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

Commented by ajfour last updated on 26/Jul/19

Find p,q,r in terms of a,b,c.

$${Find}\:{p},{q},{r}\:{in}\:{terms}\:{of}\:{a},{b},{c}. \\ $$

Commented by ajfour last updated on 27/Jul/19

do explain a bit sir! thanks.

$${do}\:{explain}\:{a}\:{bit}\:{sir}!\:{thanks}. \\ $$

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

Commented by ajfour last updated on 27/Jul/19

sin β=cos A=((b^2 +c^2 −a^2 )/(2bc))   is  straightaway true Sir!  as      β+A=π/2.

$$\mathrm{sin}\:\beta=\mathrm{cos}\:{A}=\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{bc}}\:\:\:{is} \\ $$$${straightaway}\:{true}\:{Sir}! \\ $$$${as}\:\:\:\:\:\:\beta+{A}=\pi/\mathrm{2}. \\ $$

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

R=radius of circumcircle  sin β=((AB′)/(AB)) with BB′⊥AC  AB′=((b^2 +c^2 −a^2 )/(2b))           (∗ see below)  ⇒sin β=((b^2 +c^2 −a^2 )/(2bc))  sin γ=((AC′)/(AC)) with CC′⊥AB  AC′=((c^2 +b^2 −a^2 )/(2c))  ⇒sin γ=((b^2 +c^2 −a^2 )/(2bc))=sin β  ⇒β=γ  i.e. A is midpoint of QR^(⌢)  and OA⊥QR.    ∠AOQ=2β  p=2R sin 2β=4R sin β cos β  sin β=((b^2 +c^2 −a^2 )/(2bc))  cos β=((√((2bc)^2 −(b^2 +c^2 −a^2 )^2 ))/(2bc))  =((√((2bc+b^2 +c^2 −a^2 )(2bc−b^2 −c^2 +a^2 )))/(2bc))  =((√([(b+c)^2 −a^2 ][a^2 −(b−c)^2 ]))/(2bc))  =((√((a+b+c)(−a+b+c)(a−b+c)(a+b−c)))/(2bc))  =(a/2)×((√((a+b+c)(−a+b+c)(a−b+c)(a+b−c)))/(abc))  =(a/2)×(1/R)    p=4R sin β cos β  =4R×((b^2 +c^2 −a^2 )/(2bc))×(a/2)×(1/R)  ⇒p=((a(b^2 +c^2 −a^2 ))/(bc))  similarly  ⇒q=((b(c^2 +a^2 −b^2 ))/(ca))  ⇒r=((c(a^2 +b^2 −c^2 ))/(ab))    (∗)  BB′^2 =c^2 −AB′^2 =a^2 −(b−AB′)^2   ⇒c^2 −AB′^2 =a^2 −b^2 −AB′^2 +2bAB′  ⇒b^2 +c^2 −a^2 =2bAB′  ⇒AB′=((b^2 +c^2 −a^2 )/(2b))

