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Question-53595

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Question Number 53595 by ajfour last updated on 23/Jan/19
Commented by ajfour last updated on 24/Jan/19
Commented by ajfour last updated on 24/Jan/19
Find x, given 𝛂. (two rectangles).
$${Find}\:{x},\:{given}\:\boldsymbol{\alpha}.\:\left({two}\:{rectangles}\right). \\ $$
Commented by mr W last updated on 24/Jan/19
Commented by ajfour last updated on 24/Jan/19
∠EAF is a right angle too, Sir. it′ll be the same equation; 1−x^4 =kx , if k=tan α i just tried to depict it geometricaly.
$$\angle{EAF}\:{is}\:{a}\:{right}\:{angle}\:{too},\:{Sir}. \\ $$$${it}'{ll}\:{be}\:{the}\:{same}\:{equation};\: \\ $$$$\mathrm{1}−{x}^{\mathrm{4}} ={kx}\:\:,\:{if}\:{k}=\mathrm{tan}\:\alpha \\ $$$${i}\:{just}\:{tried}\:{to}\:{depict}\:{it}\:{geometricaly}. \\ $$
Commented by mr W last updated on 24/Jan/19
great idea!
$${great}\:{idea}! \\ $$
Answered by ajfour last updated on 23/Jan/19
Commented by mr W last updated on 24/Jan/19
A(R,h) y=cx^2 with c=1 y′=2cx x_P =R+R cos θ=R(1+cos θ) y_P =h−R sin θ=cx_P ^2 =cR^2 (1+cos θ)^2 ⇒h=R[sin θ+cR(1+cos θ)^2 ] y_P ′=2cR(1+cos θ)=(1/(tan θ)) ⇒tan θ+sin θ=(1/(2cR)) ...(i) B(r,k) ⇒k=r[sin φ+cr(1+cos φ)^2 ] ⇒tan φ+sin φ=(1/(2cr)) ...(ii) h−k=(√((R+r)^2 −(R−r)^2 ))=2(√(Rr)) ⇒R[sin θ+cR(1+cos θ)^2 ]−r[sin φ+cr(1+cos φ)^2 ]=2(√(Rr)) ...(iii) ...... 3 eqn. with 3 unknowns r,φ,θ from (i): tan θ+sin θ=(1/(2cR))=(4/k)⇒k=8cR with t=tan (θ/2) ((2t)/(1−t^2 ))+((2t)/(1+t^2 ))=(4/k) t^4 +kt−1=0 ⇒t=f(k)=.... from (ii): ⇒cr=(1/(2(tan φ+sin φ))) from (iii): ⇒((k[k+4t(1+t^2 )])/(8(1+t^2 )^2 ))−(1/(tan φ+sin φ))[sin φ+(((1+cos φ)^2 )/(2(tan φ+sin φ)))]=(√(k/(tan φ+sin φ))) ⇒φ=.....
$${A}\left({R},{h}\right) \\ $$$${y}={cx}^{\mathrm{2}} \:{with}\:{c}=\mathrm{1} \\ $$$${y}'=\mathrm{2}{cx} \\ $$$${x}_{{P}} ={R}+{R}\:\mathrm{cos}\:\theta={R}\left(\mathrm{1}+\mathrm{cos}\:\theta\right) \\ $$$${y}_{{P}} ={h}−{R}\:\mathrm{sin}\:\theta={cx}_{{P}} ^{\mathrm{2}} ={cR}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{cos}\:\theta\right)^{\mathrm{2}} \\ $$$$\Rightarrow{h}={R}\left[\mathrm{sin}\:\theta+{cR}\left(\mathrm{1}+\mathrm{cos}\:\theta\right)^{\mathrm{2}} \right] \\ $$$${y}_{{P}} '=\mathrm{2}{cR}\left(\mathrm{1}+\mathrm{cos}\:\theta\right)=\frac{\mathrm{1}}{\mathrm{tan}\:\theta} \\ $$$$\Rightarrow\mathrm{tan}\:\theta+\mathrm{sin}\:\theta=\frac{\mathrm{1}}{\mathrm{2}{cR}}\:\:\:\:\:…\left({i}\right) \\ $$$${B}\left({r},{k}\right) \\ $$$$\Rightarrow{k}={r}\left[\mathrm{sin}\:\phi+{cr}\left(\mathrm{1}+\mathrm{cos}\:\phi\right)^{\mathrm{2}} \right] \\ $$$$\Rightarrow\mathrm{tan}\:\phi+\mathrm{sin}\:\phi=\frac{\mathrm{1}}{\mathrm{2}{cr}}\:\:\:…\left({ii}\right) \\ $$$$ \\ $$$${h}−{k}=\sqrt{\left({R}+{r}\right)^{\mathrm{2}} −\left({R}−{r}\right)^{\mathrm{2}} }=\mathrm{2}\sqrt{{Rr}} \\ $$$$\Rightarrow{R}\left[\mathrm{sin}\:\theta+{cR}\left(\mathrm{1}+\mathrm{cos}\:\theta\right)^{\mathrm{2}} \right]−{r}\left[\mathrm{sin}\:\phi+{cr}\left(\mathrm{1}+\mathrm{cos}\:\phi\right)^{\mathrm{2}} \right]=\mathrm{2}\sqrt{{Rr}}\:\:\:…\left({iii}\right) \\ $$$$……\:\mathrm{3}\:{eqn}.