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Question-201298




Question Number 201298 by ajfour last updated on 03/Dec/23
Commented by ajfour last updated on 03/Dec/23
Find IJ   in terms of a,b,c.
$${Find}\:{IJ}\:\:\:{in}\:{terms}\:{of}\:{a},{b},{c}. \\ $$
Commented by mr W last updated on 05/Jan/24
IJ=r(√(1−((3p^2 )/((4R+r)^2 ))))  see Q202925
$${IJ}={r}\sqrt{\mathrm{1}−\frac{\mathrm{3}{p}^{\mathrm{2}} }{\left(\mathrm{4}{R}+{r}\right)^{\mathrm{2}} }} \\ $$$${see}\:{Q}\mathrm{202925} \\ $$
Answered by ajfour last updated on 03/Dec/23
BC=ai  BA=c^� =hi+kj  BI=rcot (β/2)i+rj  BR=rcot (β/2)(((hi+kj)/( (√(h^2 +k^2 )))))  eq. of JC  r_J ^� =ai+  λ{(a−((rh)/( (√(h^2 +k^2 ))))cot (β/2))i−(((rk)/( (√(h^2 +k^2 )))))j}  eq. of JA  r_J ^� =(hi+kj)+μ{(rcot (β/2)−h)i−kj}  r_J ^� =r_J ^�   a+λ(a−((rh)/( (√(h^2 +k^2 ))))cot (β/2))    =h+μ(rcot (β/2)−h)  &  ((λr)/( (√(h^2 +k^2 ))))=μ−1  ⇒ a+λ(a−((rh)/( (√(h^2 +k^2 ))))cot (β/2))      =h+(1+((λr)/( (√(h^2 +k^2 )))))(rcot (β/2)−h)  ⇒ λ=((a−h−(rcot (β/2)−h))/((r^2 /( (√(h^2 +k^2 ))))cot (β/2)−a))  IJ=[a+{((a−h−(rcot (β/2)−h))/((r^2 /( (√(h^2 +k^2 ))))cot (β/2)−a))}{(a−((rh)/( (√(h^2 +k^2 ))))cot (β/2))−rcot (β/2)}]i      +[{((a−h−(rcot (β/2)−h))/((r^2 /( (√(h^2 +k^2 ))))cot (β/2)−a))}(r/( (√(h^2 +k^2 ))))−r]j  h=ccos β    and   k=csin β  ⇒  IJ=[a+{((a−ccos β−(rcot (β/2)−ccos β))/((r^2 /c)cot (β/2)−a))}{(a−rcos βcot (β/2))−rcot (β/2)}]i      +[{((a−ccos β−(rcot (β/2)−ccos β))/((r^2 /c)cot (β/2)−a))}(r/c)−r]j  r=((2△)/(a+b+c))
$${BC}={ai} \\ $$$${BA}=\bar {{c}}={hi}+{kj} \\ $$$${BI}={r}\mathrm{cot}\:\frac{\beta}{\mathrm{2}}{i}+{rj} \\ $$$${BR}={r}\mathrm{cot}\:\frac{\beta}{\mathrm{2}}\left(\frac{{hi}+{kj}}{\:\sqrt{{h}^{\mathrm{2}} +{k}^{\mathrm{2}} }}\right) \\ $$$${eq}.\:{of}\:{JC} \\ $$$$\bar {{r}}_{{J}} ={ai}+ \\ $$$$\lambda\left\{\left({a}−\frac{{rh}}{\:\sqrt{{h}^{\mathrm{2}} +{k}^{\mathrm{2}} }}\mathrm{cot}\:\frac{\beta}{\mathrm{2}}\right){i}−\left(\frac{{rk}}{\:\sqrt{{h}^{\mathrm{2}} +{k}^{\mathrm{2}} }}\right){j}\right\} \\ $$$${eq}.