Question Number 199922 by Mingma last updated on 11/Nov/23
Answered by deleteduser1 last updated on 11/Nov/23
![WLOG,let C be the origin and let AD coincide with the real axis;then a=a^− ;b=−b^− ;d=d^− ;e=−e^− ((b−a)/(−b−a))=((a−e)/(−e−a))⇒−be+2a^2 =+eb⇒a^2 =be ((b−a)/(−b−a))=((b−d)/(d+b))⇒b^2 −ad=−b^2 +ad⇒b^2 =ad ((b−a)/(−b−a))=((h−a)/(h^− −a))⇒bh^− −ah^− =−bh+2ab−ah ⇒h^− (b−a)=−h(b+a)+2ab...(i) (h/h^− )=((a−b)/(−b−a))⇒−h(b+a)=h^− (a−b)...(ii) (i)+(ii)⇒0=−2h(b+a)+2ab⇒h=((ab)/(b+a)) ((i−o)/(i^− −o^− ))=^? ((e−d)/(d+e))⇔((((ab)/(2b+2a))−((a+b)/2))/(((−ab)/(−2b+2a))−((a−b)/2)))=(((a^2 /b)−(b^2 /a))/((a^3 +b^3 )/(ab))) ⇔((([a^2 +b^2 +ab])/((b+a)))/(([a^2 +b^2 −ab])/((a−b))))=(((a−b)(a^2 +b^2 +ab))/((a+b)(a^2 +b^2 −ab)))⇔0=0 ⇒((i−o)/(i^− −o^− ))=((e−d)/(d+e))⇒IO⊥ED](Q199935.png)
$${WLOG},{let}\:{C}\:{be}\:{the}\:{origin}\:{and}\:{let}\:{AD}\:{coincide} \\ $$$${with}\:{the}\:{real}\:{axis};{then}\:{a}=\overset{−} {{a}};{b}=−\overset{−} {{b}};{d}=\overset{−} {{d}};{e}=−\overset{−} {{e}} \\ $$$$\frac{{b}−{a}}{−{b}−{a}}=\frac{{a}−{e}}{−{e}−{a}}\Rightarrow−{be}+\mathrm{2}{a}^{\mathrm{2}} =+{eb}\Rightarrow{a}^{\mathrm{2}} ={be} \\ $$$$\frac{{b}−{a}}{−{b}−{a}}=\frac{{b}−{d}}{{d}+{b}}\Rightarrow{b}^{\mathrm{2}} −{ad}=−{b}^{\mathrm{2}} +{ad}\Rightarrow{b}^{\mathrm{2}} ={ad} \\ $$$$\frac{{b}−{a}}{−{b}−{a}}=\frac{{h}−{a}}{\overset{−} {{h}}−{a}}\Rightarrow{b}\overset{−} {{h}}−{a}\overset{−} {{h}}=−{bh}+\mathrm{2}{ab}−{ah} \\ $$$$\Rightarrow\overset{−} {{h}}\left({b}−{a}\right)=−{h}\left({b}+{a}\right)+\mathrm{2}{ab}...\left({i}\right) \\ $$$$\frac{{h}}{\overset{−} {{h}}}=\frac{{a}−{b}}{−{b}−{a}}\Rightarrow−{h}\left({b}+{a}\right)=\overset{−} {{h}}\left({a}−{b}\right)...\left({ii}\right) \\ $$$$\left({i}\right)+\left({ii}\right)\Rightarrow\mathrm{0}=−\mathrm{2}{h}\left({b}+{a}\right)+\mathrm{2}{ab}\Rightarrow{h}=\frac{{ab}}{{b}+{a}} \\ $$$$\frac{{i}−{o}}{\overset{−} {{i}}−\overset{−} {{o}}}\overset{?} {=}\frac{{e}−{d}}{{d}+{e}}\Leftrightarrow\frac{\frac{{ab}}{\mathrm{2}{b}+\mathrm{2}{a}}−\frac{{a}+{b}}{\mathrm{2}}}{\frac{−{ab}}{−\mathrm{2}{b}+\mathrm{2}{a}}−\frac{{a}−{b}}{\mathrm{2}}}=\frac{\frac{{a}^{\mathrm{2}} }{{b}}−\frac{{b}^{\mathrm{2}} }{{a}}}{\frac{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} }{{ab}}} \\ $$$$\Leftrightarrow\frac{\frac{\left[{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{ab}\right]}{\left({b}+{a}\right)}}{\frac{\left[{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{ab}\right]}{\left({a}−{b}\right)}}=\frac{\left({a}−{b}\right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{ab}\right)}{\left({a}+{b}\right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{ab}\right)}\Leftrightarrow\mathrm{0}=\mathrm{0} \\ $$$$\Rightarrow\frac{{i}−{o}}{\overset{−} {{i}}−\overset{−} {{o}}}=\frac{{e}−{d}}{{d}+{e}}\Rightarrow{IO}\bot{ED} \\ $$
Commented by Mingma last updated on 11/Nov/23
Very elegant, sir!
Answered by ajfour last updated on 11/Nov/23
$${let}\:{H}\:{be}\:{origin}. \\ $$$${I}\left(\mathrm{0},{b}\right)\:\:\:{C}\left(\mathrm{0},\mathrm{2}{b}\right) \\ $$$${A}\left(−{r}+{a},\:\mathrm{0}\right)\:\:\:\:{B}\left({r}+{a},\:\mathrm{0}\right) \\ $$$${d}=\left(\mathrm{2}{r}\right)\left(\frac{\mathrm{2}{b}}{{r}−{a}}\right) \\ $$$${e}=\left(\mathrm{2}{r}\right)\left(\frac{\mathrm{2}{b}}{{r}+{a}}\right) \\ $$$${slope}\:{of}\:{DE} \\ $$$$\:\:{m}_{{CD}} =\frac{{d}−{e}}{\mathrm{2}{r}}=\frac{\mathrm{4}{ab}}{{r}^{\mathrm{2}} −{a}^{\mathrm{2}} } \\ $$$${but}\:\:\sqrt{\left({r}−{a}\right)\left({r}+{a}\right)}=\mathrm{2}{b} \\ $$$$\Rightarrow\:\:{m}_{{CD}} =\frac{{a}}{{b}} \\ $$$${slope}\:{of}\:{OI}\:={m}_{{OI}} =−\frac{{b}}{{a}}\:\:=−\frac{\mathrm{1}}{{m}_{{CD}} } \\ $$$$\Rightarrow\:\:{OI}\:\bot\:{CD} \\ $$
Commented by Mingma last updated on 11/Nov/23
Nice one, sir!