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Question Number 19122 by Tinkutara last updated on 05/Aug/17
Prove that r_1 + r_2 + r_3 = 4R + r
$$\mathrm{Prove}\:\mathrm{that}\:{r}_{\mathrm{1}} \:+\:{r}_{\mathrm{2}} \:+\:{r}_{\mathrm{3}} \:=\:\mathrm{4}{R}\:+\:{r} \\ $$
Answered by ajfour last updated on 05/Aug/17
2Rsin A=a r_1 =(a/(tan (B/2)+tan (C/2))) r=(a/(cot (B/2)+cot (C/2)))=((a(tan (B/2)tan (C/2)))/(tan (B/2)+tan (C/2))) so r_1 =(r/(tan (B/2)tan (C/2))) ⇒ r_1 +r_2 +r_3 =r (((Σtan (A/2))/(Πtan (A/2)))) ......(i) 4Rsin (A/2)cos (A/2)=a=r(((tan (B/2)+tan (C/2))/(tan (B/2)tan (C/2)))) ⇒((4R)/r)=(((tan (B/2)+tan (C/2))(1+tan^2 (A/2)))/(Πtan (A/2))) ((4R)/r)=((Σtan (A/2))/(Πtan (A/2)))+((tan^2 (A/2)(tan (B/2)+tan (C/2))−tan (A/2))/(Πtan (A/2))) ((4R)/r)=((r_1 +r_2 +r_3 )/r)+((tan (A/2)(Σtan (B/2)tan (C/2)−1−tan (B/2)tan (C/2)))/(Πtan (A/2))) but Σtan (B/2)tan (C/2)−1=0 , so ((4R)/r)=((r_1 +r_2 +r_3 )/r)−1 ⇒ r_1 +r_2 +r_3 =4R+r .
$$\mathrm{2Rsin}\:\mathrm{A}=\mathrm{a} \\ $$$$\mathrm{r}_{\mathrm{1}} =\frac{\mathrm{a}}{\mathrm{tan}\:\frac{\mathrm{B}}{\mathrm{2}}+\mathrm{tan}\:\frac{\mathrm{C}}{\mathrm{2}}} \\ $$$$\mathrm{r}=\frac{\mathrm{a}}{\mathrm{cot}\:\frac{\mathrm{B}}{\mathrm{2}}+\mathrm{cot}\:\frac{\mathrm{C}}{\mathrm{2}}}=\frac{\mathrm{a}\left(\mathrm{tan}\:\frac{\mathrm{B}}{\mathrm{2}}\mathrm{tan}\:\frac{\mathrm{C}}{\mathrm{2}}\right)}{\mathrm{tan}\:\frac{\mathrm{B}}{\mathrm{2}}+\mathrm{tan}\:\frac{\mathrm{C}}{\mathrm{2}}} \\ $$$$\mathrm{so}\:\:\mathrm{r}_{\mathrm{1}} =\frac{\mathrm{r}}{\mathrm{tan}\:\frac{\mathrm{B}}{\mathrm{2}}\mathrm{tan}\:\frac{\mathrm{C}}{\mathrm{2}}} \\ $$$$\Rightarrow\:\mathrm{r}_{\mathrm{1}} +\mathrm{r}_{\mathrm{2}} +\mathrm{r}_{\mathrm{3}} =\mathrm{r}\:\left(\frac{\Sigma\mathrm{tan}\:\frac{\mathrm{A}}{\mathrm{2}}}{\Pi\mathrm{tan}\:\frac{\mathrm{A}}{\mathrm{2}}}\right)\:\:\:……\left(\mathrm{i}\right) \\ $$$$\mathrm{4Rsin}\:\frac{\mathrm{A}}{\mathrm{2}}\mathrm{cos}\:\frac{\mathrm{A}}{\mathrm{2}}=\mathrm{a}=\mathrm{r}\left(\frac{\mathrm{tan}\:\frac{\mathrm{B}}{\mathrm{2}}+\mathrm{tan}\:\frac{\mathrm{C}}{\mathrm{2}}}{\mathrm{tan}\:\frac{\mathrm{B}}{\mathrm{2}}\mathrm{tan}\:\frac{\mathrm{C}}{\mathrm{2}}}\right) \\ $$$$\Rightarrow\frac{\mathrm{4R}}{\mathrm{r}}=\frac{\left(\mathrm{tan}\:\frac{\mathrm{B}}{\mathrm{2}}+\mathrm{tan}\:\frac{\mathrm{C}}{\mathrm{2}}\right)\left(\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \frac{\mathrm{A}}{\mathrm{2}}\right)}{\Pi\mathrm{tan}\:\frac{\mathrm{A}}{\mathrm{2}}} \\ $$$$\frac{\mathrm{4R}}{\mathrm{r}}=\frac{\Sigma\mathrm{tan}\:\frac{\mathrm{A}}{\mathrm{2}}}{\Pi\mathrm{tan}\:\frac{\mathrm{A}}{\mathrm{2}}}+\frac{\mathrm{tan}\:^{\mathrm{2}} \frac{\mathrm{A}}{\mathrm{2}}\left(\mathrm{tan}\:\frac{\mathrm{B}}{\mathrm{2}}+\mathrm{tan}\:\frac{\mathrm{C}}{\mathrm{2}}\right)−\mathrm{tan}\:\frac{\mathrm{A}}{\mathrm{2}}}{\Pi\mathrm{tan}\:\frac{\mathrm{A}}{\mathrm{2}}} \\ $$$$\frac{\mathrm{4R}}{\mathrm{r}}=\frac{\mathrm{r}_{\mathrm{1}} +\mathrm{r}_{\mathrm{2}} +\mathrm{r}_{\mathrm{3}} }{\mathrm{r}}+\frac{\mathrm{tan}\:\frac{\mathrm{A}}{\mathrm{2}}\left(\Sigma\mathrm{tan}\:\frac{\mathrm{B}}{\mathrm{2}}\mathrm{tan}\:\frac{\mathrm{C}}{\mathrm{2}}−\mathrm{1}−\mathrm{tan}\:\frac{\mathrm{B}}{\mathrm{2}}\mathrm{tan}\:\frac{\mathrm{C}}{\mathrm{2}}\right)}{\Pi\mathrm{tan}\:\frac{\mathrm{A}}{\mathrm{2}}} \\ $$$$\mathrm{but}\:\Sigma\mathrm{tan}\:\frac{\mathrm{B}}{\mathrm{2}}\mathrm{tan}\:\frac{\mathrm{C}}{\mathrm{2}}−\mathrm{1}=\mathrm{0}\:\:\:,\:\mathrm{so} \\ $$$$\frac{\mathrm{4R}}{\mathrm{r}}=\frac{\mathrm{r}_{\mathrm{1}} +\mathrm{r}_{\mathrm{2}} +\mathrm{r}_{\mathrm{3}} }{\mathrm{r}}−\mathrm{1} \\ $$$$\Rightarrow\:\:\mathrm{r}_{\mathrm{1}} +\mathrm{r}_{\mathrm{2}} +\mathrm{r}_{\mathrm{3}} =\mathrm{4R}+\mathrm{r}\:. \\ $$
Commented by Tinkutara last updated on 06/Aug/17
Thank you very much Sir! I really appreciate.
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}!\:\mathrm{I}\:\mathrm{really} \\ $$$$\mathrm{appreciate}. \\ $$
Commented by ajfour last updated on 05/Aug/17
Commented by ajfour last updated on 05/Aug/17
from figure: Rsin A=(a/2) r(cot (B/2)+cot (C/2))=a ⇒ r=a(((tan (B/2)tan (C/2))/(tan (B/2)+tan (C/2)))) and r_1 (tan (B/2)+tan (C/2))=a or r_1 =(r/(tan (B/2)tan (C/2))) .
$$\mathrm{from}\:\mathrm{figure}: \\ $$$$\mathrm{Rsin}\:\mathrm{A}=\frac{\mathrm{a}}{\mathrm{2}} \\ $$$$\:\:\:\mathrm{r}\left(\mathrm{cot}\:\frac{\mathrm{B}}{\mathrm{2}}+\mathrm{cot}\:\frac{\mathrm{C}}{\mathrm{2}}\right)=\mathrm{a} \\ $$$$\Rightarrow\:\:\mathrm{r}=\mathrm{a}\left(\frac{\mathrm{tan}\:\frac{\mathrm{B}}{\mathrm{2}}\mathrm{tan}\:\frac{\mathrm{C}}{\mathrm{2}}}{\mathrm{tan}\:\frac{\mathrm{B}}{\mathrm{2}}+\mathrm{tan}\:\frac{\mathrm{C}}{\mathrm{2}}}\right) \\ $$$$\mathrm{and}\:\:\mathrm{r}_{\mathrm{1}} \left(\mathrm{tan}\:\frac{\mathrm{B}}{\mathrm{2}}+\mathrm{tan}\:\frac{\mathrm{C}}{\mathrm{2}}\right)=\mathrm{a} \\ $$$$\mathrm{or}\:\:\mathrm{r}_{\mathrm{1}} =\frac{\mathrm{r}}{\mathrm{tan}\:\frac{\mathrm{B}}{\mathrm{2}}\mathrm{tan}\:\frac{\mathrm{C}}{\mathrm{2}}}\:. \\ $$

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