\:{of}\:{JA} \\ $$$$\bar {{r}}_{{J}} =\left({hi}+{kj}\right)+\mu\left\{\left({r}\mathrm{cot}\:\frac{\beta}{\mathrm{2}}−{h}\right){i}−{kj}\right\} \\ $$$$\bar {{r}}_{{J}} =\bar {{r}}_{{J}} \\ $$$${a}+\lambda\left({a}−\frac{{rh}}{\:\sqrt{{h}^{\mathrm{2}} +{k}^{\mathrm{2}} }}\mathrm{cot}\:\frac{\beta}{\mathrm{2}}\right) \\ $$$$\:\:={h}+\mu\left({r}\mathrm{cot}\:\frac{\beta}{\mathrm{2}}−{h}\right) \\ $$$$\&\:\:\frac{\lambda{r}}{\:\sqrt{{h}^{\mathrm{2}} +{k}^{\mathrm{2}} }}=\mu−\mathrm{1} \\ $$$$\Rightarrow\:{a}+\lambda\left({a}−\frac{{rh}}{\:\sqrt{{h}^{\mathrm{2}} +{k}^{\mathrm{2}} }}\mathrm{cot}\:\frac{\beta}{\mathrm{2}}\right) \\ $$$$\:\:\:\:={h}+\left(\mathrm{1}+\frac{\lambda{r}}{\:\sqrt{{h}^{\mathrm{2}} +{k}^{\mathrm{2}} }}\right)\left({r}\mathrm{cot}\:\frac{\beta}{\mathrm{2}}−{h}\right) \\ $$$$\Rightarrow\:\lambda=\frac{{a}−{h}−\left({r}\mathrm{cot}\:\frac{\beta}{\mathrm{2}}−{h}\right)}{\frac{{r}^{\mathrm{2}} }{\:\sqrt{{h}^{\mathrm{2}} +{k}^{\mathrm{2}} }}\mathrm{cot}\:\frac{\beta}{\mathrm{2}}−{a}} \\ $$$${IJ}=\left[{a}+\left\{\frac{{a}−{h}−\left({r}\mathrm{cot}\:\frac{\beta}{\mathrm{2}}−{h}\right)}{\frac{{r}^{\mathrm{2}} }{\:\sqrt{{h}^{\mathrm{2}} +{k}^{\mathrm{2}} }}\mathrm{cot}\:\frac{\beta}{\mathrm{2}}−{a}}\right\}\left\{\left({a}−\frac{{rh}}{\:\sqrt{{h}^{\mathrm{2}} +{k}^{\mathrm{2}} }}\mathrm{cot}\:\frac{\beta}{\mathrm{2}}\right)−{r}\mathrm{cot}\:\frac{\beta}{\mathrm{2}}\right\}\right]{i} \\ $$$$\:\:\:\:+\left[\left\{\frac{{a}−{h}−\left({r}\mathrm{cot}\:\frac{\beta}{\mathrm{2}}−{h}\right)}{\frac{{r}^{\mathrm{2}} }{\:\sqrt{{h}^{\mathrm{2}} +{k}^{\mathrm{2}} }}\mathrm{cot}\:\frac{\beta}{\mathrm{2}}−{a}}\right\}\frac{{r}}{\:\sqrt{{h}^{\mathrm{2}} +{k}^{\mathrm{2}} }}−{r}\right]{j} \\ $$$${h}={c}\mathrm{cos}\:\beta\:\:\:\:{and}\:\:\:{k}={c}\mathrm{sin}\:\beta \\ $$$$\Rightarrow \\ $$$${IJ}=\left[{a}+\left\{\frac{{a}−{c}\mathrm{cos}\:\beta−\left({r}\mathrm{cot}\:\frac{\beta}{\mathrm{2}}−{c}\mathrm{cos}\:\beta\right)}{\frac{{r}^{\mathrm{2}} }{{c}}\mathrm{cot}\:\frac{\beta}{\mathrm{2}}−{a}}\right\}\left\{\left({a}−{r}\mathrm{cos}\:\beta\mathrm{cot}\:\frac{\beta}{\mathrm{2}}\right)−{r}\mathrm{cot}\:\frac{\beta}{\mathrm{2}}\right\}\right]{i} \\ $$$$\:\:\:\:+\left[\left\{\frac{{a}−{c}\mathrm{cos}\:\beta−\left({r}\mathrm{cot}\:\frac{\beta}{\mathrm{2}}−{c}\mathrm{cos}\:\beta\right)}{\frac{{r}^{\mathrm{2}} }{{c}}\mathrm{cot}\:\frac{\beta}{\mathrm{2}}−{a}}\right\}\frac{{r}}{{c}}−{r}\right]{j} \\ $$$${r}=\frac{\mathrm{2}\bigtriangleup}{{a}+{b}+{c}} \\ $$
Commented by mr W last updated on 04/Dec/23
very nice!
$${very}\:{nice}! \\ $$

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