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Question Number 22576 by ajfour last updated on 20/Oct/17
Commented by ajfour last updated on 20/Oct/17
Q. 22536 (Alternate solution)
$${Q}.\:\mathrm{22536}\:\left({Alternate}\:{solution}\right) \\ $$
Answered by ajfour last updated on 20/Oct/17
△PQI∼△ABC (first diagram) So ((PQ)/(AB))=(r/(ACsin A)) ⇒ r(cot A+cot B)=((cr)/(bsin A)) or cot A+cot B=(c/(bsin A)) we know ((sin A)/a)=((sin B)/b)=((sin C)/c)=(1/(2R)) so cot A+cot B=((2cR)/(ab)) 2Σcot A=2R((c/(ab))+(a/(bc))+(b/(ca))) or Σcot A=((R(a^2 +b^2 +c^2 ))/(abc)) ...(i) From second diagram: 2×(1/2)×r×rΣcot (A/2)=△ ⇒ r^2 Σcot (A/2)=△ ......(ii) and (1/2)c×bsin A=△ or ((bc)/2)×(a/(2R))=△ ⇒ ((abc)/(4R))=△ ..(iii) and rΣcot (A/2)=x+y+z ⇒ rΣcot (A/2)=((a+b+c)/2) ....(iv) squaring (iv)and dividing by ΣcotA , ⇒ ((r^2 (Σcot (A/2))^2 )/(Σcot A))=(((a+b+c)^2 )/(4Σcot A)) using (ii) and (i) in eqn. above (((Σcot (A/2))△)/(Σcot A))=(((a+b+c)^2 )/((4R(a^2 +b^2 +c^2 ))/(abc))) Now using eqn.(iii): abc=4△R ((Σcot (A/2))/(Σcot A)) =(((a+b+c)^2 )/(a^2 +b^2 +c^2 )) .
$$\bigtriangleup{PQI}\sim\bigtriangleup{ABC}\:\:\left({first}\:{diagram}\right) \\ $$$${So}\:\frac{{PQ}}{{AB}}=\frac{{r}}{{AC}\mathrm{sin}\:{A}} \\ $$$$\Rightarrow\:\:{r}\left(\mathrm{cot}\:{A}+\mathrm{cot}\:{B}\right)=\frac{{cr}}{{b}\mathrm{sin}\:{A}} \\ $$$${or}\:\:\:\mathrm{cot}\:{A}+\mathrm{cot}\:{B}=\frac{{c}}{{b}\mathrm{sin}\:{A}} \\ $$$${we}\:{know}\:\frac{\mathrm{sin}\:{A}}{{a}}=\frac{\mathrm{sin}\:{B}}{{b}}=\frac{\mathrm{sin}\:{C}}{{c}}=\frac{\mathrm{1}}{\mathrm{2}{R}} \\ $$$${so}\:\:\mathrm{cot}\:{A}+\mathrm{cot}\:{B}=\frac{\mathrm{2}{cR}}{{ab}} \\ $$$$\:\:\:\:\mathrm{2}\Sigma\mathrm{cot}\:{A}=\mathrm{2}{R}\left(\frac{{c}}{{ab}}+\frac{{a}}{{bc}}+\frac{{b}}{{ca}}\right) \\ $$$${or}\:\:\:\Sigma\mathrm{cot}\:{A}=\frac{{R}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)}{{abc}}\:…\left({i}\right) \\ $$$$\:\:{From}\:{second}\:{diagram}: \\ $$$$\:\:\:\:\:\:\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}×{r}×{r}\Sigma\mathrm{cot}\:\frac{{A}}{\mathrm{2}}=\bigtriangleup \\ $$$$\Rightarrow\:\:\:\:\:{r}^{\mathrm{2}} \Sigma\mathrm{cot}\:\frac{{A}}{\mathrm{2}}=\bigtriangleup\:\:\:……\left({ii}\right) \\ $$$${and}\:\:\frac{\mathrm{1}}{\mathrm{2}}{c}×{b}\mathrm{sin}\:{A}=\bigtriangleup \\ $$$${or}\:\:\:\frac{{bc}}{\mathrm{2}}×\frac{{a}}{\mathrm{2}{R}}=\bigtriangleup\:\:\Rightarrow\:\:\frac{{abc}}{\mathrm{4}{R}}=\bigtriangleup\:..\left({iii}\right) \\ $$$$\:{and}\:\:\:\:\:{r}\Sigma\mathrm{cot}\:\frac{{A}}{\mathrm{2}}={x}+{y}+{z} \\ $$$$\Rightarrow\:\:\:\:\:{r}\Sigma\mathrm{cot}\:\frac{{A}}{\mathrm{2}}=\frac{{a}+{b}+{c}}{\mathrm{2}}\:\:\:\:….\left({iv}\right) \\ $$$${squaring}\:\left({iv}\right){and}\:{dividing}\:{by}\:\Sigma{cotA}\:, \\ $$$$\Rightarrow\:\:\:\frac{{r}^{\mathrm{2}} \left(\Sigma\mathrm{cot}\:\frac{{A}}{\mathrm{2}}\right)^{\mathrm{2}} }{\Sigma\mathrm{cot}\:{A}}=\frac{\left({a}+{b}+{c}\right)^{\mathrm{2}} }{\mathrm{4}\Sigma\mathrm{cot}\:{A}} \\ $$$${using}\:\left({ii}\right)\:{and}\:\left({i}\right)\:{in}\:{eqn}.\:{above} \\ $$$$\:\:\:\:\:\:\:\frac{\left(\Sigma\mathrm{cot}\:\frac{{A}}{\mathrm{2}}\right)\bigtriangleup}{\Sigma\mathrm{cot}\:{A}}=\frac{\left({a}+{b}+{c}\right)^{\mathrm{2}} }{\frac{\mathrm{4}{R}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)}{{abc}}} \\ $$$$\:{Now}\:{using}\:{eqn}.\left({iii}\right):\:\:{abc}=\mathrm{4}\bigtriangleup{R} \\ $$$$\:\:\:\frac{\Sigma\mathrm{cot}\:\frac{{A}}{\mathrm{2}}}{\Sigma\mathrm{cot}\:{A}}\:=\frac{\left(\boldsymbol{{a}}+\boldsymbol{{b}}+\boldsymbol{{c}}\right)^{\mathrm{2}} }{\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{b}}^{\mathrm{2}} +\boldsymbol{{c}}^{\mathrm{2}} }\:\:. \\ $$
Commented by NECx last updated on 20/Oct/17
thanks boss
$${thanks}\:{boss} \\ $$
Commented by math solver last updated on 20/Oct/17
+1 :)
$$\left.+\mathrm{1}\::\right)\: \\ $$

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