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Question Number 89281 by ajfour last updated on 16/Apr/20

Commented by ajfour last updated on 17/Apr/20

Both parabolas have same shape.  If equation of one is y=x^2 ,  and the inscribed triangle is  eqiuilateral, find the eq. of the  other parabola.  Find also the triangle side.

$${Both}\:{parabolas}\:{have}\:{same}\:{shape}. \\ $$$${If}\:{equation}\:{of}\:{one}\:{is}\:{y}={x}^{\mathrm{2}} , \\ $$$${and}\:{the}\:{inscribed}\:{triangle}\:{is} \\ $$$${eqiuilateral},\:{find}\:{the}\:{eq}.\:{of}\:{the} \\ $$$${other}\:{parabola}. \\ $$$${Find}\:{also}\:{the}\:{triangle}\:{side}. \\ $$

Commented by ajfour last updated on 17/Apr/20

MjS Sir, mrW Sir , please solve.

$${MjS}\:{Sir},\:{mrW}\:{Sir}\:,\:{please}\:{solve}. \\ $$

Answered by mr W last updated on 17/Apr/20

say the side length of the equilateral  triangle is s.  for a given s, the coordinates of points  A,B, C can be determined according  to the method from ajfour sir in  Q62938.

$${say}\:{the}\:{side}\:{length}\:{of}\:{the}\:{equilateral} \\ $$$${triangle}\:{is}\:{s}. \\ $$$${for}\:{a}\:{given}\:{s},\:{the}\:{coordinates}\:{of}\:{points} \\ $$$${A},{B},\:{C}\:{can}\:{be}\:{determined}\:{according} \\ $$$${to}\:{the}\:{method}\:{from}\:{ajfour}\:{sir}\:{in} \\ $$$${Q}\mathrm{62938}. \\ $$

Commented by mr W last updated on 24/Apr/20

coordinates of the centroid G(h,k):  h=(√((s^2 /(48))−(1/4)))  k=((3s^2 )/(16))−(1/4)  (s^2 /(48))−(1/4)≥0 ⇒s≥2(√3)    the x−coordinates of points A,B,C  can be calculated as roots of following  equation:  x^3 +ax^2 +bx+c=0  with  a=−3h  b=(3/2)(3h^2 −k)  c=(1/(3h))(h^2 +k^2 −(s^2 /3))    the equation can be reformed to:  z^3 +3pz+2q=0  with  x=z+h  p=((h^2 −k)/2)  q=((h(5h^2 −3k))/4)+((3h^2 +3k^2 −s^2 )/(18h))    the roots are  x_i =2(√(−p)) sin ((1/3)sin^(−1) (q/(√(−p^3 )))+((2iπ)/3))+h  (i=0,1,2 for point A,B,C respectively)    x_A =x_0 =2(√(−p)) sin ((1/3)sin^(−1) (q/(√(−p^3 ))))+h  x_C =x_2 =2(√(−p)) sin ((1/3)sin^(−1) (q/(√(−p^3 )))+((4π)/3))+h