$${R}={radius}\:{of}\:{circumcircle} \\ $$$$\mathrm{sin}\:\beta=\frac{{AB}'}{{AB}}\:{with}\:{BB}'\bot{AC} \\ $$$${AB}'=\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{b}}\:\:\:\:\:\:\:\:\:\:\:\left(\ast\:{see}\:{below}\right) \\ $$$$\Rightarrow\mathrm{sin}\:\beta=\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{bc}} \\ $$$$\mathrm{sin}\:\gamma=\frac{{AC}'}{{AC}}\:{with}\:{CC}'\bot{AB} \\ $$$${AC}'=\frac{{c}^{\mathrm{2}} +{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{c}} \\ $$$$\Rightarrow\mathrm{sin}\:\gamma=\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{bc}}=\mathrm{sin}\:\beta \\ $$$$\Rightarrow\beta=\gamma \\ $$$${i}.{e}.\:{A}\:{is}\:{midpoint}\:{of}\:\overset{\frown} {{QR}}\:{and}\:{OA}\bot{QR}. \\ $$$$ \\ $$$$\angle{AOQ}=\mathrm{2}\beta \\ $$$${p}=\mathrm{2}{R}\:\mathrm{sin}\:\mathrm{2}\beta=\mathrm{4}{R}\:\mathrm{sin}\:\beta\:\mathrm{cos}\:\beta \\ $$$$\mathrm{sin}\:\beta=\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{bc}} \\ $$$$\mathrm{cos}\:\beta=\frac{\sqrt{\left(\mathrm{2}{bc}\right)^{\mathrm{2}} −\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)^{\mathrm{2}} }}{\mathrm{2}{bc}} \\ $$$$=\frac{\sqrt{\left(\mathrm{2}{bc}+{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)\left(\mathrm{2}{bc}−{b}^{\mathrm{2}} −{c}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)}}{\mathrm{2}{bc}} \\ $$$$=\frac{\sqrt{\left[\left({b}+{c}\right)^{\mathrm{2}} −{a}^{\mathrm{2}} \right]\left[{a}^{\mathrm{2}} −\left({b}−{c}\right)^{\mathrm{2}} \right]}}{\mathrm{2}{bc}} \\ $$$$=\frac{\sqrt{\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)\left({a}−{b}+{c}\right)\left({a}+{b}−{c}\right)}}{\mathrm{2}{bc}} \\ $$$$=\frac{{a}}{\mathrm{2}}×\frac{\sqrt{\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)\left({a}−{b}+{c}\right)\left({a}+{b}−{c}\right)}}{{abc}} \\ $$$$=\frac{{a}}{\mathrm{2}}×\frac{\mathrm{1}}{{R}} \\ $$$$ \\ $$$${p}=\mathrm{4}{R}\:\mathrm{sin}\:\beta\:\mathrm{cos}\:\beta \\ $$$$=\mathrm{4}{R}×\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{bc}}×\frac{{a}}{\mathrm{2}}×\frac{\mathrm{1}}{{R}} \\ $$$$\Rightarrow{p}=\frac{{a}\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)}{{bc}} \\ $$$${similarly} \\ $$$$\Rightarrow{q}=\frac{{b}\left({c}^{\mathrm{2}} +{a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)}{{ca}} \\ $$$$\Rightarrow{r}=\frac{{c}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)}{{ab}} \\ $$$$ \\ $$$$\left(\ast\right) \\ $$$${BB}'^{\mathrm{2}} ={c}^{\mathrm{2}} −{AB}'^{\mathrm{2}} ={a}^{\mathrm{2}} −\left({b}−{AB}'\right)^{\mathrm{2}} \\ $$$$\Rightarrow{c}^{\mathrm{2}} −{AB}'^{\mathrm{2}} ={a}^{\mathrm{2}} −{b}^{\mathrm{2}} −{AB}'^{\mathrm{2}} +\mathrm{2}{bAB}' \\ $$$$\Rightarrow{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} =\mathrm{2}{bAB}' \\ $$$$\Rightarrow{AB}'=\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{b}} \\ $$

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

thanks sir! indeed much could be  simplified.

$${thanks}\:{sir}!\:{indeed}\:{much}\:{could}\:{be} \\ $$$${simplified}. \\ $$

Commented by ajfour last updated on 27/Jul/19

And cos β=sin A  and since  ((sin A)/a)=(1/(2R))       cos β=(a/(2R))  hence  p=2Rsin 2β     p=4R(((b^2 +c^2 −a^2 )/(2bc)))((a/(2R)))  ⇒  p=((a(b^2 +c^2 −a^2 ))/(bc)) .  brilliant Sir and it′s all quite  easy and straight, then.  Thanks.

$${And}\:\mathrm{cos}\:\beta=\mathrm{sin}\:{A} \\ $$$${and}\:{since}\:\:\frac{\mathrm{sin}\:{A}}{{a}}=\frac{\mathrm{1}}{\mathrm{2}{R}} \\ $$$$\:\:\:\:\:\mathrm{cos}\:\beta=\frac{{a}}{\mathrm{2}{R}} \\ $$$${hence}\:\:{p}=\mathrm{2}{R}\mathrm{sin}\:\mathrm{2}\beta \\ $$$$\:\:\:{p}=\mathrm{4}{R}\left(\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{bc}}\right)\left(\frac{{a}}{\mathrm{2}{R}}\right) \\ $$$$\Rightarrow\:\:{p}=\frac{{a}\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)}{{bc}}\:. \\ $$$${brilliant}\:{Sir}\:{and}\:{it}'{s}\:{all}\:{quite} \\ $$$${easy}\:{and}\:{straight},\:{then}. \\ $$$$\mathcal{T}{hanks}. \\ $$

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