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Question-165217




Question Number 165217 by ajfour last updated on 27/Jan/22
Commented by mr W last updated on 28/Jan/22
it should be possible for any a>0.
$${it}\:{should}\:{be}\:{possible}\:{for}\:{any}\:{a}>\mathrm{0}. \\ $$
Commented by mr W last updated on 28/Jan/22
nice question sir!  it seems to have no unique solution.  that means it is possible for a range  of values of a, not a single one.
$${nice}\:{question}\:{sir}! \\ $$$${it}\:{seems}\:{to}\:{have}\:{no}\:{unique}\:{solution}. \\ $$$${that}\:{means}\:{it}\:{is}\:{possible}\:{for}\:{a}\:{range} \\ $$$${of}\:{values}\:{of}\:{a},\:{not}\:{a}\:{single}\:{one}. \\ $$
Commented by mr W last updated on 28/Jan/22
Commented by mr W last updated on 28/Jan/22
Commented by ajfour last updated on 28/Jan/22
Lets have the range then..  Thanks Sir!
$${Lets}\:{have}\:{the}\:{range}\:{then}.. \\ $$$${Thanks}\:{Sir}! \\ $$
Answered by mr W last updated on 28/Jan/22
Commented by mr W last updated on 28/Jan/22
we take O as origin.  let m=tan θ, k=tan ϕ  A(0,1)  B(a,0)  C_1 (a−r,1−r)  C_2 (h,r)  eqn. of OE:  mx−y=0  eqn. of AD:  kx+y−1=0    r(√(m^2 +1))=(a−r)m−(1−r)  r(√(m^2 +1))=hm−r  (a−r)m−(1−r)=hm−r  ⇒h=a−r−((1−2r)/m) ✓  r(√(m^2 +1))=(a−r)m−(1−r)  ⇒r=((am−1)/(m+(√(m^2 −1))−1)) ✓    r(√(k^2 +1))=(a−r)k+(1−r)−1  r(√(k^2 +1))=−(hk+r−1)  (a−r)k+(1−r)−1=−(hk+r−1)  1−(a−r)k=hk  ⇒h=(1/k)−(a−r) ✓  r(√(k^2 +1))=−[1−(a−r)k+r−1]  ⇒r=((ak)/( 1+k+(√(k^2 +1)))) ✓    (1/k)−a+r=a−r−((1−2r)/m)  (1/k)=2(a−r)−((1−2r)/m)  (1/k)=2a−(1/m)+2((1/m)−1)r  ⇒k=(1/(2a−(1/m)+2((1/m)−1)r)) ✓    for any given a we can obtain m from  ((am−1)/(m+(√(m^2 −1))−1))=((ak)/( 1+k+(√(k^2 +1))))
$${we}\:{take}\:{O}\:{as}\:{origin}. \\ $$$${let}\:{m}=\mathrm{tan}\:\theta,\:{k}=\mathrm{tan}\:\varphi \\ $$$${A}\left(\mathrm{0},\mathrm{1}\right) \\ $$$${B}\left({a},\mathrm{0}\right) \\ $$$${C}_{\mathrm{1}} \left({a}−{r},\mathrm{1}−{r}\right) \\ $$$${C}_{\mathrm{2}} \left({h},{r}\right) \\ $$$${eqn}.\:{of}\:{OE}: \\ $$$${mx}−{y}=\mathrm{0} \\ $$$${eqn}.\:{of}\:{AD}: \\ $$$${kx}+{y}−\mathrm{1}=\mathrm{0} \\ $$$$ \\ $$$${r}\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}=\left({a}−{r}\right){m}−\left(\mathrm{1}−{r}\right) \\ $$$${r}\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}={hm}−{r} \\ $$$$\left({a}−{r}\right){m}−\left(\mathrm{1}−{r}\right)={hm}−{r} \\ $$$$\Rightarrow{h}={a}−{r}−\frac{\mathrm{1}−\mathrm{2}{r}}{{m}}\:\checkmark \\ $$$${r}\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}=\left({a}−{r}\right){m}−\left(\mathrm{1}−{r}\right) \\ $$$$\Rightarrow{r}=\frac{{am}−\mathrm{1}}{{m}+\sqrt{{m}^{\mathrm{2}} −\mathrm{1}}−\mathrm{1}}\:\checkmark \\ $$$$ \\ $$$${r}\sqrt{{k}^{\mathrm{2}} +\mathrm{1}}=\left({a}−{r}\right){k}+\left(\mathrm{1}−{r}\right)−\mathrm{1} \\ $$$${r}\sqrt{{k}^{\mathrm{2}} +\mathrm{1}}=−\left({hk}+{r}−\mathrm{1}\right) \\ $$$$\left({a}−{r}\right){k}+\left(\mathrm{1}−{r}\right)−\mathrm{1}=−\left({hk}+{r}−\mathrm{1}\right) \\ $$$$\mathrm{1}−\left({a}−{r}\right){k}={hk} \\ $$$$\Rightarrow{h}=\frac{\mathrm{1}}{{k}}−\left({a}−{r}\right)\:\checkmark \\ $$$${r}\sqrt{{k}^{\mathrm{2}} +\mathrm{1}}=−\left[\mathrm{1}−\left({a}−{r}\right){k}+{r}−\mathrm{1}\right] \\ $$$$\Rightarrow{r}=\frac{{ak}}{\:\mathrm{1}+{k}+\sqrt{{k}^{\mathrm{2}} +\mathrm{1}}}\:\checkmark \\ $$$$ \\ $$$$\frac{\mathrm{1}}{{k}}−{a}+{r}={a}−{r}−\frac{\mathrm{1}−\mathrm{2}{r}}{{m}} \\ $$$$\frac{\mathrm{1}}{{k}}=\mathrm{2}\left({a}−{r}\right)−\frac{\mathrm{1}−\mathrm{2}{r}}{{m}} \\ $$$$\frac{\mathrm{1}}{{k}}=\mathrm{2}{a}−\frac{\mathrm{1}}{{m}}+\mathrm{2}\left(\frac{\mathrm{1}}{{m}}−\mathrm{1}\right){r} \\ $$$$\Rightarrow{k}=\frac{\mathrm{1}}{\mathrm{2}{a}−\frac{\mathrm{1}}{{m}}+\mathrm{2}\left(\frac{\mathrm{1}}{{m}}−\mathrm{1}\right){r}}\:\checkmark \\ $$$$ \\ $$$${for}\:{any}\:{given}\:{a}\:{we}\:{can}\:{obtain}\:{m}\:{from} \\ $$$$\frac{{am}−\mathrm{1}}{{m}+\sqrt{{m}^{\mathrm{2}} −\mathrm{1}}−\mathrm{1}}=\frac{{ak}}{\:\mathrm{1}+{k}+\sqrt{{k}^{\mathrm{2}} +\mathrm{1}}}\: \\ $$
Commented by mr W last updated on 28/Jan/22
Commented by mr W last updated on 28/Jan/22
Commented by mr W last updated on 28/Jan/22
Commented by ajfour last updated on 28/Jan/22
Thanks sir, i will go through..
$${Thanks}\:{sir},\:{i}\:{will}\:{go}\:{through}.. \\ $$
Commented by Tawa11 last updated on 28/Jan/22
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Answered by ajfour last updated on 28/Jan/22
Commented by ajfour last updated on 28/Jan/22
tan φ=(r/(b+r))  ;   tan θ=(r/(a−r))  cos 2θ=r+rcosec φsin (2θ+φ)  (a+b)(1)=cot 2φ+2rcosec 2φ  .....
$$\mathrm{tan}\:\phi=\frac{{r}}{{b}+{r}}\:\:;\:\:\:\mathrm{tan}\:\theta=\frac{{r}}{{a}−{r}} \\ $$$$\mathrm{cos}\:\mathrm{2}\theta={r}+{r}\mathrm{cosec}\:\phi\mathrm{sin}\:\left(\mathrm{2}\theta+\phi\right) \\ $$$$\left({a}+{b}\right)\left(\mathrm{1}\right)=\mathrm{cot}\:\mathrm{2}\phi+\mathrm{2}{r}\mathrm{cosec}\:\mathrm{2}\phi \\ $$$$….. \\ $$

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