$${coordinates}\:{of}\:{the}\:{centroid}\:{G}\left({h},{k}\right): \\ $$$${h}=\sqrt{\frac{{s}^{\mathrm{2}} }{\mathrm{48}}−\frac{\mathrm{1}}{\mathrm{4}}} \\ $$$${k}=\frac{\mathrm{3}{s}^{\mathrm{2}} }{\mathrm{16}}−\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\frac{{s}^{\mathrm{2}} }{\mathrm{48}}−\frac{\mathrm{1}}{\mathrm{4}}\geqslant\mathrm{0}\:\Rightarrow\boldsymbol{{s}}\geqslant\mathrm{2}\sqrt{\mathrm{3}} \\ $$$$ \\ $$$${the}\:{x}−{coordinates}\:{of}\:{points}\:{A},{B},{C} \\ $$$${can}\:{be}\:{calculated}\:{as}\:{roots}\:{of}\:{following} \\ $$$${equation}: \\ $$$${x}^{\mathrm{3}} +{ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0} \\ $$$${with} \\ $$$${a}=−\mathrm{3}{h} \\ $$$${b}=\frac{\mathrm{3}}{\mathrm{2}}\left(\mathrm{3}{h}^{\mathrm{2}} −{k}\right) \\ $$$${c}=\frac{\mathrm{1}}{\mathrm{3}{h}}\left({h}^{\mathrm{2}} +{k}^{\mathrm{2}} −\frac{{s}^{\mathrm{2}} }{\mathrm{3}}\right) \\ $$$$ \\ $$$${the}\:{equation}\:{can}\:{be}\:{reformed}\:{to}: \\ $$$${z}^{\mathrm{3}} +\mathrm{3}{pz}+\mathrm{2}{q}=\mathrm{0} \\ $$$${with} \\ $$$${x}={z}+{h} \\ $$$${p}=\frac{{h}^{\mathrm{2}} −{k}}{\mathrm{2}} \\ $$$${q}=\frac{{h}\left(\mathrm{5}{h}^{\mathrm{2}} −\mathrm{3}{k}\right)}{\mathrm{4}}+\frac{\mathrm{3}{h}^{\mathrm{2}} +\mathrm{3}{k}^{\mathrm{2}} −{s}^{\mathrm{2}} }{\mathrm{18}{h}} \\ $$$$ \\ $$$${the}\:{roots}\:{are} \\ $$$${x}_{{i}} =\mathrm{2}\sqrt{−{p}}\:\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \frac{{q}}{\sqrt{−{p}^{\mathrm{3}} }}+\frac{\mathrm{2}{i}\pi}{\mathrm{3}}\right)+{h} \\ $$$$\left({i}=\mathrm{0},\mathrm{1},\mathrm{2}\:{for}\:{point}\:{A},{B},{C}\:{respectively}\right) \\ $$$$ \\ $$$${x}_{{A}} ={x}_{\mathrm{0}} =\mathrm{2}\sqrt{−{p}}\:\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \frac{{q}}{\sqrt{−{p}^{\mathrm{3}} }}\right)+{h} \\ $$$${x}_{{C}} ={x}_{\mathrm{2}} =\mathrm{2}\sqrt{−{p}}\:\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \frac{{q}}{\sqrt{−{p}^{\mathrm{3}} }}+\frac{\mathrm{4}\pi}{\mathrm{3}}\right)+{h} \\ $$

Commented by ajfour last updated on 18/Apr/20

Sir, Q.89415 please , can you  confirm just the answer..

$${Sir},\:{Q}.\mathrm{89415}\:{please}\:,\:{can}\:{you} \\ $$$${confirm}\:{just}\:{the}\:{answer}.. \\ $$

Commented by mr W last updated on 17/Apr/20

Commented by mr W last updated on 18/Apr/20

this solution was only possible due to  your earlier wonderful work. therefore  big thanks to you for the question and  for the contribution to the solution!

$${this}\:{solution}\:{was}\:{only}\:{possible}\:{due}\:{to} \\ $$$${your}\:{earlier}\:{wonderful}\:{work}.\:{therefore} \\ $$$${big}\:{thanks}\:{to}\:{you}\:{for}\:{the}\:{question}\:{and} \\ $$$${for}\:{the}\:{contribution}\:{to}\:{the}\:{solution}! \\ $$

Commented by mr W last updated on 17/Apr/20

Commented by mr W last updated on 18/Apr/20

since both parabolas have the same  shape and the triangle is equilateral,  the correlation between the second  parabola and the triangle is the same  as the correlation between the first  parabola and the triangle. the second  parabola is just the first parabola  after a rotation of 120° about the  centroid G of the triangle. and the  point A is rotated to the position  of point C, point B to the position of  point A, point C to the position of  point B.    from Q55613 we know: when a curve  y=f(x) is rotated about the point (h,k)  by an angle θ (clockwise +), it becomes  the curce v=f(u) with  u=(x−h) cos θ−(y−k) sin θ+h  v=(x−h) sin θ+(y−k) cos θ+k    in current case:  θ=120°  from y=x^2  we get then  ((√3)/2)(x−h)−(1/2)(y−k)+k =                    [−(1/2)(x−h)−((√3)/2)(y−k)+h]^2   or  2(√3)(x−h)−2y+3k=[x−2h+(√3)(y−k)]^2     this is the equation of the second  parabola with  h=(√((s^2 /(48))−(1/4)))  k=((3s^2 )/(16))−(1/4)    but there is an other condition:  both parabolas should touch each other  at point C.

