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Question-64112




Question Number 64112 by mr W last updated on 13/Jul/19
Commented by mr W last updated on 13/Jul/19
Find the radius of the n−th inscribed  circle within the parabola.
$${Find}\:{the}\:{radius}\:{of}\:{the}\:{n}−{th}\:{inscribed} \\ $$$${circle}\:{within}\:{the}\:{parabola}. \\ $$
Answered by ajfour last updated on 13/Jul/19
x^2 +(y−r_1 )^2 =r_1 ^2   x+(dy/dx)(y−r_1 )=0  (dy/dx)=(x/(r_1 −y))  ⇒  (d^2 y/dx^2 )=((r_1 −y+x(dy/dx))/((r_1 −y)^2 ))  At origin we then have      (1/r_1 )=2    ⇒   r_1 =1/2  let centre of n^(th)  circle be (0,c_n )     c_n =2(r_1 +r_2 +...+r_(n−1) )+r_n   let this circle touch parabola at  (h,k).  h=r_n sin θ  k=c_n −r_n cos θ  2r_n sin θ=tan θ  ⇒  r_n cos θ=(1/2)     c_n −(1/2)=r_n ^2 −(1/4)  ⇒   c_n =r_n ^2 +(1/4)     c_n =2(r_1 +r_2 +...+r_(n−1) )+r_n      (1/2)(r_n ^2 −r_n +(1/4))=r_1 +r_2 +...+r_(n−1)     (1/2)(r_n −(1/2))^2 =r_1 +r_2 +...+r_(n−1)      (1/2)(r_(n+1) −(1/2))=r_1 +r_2 +..+r_n   ⇒ (1/2)(r_(n+1) −r_n )(r_(n+1) +r_n −1)                     = r_n −r_(n−1)   ⇒ r_(n+1) ^2 −r_(n+1) +r_n −r_n ^2 =2r_n −2r_(n−1)   ⇒ r_(n+1) ^2 −r_(n+1) =r_n ^2 +r_n −2r_(n−1)   or   (r_(n+1) −(1/2))^2 =(r_n +(1/2))^2 −2r_(n−1)        with  r_1 =(1/2) .
$${x}^{\mathrm{2}} +\left({y}−{r}_{\mathrm{1}} \right)^{\mathrm{2}} ={r}_{\mathrm{1}} ^{\mathrm{2}} \\ $$$${x}+\frac{{dy}}{{dx}}\left({y}−{r}_{\mathrm{1}} \right)=\mathrm{0} \\ $$$$\frac{{dy}}{{dx}}=\frac{{x}}{{r}_{\mathrm{1}} −{y}}\:\:\Rightarrow\:\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\frac{{r}_{\mathrm{1}} −{y}+{x}\frac{{dy}}{{dx}}}{\left({r}_{\mathrm{1}} −{y}\right)^{\mathrm{2}} } \\ $$$${At}\:{origin}\:{we}\:{then}\:{have} \\ $$$$\:\:\:\:\frac{\mathrm{1}}{{r}_{\mathrm{1}} }=\mathrm{2}\:\:\:\:\Rightarrow\:\:\:{r}_{\mathrm{1}} =\mathrm{1}/\mathrm{2} \\ $$$${let}\:{centre}\:{of}\:{n}^{{th}} \:{circle}\:{be}\:\left(\mathrm{0},{c}_{{n}} \right) \\ $$$$\:\:\:{c}_{{n}} =\mathrm{2}\left({r}_{\mathrm{1}} +{r}_{\mathrm{2}} +…+{r}_{{n}−\mathrm{1}} \right)+{r}_{{n}} \\ $$$${let}\:{this}\:{circle}\:{touch}\:{parabola}\:{at} \\ $$$$\left({h},{k}\right). \\ $$$${h}={r}_{{n}} \mathrm{sin}\:\theta \\ $$$${k}={c}_{{n}} −{r}_{{n}} \mathrm{cos}\:\theta \\ $$$$\mathrm{2}{r}_{{n}} \mathrm{sin}\:\theta=\mathrm{tan}\:\theta \\ $$$$\Rightarrow\:\:{r}_{{n}} \mathrm{cos}\:\theta=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:\:{c}_{{n}} −\frac{\mathrm{1}}{\mathrm{2}}={r}_{{n}} ^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Rightarrow\:\:\:{c}_{{n}} ={r}_{{n}} ^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\:\:\:{c}_{{n}} =\mathrm{2}\left({r}_{\mathrm{1}} +{r}_{\mathrm{2}} +…+{r}_{{n}−\mathrm{1}} \right)+{r}_{{n}} \\ $$$$\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\left({r}_{{n}} ^{\mathrm{2}} −{r}_{{n}} +\frac{\mathrm{1}}{\mathrm{4}}\right)={r}_{\mathrm{1}} +{r}_{\mathrm{2}} +…+{r}_{{n}−\mathrm{1}} \\ $$$$\:\:\frac{\mathrm{1}}{\mathrm{2}}\left({r}_{{n}} −\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} ={r}_{\mathrm{1}} +{r}_{\mathrm{2}} +…+{r}_{{n}−\mathrm{1}} \\ $$$$\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\left({r}_{{n}+\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{2}}\right)={r}_{\mathrm{1}} +{r}_{\mathrm{2}} +..+{r}_{{n}} \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{\mathrm{2}}\left({r}_{{n}+\mathrm{1}} −{r}_{{n}} \right)\left({r}_{{n}+\mathrm{1}} +{r}_{{n}} −\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:{r}_{{n}} −{r}_{{n}−\mathrm{1}} \\ $$$$\Rightarrow\:{r}_{{n}+\mathrm{1}} ^{\mathrm{2}} −{r}_{{n}+\mathrm{1}} +{r}_{{n}} −{r}_{{n}} ^{\mathrm{2}} =\mathrm{2}{r}_{{n}} −\mathrm{2}{r}_{{n}−\mathrm{1}} \\ $$$$\Rightarrow\:{r}_{{n}+\mathrm{1}} ^{\mathrm{2}} −{r}_{{n}+\mathrm{1}} ={r}_{{n}} ^{\mathrm{2}} +{r}_{{n}} −\mathrm{2}{r}_{{n}−\mathrm{1}} \\ $$$${or}\:\:\:\left({r}_{{n}+\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} =\left({r}_{{n}} +\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\mathrm{2}{r}_{{n}−\mathrm{1}} \\ $$$$\:\:\:\:\:{with}\:\:{r}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\:. \\ $$
Commented by mr W last updated on 13/Jul/19
thank you sir!   acc. to drawing it seems that  r_n =n−(1/2)  but this doesn′t match   (r_(n+1) −(1/2))^2 =(r_n +(1/2))^2 −2r_(n−1)   can you please recheck?
$${thank}\:{you}\:{sir}!\: \\ $$$${acc}.\:{to}\:{drawing}\:{it}\:{seems}\:{that} \\ $$$${r}_{{n}} ={n}−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${but}\:{this}\:{doesn}'{t}\:{match}\: \\ $$$$\left({r}_{{n}+\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} =\left({r}_{{n}} +\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\mathrm{2}{r}_{{n}−\mathrm{1}} \\ $$$${can}\:{you}\:{please}\:{recheck}? \\ $$
Answered by mr W last updated on 13/Jul/19
first circle touches the parabola at (0,0)  y′=2x  y′′=2 ⇒R=(1/2)  ⇒radius of the first circle r_1 =(1/2)  let r_n =radius of nth circle  d_n =diameter of nth circle, d_n =2r_n   the nth circle touches the n−1th circle  at (0,d_1 +d_2 +...d_(n−1) )  let a_n =d_1 +d_2 +...d_(n−1) =2(r_1 +r_2 +...+r_(n−1) )  eqn. of nth circle:  x^2 +(y−a_n −r_n )^2 =r_n ^2   intersection with parabola y=x^2 :  y+(y−a_n −r_n )^2 =r_n ^2   ⇒y^2 −[2(a_n +r_n )−1]y+a_n (a_n +2r_n )=0  due to tangency:  Δ=[2(a_n +r_n )−1]^2 −4a_n (a_n +2r_n )=0  4(a_n ^2 +2a_n r_n +r_n ^2 )−4(a_n +r_n )+1−4a_n ^2 −8a_n r_n =0  r_n ^2 −r_n −(a_n −(1/4))=0  ⇒r_n =(1/2)+(√a_n )  ⇒r_n =(1/2)+(√(2(r_1 +r_2 +...+r_(n−1) )))  (r_n −(1/2))^2 =2(r_1 +r_2 +...+r_(n−1) )  ⇒r_n ^2 −r_n +(1/4)=2(r_1 +r_2 +...+r_(n−1) )   ...(i)  ⇒r_(n−1) ^2 −r_(n−1) +(1/4)=2(r_1 +r_2 +...+r_(n−2) )  ⇒r_(n−1) ^2 +r_(n−1) +(1/4)=2(r_1 +r_2 +...+r_(n−1) )   ...(ii)  (i)−(ii):  (r_n +r_(n−1) )(r_n −r_(n−1) −1)=0  ⇒r_n −r_(n−1) −1=0  ⇒r_n −r_(n−1) =1  ⇒r_1 ,r_2 ,r_3 ,..,r_n  are A.P. with common  difference 1, since r_1 =(1/2),  ⇒r_n =(1/2)+(n−1)×1=n−(1/2)
