times \frac{\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}{abc}$$$$=\frac{a}{2}\times \frac{1}{R}$$$$$$$$p=4R\sin \beta \cos \beta$$$$=4R\times \frac{b^2+c^2-a^2}{2bc}\times \frac{a}{2}\times \frac{1}{R}$$$$\Rightarrow p=\frac{a(b^2+c^2-a^2)}{bc}$$$$similarly$$$$\Rightarrow q=\frac{b(c^2+a^2-b^2)}{ca}$$$$\Rightarrow r=\frac{c(a^2+b^2-c^2)}{ab}$$$$$$$$(\ast )$$$${BB}^{\prime 2}=c^2-{AB}^{\prime 2}=a^2-{(b-{AB}^{\prime })}^2$$$$\Rightarrow c^2-{AB}^{\prime 2}=a^2-b^2-{AB}^{\prime 2}+2{bAB}^{\prime }$$$$\Rightarrow b^2+c^2-a^2=2{bAB}^{\prime }$$$$\Rightarrow {AB}^{\prime }=\frac{b^2+c^2-a^2}{2b}$$

Commented by mr W last updated on 27/Jul/19

$$thankssir!indeedmuchcouldbe$$$$simplified.$$

Commented by ajfour last updated on 27/Jul/19

$$And\cos \beta =\sin A$$$$andsince\frac{\sin A}{a}=\frac{1}{2R}$$$$\cos \beta =\frac{a}{2R}$$$$hencep=2R\sin 2\beta$$$$p=4R\left(\frac{b^2+c^2-a^2}{2bc}\right)\left(\frac{a}{2R}\right)$$$$\Rightarrow p=\frac{a(b^2+c^2-a^2)}{bc}.$$$$brilliantSirand{it}^{\prime }sallquite$$$$easyandstraight,then.$$$$\mathcal{T}hanks.$$

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