\:{with}\:\mathrm{3}\:{unknowns}\:{r},\phi,\theta \\ $$$${from}\:\left({i}\right): \\ $$$$\mathrm{tan}\:\theta+\mathrm{sin}\:\theta=\frac{\mathrm{1}}{\mathrm{2}{cR}}=\frac{\mathrm{4}}{{k}}\Rightarrow{k}=\mathrm{8}{cR} \\ $$$${with}\:{t}=\mathrm{tan}\:\frac{\theta}{\mathrm{2}} \\ $$$$\frac{\mathrm{2}{t}}{\mathrm{1}−{t}^{\mathrm{2}} }+\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }=\frac{\mathrm{4}}{{k}} \\ $$$${t}^{\mathrm{4}} +{kt}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{t}={f}\left({k}\right)=…. \\ $$$${from}\:\left({ii}\right): \\ $$$$\Rightarrow{cr}=\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{tan}\:\phi+\mathrm{sin}\:\phi\right)} \\ $$$${from}\:\left({iii}\right): \\ $$$$\Rightarrow\frac{{k}\left[{k}+\mathrm{4}{t}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\right]}{\mathrm{8}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{tan}\:\phi+\mathrm{sin}\:\phi}\left[\mathrm{sin}\:\phi+\frac{\left(\mathrm{1}+\mathrm{cos}\:\phi\right)^{\mathrm{2}} }{\mathrm{2}\left(\mathrm{tan}\:\phi+\mathrm{sin}\:\phi\right)}\right]=\sqrt{\frac{{k}}{\mathrm{tan}\:\phi+\mathrm{sin}\:\phi}} \\ $$$$\Rightarrow\phi=….. \\ $$
Commented by ajfour last updated on 01/Feb/19
Thats what sir, we cant even get centre of circles, given the radii, biquadratic problem..
$${Thats}\:{what}\:{sir},\:{we}\:{cant}\:{even}\:{get} \\ $$$${centre}\:{of}\:{circles},\:{given}\:{the}\:{radii}, \\ $$$${biquadratic}\:{problem}..\:\:\:\:\:\:\: \\ $$
Commented by mr W last updated on 24/Jan/19
t^4 +kt−1=0 should be solvable. can MJS sir help?
$${t}^{\mathrm{4}} +{kt}−\mathrm{1}=\mathrm{0} \\ $$$${should}\:{be}\:{solvable}.\:{can}\:{MJS}\:{sir}\:{help}? \\ $$
Commented by MJS last updated on 24/Jan/19
no “beautiful” solution t_(1, 2) =α±(√β) ∈R t_(3, 4) =γ±(√δ) ∉R α=−γ β=−γ^2 +(k/(4γ)) δ=−γ^2 −(k/(4γ)) γ=(1/( ((1152))^(1/6) ))(√(((9k^2 +(√(81k^4 +768))))^(1/3) −((−9k^2 +(√(81k^4 +768))))^(1/3) )) ∀k∈R\{0} k=0 ⇒ t=±1
$$\mathrm{no}\:“\mathrm{beautiful}''\:\mathrm{solution} \\ $$$${t}_{\mathrm{1},\:\mathrm{2}} =\alpha\pm\sqrt{\beta}\:\in\mathbb{R} \\ $$$${t}_{\mathrm{3},\:\mathrm{4}} =\gamma\pm\sqrt{\delta}\:\notin\mathbb{R} \\ $$$$\alpha=−\gamma \\ $$$$\beta=−\gamma^{\mathrm{2}} +\frac{{k}}{\mathrm{4}\gamma} \\ $$$$\delta=−\gamma^{\mathrm{2}} −\frac{{k}}{\mathrm{4}\gamma} \\ $$$$\gamma=\frac{\mathrm{1}}{\:\sqrt[{\mathrm{6}}]{\mathrm{1152}}}\sqrt{\sqrt[{\mathrm{3}}]{\mathrm{9}{k}^{\mathrm{2}} +\sqrt{\mathrm{81}{k}^{\mathrm{4}} +\mathrm{768}}}−\sqrt[{\mathrm{3}}]{−\mathrm{9}{k}^{\mathrm{2}} +\sqrt{\mathrm{81}{k}^{\mathrm{4}} +\mathrm{768}}}} \\ $$$$\forall{k}\in\mathbb{R}\backslash\left\{\mathrm{0}\right\} \\ $$$${k}=\mathrm{0}\:\Rightarrow\:{t}=\pm\mathrm{1} \\ $$
Commented by mr W last updated on 24/Jan/19
sir, it′s a beautiful solution!
$${sir},\:{it}'{s}\:{a}\:\mathrm{beautiful}\:\mathrm{solution}! \\ $$
Commented by ajfour last updated on 24/Jan/19
thanks afterall Sir!