$${first}\:{circle}\:{touches}\:{the}\:{parabola}\:{at}\:\left(\mathrm{0},\mathrm{0}\right) \\ $$$${y}'=\mathrm{2}{x} \\ $$$${y}''=\mathrm{2}\:\Rightarrow{R}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow{radius}\:{of}\:{the}\:{first}\:{circle}\:{r}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${let}\:{r}_{{n}} ={radius}\:{of}\:{nth}\:{circle} \\ $$$${d}_{{n}} ={diameter}\:{of}\:{nth}\:{circle},\:{d}_{{n}} =\mathrm{2}{r}_{{n}} \\ $$$${the}\:{nth}\:{circle}\:{touches}\:{the}\:{n}−\mathrm{1}{th}\:{circle} \\ $$$${at}\:\left(\mathrm{0},{d}_{\mathrm{1}} +{d}_{\mathrm{2}} +…{d}_{{n}−\mathrm{1}} \right) \\ $$$${let}\:{a}_{{n}} ={d}_{\mathrm{1}} +{d}_{\mathrm{2}} +…{d}_{{n}−\mathrm{1}} =\mathrm{2}\left({r}_{\mathrm{1}} +{r}_{\mathrm{2}} +…+{r}_{{n}−\mathrm{1}} \right) \\ $$$${eqn}.\:{of}\:{nth}\:{circle}: \\ $$$${x}^{\mathrm{2}} +\left({y}−{a}_{{n}} −{r}_{{n}} \right)^{\mathrm{2}} ={r}_{{n}} ^{\mathrm{2}} \\ $$$${intersection}\:{with}\:{parabola}\:{y}={x}^{\mathrm{2}} : \\ $$$${y}+\left({y}−{a}_{{n}} −{r}_{{n}} \right)^{\mathrm{2}} ={r}_{{n}} ^{\mathrm{2}} \\ $$$$\Rightarrow{y}^{\mathrm{2}} −\left[\mathrm{2}\left({a}_{{n}} +{r}_{{n}} \right)−\mathrm{1}\right]{y}+{a}_{{n}} \left({a}_{{n}} +\mathrm{2}{r}_{{n}} \right)=\mathrm{0} \\ $$$${due}\:{to}\:{tangency}: \\ $$$$\Delta=\left[\mathrm{2}\left({a}_{{n}} +{r}_{{n}} \right)−\mathrm{1}\right]^{\mathrm{2}} −\mathrm{4}{a}_{{n}} \left({a}_{{n}} +\mathrm{2}{r}_{{n}} \right)=\mathrm{0} \\ $$$$\mathrm{4}\left({a}_{{n}} ^{\mathrm{2}} +\mathrm{2}{a}_{{n}} {r}_{{n}} +{r}_{{n}} ^{\mathrm{2}} \right)−\mathrm{4}\left({a}_{{n}} +{r}_{{n}} \right)+\mathrm{1}−\mathrm{4}{a}_{{n}} ^{\mathrm{2}} −\mathrm{8}{a}_{{n}} {r}_{{n}} =\mathrm{0} \\ $$$${r}_{{n}} ^{\mathrm{2}} −{r}_{{n}} −\left({a}_{{n}} −\frac{\mathrm{1}}{\mathrm{4}}\right)=\mathrm{0} \\ $$$$\Rightarrow{r}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}+\sqrt{{a}_{{n}} } \\ $$$$\Rightarrow{r}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}+\sqrt{\mathrm{2}\left({r}_{\mathrm{1}} +{r}_{\mathrm{2}} +…+{r}_{{n}−\mathrm{1}} \right)} \\ $$$$\left({r}_{{n}} −\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{2}\left({r}_{\mathrm{1}} +{r}_{\mathrm{2}} +…+{r}_{{n}−\mathrm{1}} \right) \\ $$$$\Rightarrow{r}_{{n}} ^{\mathrm{2}} −{r}_{{n}} +\frac{\mathrm{1}}{\mathrm{4}}=\mathrm{2}\left({r}_{\mathrm{1}} +{r}_{\mathrm{2}} +…+{r}_{{n}−\mathrm{1}} \right)\:\:\:…\left({i}\right) \\ $$$$\Rightarrow{r}_{{n}−\mathrm{1}} ^{\mathrm{2}} −{r}_{{n}−\mathrm{1}} +\frac{\mathrm{1}}{\mathrm{4}}=\mathrm{2}\left({r}_{\mathrm{1}} +{r}_{\mathrm{2}} +…+{r}_{{n}−\mathrm{2}} \right) \\ $$$$\Rightarrow{r}_{{n}−\mathrm{1}} ^{\mathrm{2}} +{r}_{{n}−\mathrm{1}} +\frac{\mathrm{1}}{\mathrm{4}}=\mathrm{2}\left({r}_{\mathrm{1}} +{r}_{\mathrm{2}} +…+{r}_{{n}−\mathrm{1}} \right)\:\:\:…\left({ii}\right) \\ $$$$\left({i}\right)−\left({ii}\right): \\ $$$$\left({r}_{{n}} +{r}_{{n}−\mathrm{1}} \right)\left({r}_{{n}} −{r}_{{n}−\mathrm{1}} −\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow{r}_{{n}} −{r}_{{n}−\mathrm{1}} −\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{r}_{{n}} −{r}_{{n}−\mathrm{1}} =\mathrm{1} \\ $$$$\Rightarrow{r}_{\mathrm{1}} ,{r}_{\mathrm{2}} ,{r}_{\mathrm{3}} ,..,{r}_{{n}} \:{are}\:{A}.{P}.\:{with}\:{common} \\ $$$${difference}\:\mathrm{1},\:{since}\:{r}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}, \\ $$$$\Rightarrow{r}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}+\left({n}−\mathrm{1}\right)×\mathrm{1}={n}−\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by ajfour last updated on 13/Jul/19
Nice thought sir, getting rid of  r_(n−2)  !  You had solved it before, though  not this rigorous way.
$${Nice}\:{thought}\:{sir},\:{getting}\:{rid}\:{of}\:\:{r}_{{n}−\mathrm{2}} \:! \\ $$$${You}\:{had}\:{solved}\:{it}\:{before},\:{though} \\ $$$${not}\:{this}\:{rigorous}\:{way}. \\ $$
Commented by mr W last updated on 13/Jul/19
to be honest i found at first only the  relation r=(1/2)+(√a) and drew the circles  one after one. after i have posted the question,  i found that the circles touch each other   always at perfect squares 1, 4, 9, 16...  that means the radii are an A.P.  after i have read your post i tried to find  a way to prove this. i found a way after   some trials.
$${to}\:{be}\:{honest}\:{i}\:{found}\:{at}\:{first}\:{only}\:{the} \\ $$$${relation}\:{r}=\frac{\mathrm{1}}{\mathrm{2}}+\sqrt{{a}}\:{and}\:{drew}\:{the}\:{circles} \\ $$$${one}\:{after}\:{one}.\:{after}\:{i}\:{have}\:{posted}\:{the}\:{question}, \\ $$$${i}\:{found}\:{that}\:{the}\:{circles}\:{touch}\:{each}\:{other}\: \\ $$$${always}\:{at}\:{perfect}\:{squares}\:\mathrm{1},\:\mathrm{4},\:\mathrm{9},\:\mathrm{16}… \\ $$$${that}\:{means}\:{the}\:{radii}\:{are}\:{an}\:{A}.{P}. \\ $$$${after}\:{i}\:{have}\:{read}\:{your}\:{post}\:{i}\:{tried}\:{to}\:{find} \\ $$$${a}\:{way}\:{to}\:{prove}\:{this}.\:{i}\:{found}\:{a}\:{way}\:{after}\: \\ $$$${some}\:{trials}. \\ $$
Commented by Tawa11 last updated on 04/Jan/22
I just see this. God bless you sir.
$$\mathrm{I}\:\mathrm{just}\:\mathrm{see}\:\mathrm{this}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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