$${since}\:{both}\:{parabolas}\:{have}\:{the}\:{same} \\ $$$${shape}\:{and}\:{the}\:{triangle}\:{is}\:{equilateral}, \\ $$$${the}\:{correlation}\:{between}\:{the}\:{second} \\ $$$${parabola}\:{and}\:{the}\:{triangle}\:{is}\:{the}\:{same} \\ $$$${as}\:{the}\:{correlation}\:{between}\:{the}\:{first} \\ $$$${parabola}\:{and}\:{the}\:{triangle}.\:{the}\:{second} \\ $$$${parabola}\:{is}\:{just}\:{the}\:{first}\:{parabola} \\ $$$${after}\:{a}\:{rotation}\:{of}\:\mathrm{120}°\:{about}\:{the} \\ $$$${centroid}\:{G}\:{of}\:{the}\:{triangle}.\:{and}\:{the} \\ $$$${point}\:{A}\:{is}\:{rotated}\:{to}\:{the}\:{position} \\ $$$${of}\:{point}\:{C},\:{point}\:{B}\:{to}\:{the}\:{position}\:{of} \\ $$$${point}\:{A},\:{point}\:{C}\:{to}\:{the}\:{position}\:{of} \\ $$$${point}\:{B}. \\ $$$$ \\ $$$${from}\:{Q}\mathrm{55613}\:{we}\:{know}:\:{when}\:{a}\:{curve} \\ $$$${y}={f}\left({x}\right)\:{is}\:{rotated}\:{about}\:{the}\:{point}\:\left({h},{k}\right) \\ $$$${by}\:{an}\:{angle}\:\theta\:\left({clockwise}\:+\right),\:{it}\:{becomes} \\ $$$${the}\:{curce}\:{v}={f}\left({u}\right)\:{with} \\ $$$${u}=\left({x}−{h}\right)\:\mathrm{cos}\:\theta−\left({y}−{k}\right)\:\mathrm{sin}\:\theta+{h} \\ $$$${v}=\left({x}−{h}\right)\:\mathrm{sin}\:\theta+\left({y}−{k}\right)\:\mathrm{cos}\:\theta+{k} \\ $$$$ \\ $$$${in}\:{current}\:{case}: \\ $$$$\theta=\mathrm{120}° \\ $$$${from}\:{y}={x}^{\mathrm{2}} \:{we}\:{get}\:{then} \\ $$$$\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\left({x}−{h}\right)−\frac{\mathrm{1}}{\mathrm{2}}\left({y}−{k}\right)+{k}\:= \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[−\frac{\mathrm{1}}{\mathrm{2}}\left({x}−{h}\right)−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\left({y}−{k}\right)+{h}\right]^{\mathrm{2}} \\ $$$${or} \\ $$$$\mathrm{2}\sqrt{\mathrm{3}}\left({x}−{h}\right)−\mathrm{2}{y}+\mathrm{3}{k}=\left[{x}−\mathrm{2}{h}+\sqrt{\mathrm{3}}\left({y}−{k}\right)\right]^{\mathrm{2}} \\ $$$$ \\ $$$${this}\:{is}\:{the}\:{equation}\:{of}\:{the}\:{second} \\ $$$${parabola}\:{with} \\ $$$${h}=\sqrt{\frac{{s}^{\mathrm{2}} }{\mathrm{48}}−\frac{\mathrm{1}}{\mathrm{4}}} \\ $$$${k}=\frac{\mathrm{3}{s}^{\mathrm{2}} }{\mathrm{16}}−\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$ \\ $$$${but}\:{there}\:{is}\:{an}\:{other}\:{condition}: \\ $$$${both}\:{parabolas}\:{should}\:{touch}\:{each}\:{other} \\ $$$${at}\:{point}\:{C}. \\ $$

Commented by mr W last updated on 17/Apr/20

Commented by ajfour last updated on 18/Apr/20

hence side of triangle, remains to be obtained.. side of triangle under the tangency condition, must be unique.