$${thanks}\:{afterall}\:{Sir}! \\ $$
Commented by MJS last updated on 24/Jan/19
you′re welcome
$$\mathrm{you}'\mathrm{re}\:\mathrm{welcome} \\ $$
Commented by MJS last updated on 24/Jan/19
trying something else: parabola: y=ax^2 normal in P= ((p),((ap^2 )) ): y=−(1/(2ap))x+((2a^2 p^2 +1)/(2a)) horizontal: y=h normal∩horizontal = C= ((x_C ),(y_C ) ) C= (((p(2p^2 +1−2h))),(h) ) ∣CP∣=x_C ⇒ we can calculate the 3^(rd) parameter when 2 are given a=((h+(√(4h^2 −3p^2 )))/(3p^2 )) h=p(−ap+(√(4a^2 p^2 +1))) p=((√(2ah−1+(√(16a^2 h^2 −4ah+1))))/( (√6)a)) on the other hand two circles both touching the y−axis and each others can be calculated c_1 : (x−r_1 )^2 +(y−h_1 )^2 =r_1 ^2 c_2 : (x−r_2 )^2 +(y−h_2 )^2 =r_2 ^2 c_1 ∩c_2 =exactly 1 solution ⇒ ⇒ 4r_1 r_2 =(h_1 −h_2 )^2 ...not yet sure where this will lead to...
$$\mathrm{trying}\:\mathrm{something}\:\mathrm{else}: \\ $$$$\mathrm{parabola}:\:{y}={ax}^{\mathrm{2}} \\ $$$$\mathrm{normal}\:\mathrm{in}\:{P}=\begin{pmatrix}{{p}}\\{{ap}^{\mathrm{2}} }\end{pmatrix}:\:{y}=−\frac{\mathrm{1}}{\mathrm{2}{ap}}{x}+\frac{\mathrm{2}{a}^{\mathrm{2}} {p}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}{a}} \\ $$$$\mathrm{horizontal}:\:{y}={h} \\ $$$$\mathrm{normal}\cap\mathrm{horizontal}\:=\:{C}=\begin{pmatrix}{{x}_{{C}} }\\{{y}_{{C}} }\end{pmatrix} \\ $$$${C}=\begin{pmatrix}{{p}\left(\mathrm{2}{p}^{\mathrm{2}} +\mathrm{1}−\mathrm{2}{h}\right)}\\{{h}}\end{pmatrix} \\ $$$$\mid{CP}\mid={x}_{{C}} \:\Rightarrow\:\mathrm{we}\:\mathrm{can}\:\mathrm{calculate}\:\mathrm{the}\:\mathrm{3}^{\mathrm{rd}} \:\mathrm{parameter} \\ $$$$\mathrm{when}\:\mathrm{2}\:\mathrm{are}\:\mathrm{given} \\ $$$${a}=\frac{{h}+\sqrt{\mathrm{4}{h}^{\mathrm{2}} −\mathrm{3}{p}^{\mathrm{2}} }}{\mathrm{3}{p}^{\mathrm{2}} } \\ $$$${h}={p}\left(−{ap}+\sqrt{\mathrm{4}{a}^{\mathrm{2}} {p}^{\mathrm{2}} +\mathrm{1}}\right) \\ $$$${p}=\frac{\sqrt{\mathrm{2}{ah}−\mathrm{1}+\sqrt{\mathrm{16}{a}^{\mathrm{2}} {h}^{\mathrm{2}} −\mathrm{4}{ah}+\mathrm{1}}}}{\:\sqrt{\mathrm{6}}{a}} \\ $$$$ \\ $$$$\mathrm{on}\:\mathrm{the}\:\mathrm{other}\:\mathrm{hand}\:\mathrm{two}\:\mathrm{circles}\:\mathrm{both}\:\mathrm{touching} \\ $$$$\mathrm{the}\:{y}−\mathrm{axis}\:\mathrm{and}\:\mathrm{each}\:\mathrm{others}\:\mathrm{can}\:\mathrm{be}\:\mathrm{calculated} \\ $$$${c}_{\mathrm{1}} :\:\left({x}−{r}_{\mathrm{1}} \right)^{\mathrm{2}} +\left({y}−{h}_{\mathrm{1}} \right)^{\mathrm{2}} ={r}_{\mathrm{1}} ^{\mathrm{2}} \\ $$$${c}_{\mathrm{2}} :\:\left({x}−{r}_{\mathrm{2}} \right)^{\mathrm{2}} +\left({y}−{h}_{\mathrm{2}} \right)^{\mathrm{2}} ={r}_{\mathrm{2}} ^{\mathrm{2}} \\ $$$${c}_{\mathrm{1}} \cap{c}_{\mathrm{2}} =\mathrm{exactly}\:\mathrm{1}\:\mathrm{solution}\:\Rightarrow \\ $$$$\Rightarrow\:\mathrm{4}{r}_{\mathrm{1}} {r}_{\mathrm{2}} =\left({h}_{\mathrm{1}} −{h}_{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$ \\ $$$$…\mathrm{not}\:\mathrm{yet}\:\mathrm{sure}\:\mathrm{where}\:\mathrm{this}\:\mathrm{will}\:\mathrm{lead}\:\mathrm{to}… \\ $$

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