Commented by mr W last updated on 24/Apr/20

that means: when the tangent line  of parabola at point A is rotated by  120°, it should coincide with the  tangent line of the parabola at point C.  i.e. θ_A =θ_C +120°    at point A:  tan θ_A =(dy/dx)=2x_A   at point C:  tan θ_C =(dy/dx)=2x_C   tan θ_A =tan (θ_C +120°)=((tan θ_C −(√3))/(1+(√3) tan θ_C ))  ⇒ 2x_A =((2x_C −(√3))/(1+2(√3) x_C ))       ...(I)  with  x_A =2(√(−p)) sin ((1/3)sin^(−1) (q/(√(−p^3 ))))+h  x_C =2(√(−p)) sin ((1/3)sin^(−1) (q/(√(−p^3 )))+((4π)/3))+h  p=((h^2 −k)/2)  q=((h(5h^2 −3k))/4)+((3h^2 +3k^2 −s^2 )/(18h))    we get from eqn. (I):  s≈3.779421

$${that}\:{means}:\:{when}\:{the}\:{tangent}\:{line} \\ $$$${of}\:{parabola}\:{at}\:{point}\:{A}\:{is}\:{rotated}\:{by} \\ $$$$\mathrm{120}°,\:{it}\:{should}\:{coincide}\:{with}\:{the} \\ $$$${tangent}\:{line}\:{of}\:{the}\:{parabola}\:{at}\:{point}\:{C}. \\ $$$${i}.{e}.\:\theta_{{A}} =\theta_{{C}} +\mathrm{120}° \\ $$$$ \\ $$$${at}\:{point}\:{A}: \\ $$$$\mathrm{tan}\:\theta_{{A}} =\frac{{dy}}{{dx}}=\mathrm{2}{x}_{{A}} \\ $$$${at}\:{point}\:{C}: \\ $$$$\mathrm{tan}\:\theta_{{C}} =\frac{{dy}}{{dx}}=\mathrm{2}{x}_{{C}} \\ $$$$\mathrm{tan}\:\theta_{{A}} =\mathrm{tan}\:\left(\theta_{{C}} +\mathrm{120}°\right)=\frac{\mathrm{tan}\:\theta_{{C}} −\sqrt{\mathrm{3}}}{\mathrm{1}+\sqrt{\mathrm{3}}\:\mathrm{tan}\:\theta_{{C}} } \\ $$$$\Rightarrow\:\mathrm{2}{x}_{{A}} =\frac{\mathrm{2}{x}_{{C}} −\sqrt{\mathrm{3}}}{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{3}}\:{x}_{{C}} }\:\:\:\:\:\:\:...\left({I}\right) \\ $$$${with} \\ $$$${x}_{{A}} =\mathrm{2}\sqrt{−{p}}\:\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \frac{{q}}{\sqrt{−{p}^{\mathrm{3}} }}\right)+{h} \\ $$$${x}_{{C}} =\mathrm{2}\sqrt{−{p}}\:\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \frac{{q}}{\sqrt{−{p}^{\mathrm{3}} }}+\frac{\mathrm{4}\pi}{\mathrm{3}}\right)+{h} \\ $$$${p}=\frac{{h}^{\mathrm{2}} −{k}}{\mathrm{2}} \\ $$$${q}=\frac{{h}\left(\mathrm{5}{h}^{\mathrm{2}} −\mathrm{3}{k}\right)}{\mathrm{4}}+\frac{\mathrm{3}{h}^{\mathrm{2}} +\mathrm{3}{k}^{\mathrm{2}} −{s}^{\mathrm{2}} }{\mathrm{18}{h}} \\ $$$$ \\ $$$${we}\:{get}\:{from}\:{eqn}.\:\left({I}\right): \\ $$$${s}\approx\mathrm{3}.\mathrm{779421} \\ $$

Commented by mr W last updated on 18/Apr/20

Commented by ajfour last updated on 18/Apr/20

But afterall its you who solved  it really, i was very puzzled, so  thank you enormously Sir.

$${But}\:{afterall}\:{its}\:{you}\:{who}\:{solved} \\ $$$${it}\:{really},\:{i}\:{was}\:{very}\:{puzzled},\:{so} \\ $$$${thank}\:{you}\:{enormously}\:{Sir}. \\ $$

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