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Cubic equation with a unit radius circle A cubic equation and circle (unit radius) has intersection at A,B,C,D. ABCD is a square. Find the angle $\theta$.

I tried:

  1. $(0,0)$ is a solution so constant term is $0$

  2. Substituting A(x,y) and C(-x,-y) and adding them gives coefficient of $x^2$ is 0.

Then the cubic becomes f(x) = $ax^3+bx$.

3.Substituting A and B and added the two equations.

I found it interesting-for n points given we can find a unique n+1 degree polynomial

Also - Can complex number be used here?

Please note: I am not sure whether we can find the angle(integer) without knowing the coefficients of the cubic.

EDIT: From the answers
1.putting A $(cos\theta,sin\theta)$ in f(x) : $acos^3\theta + b cos\theta = sin\theta$

2.putting B $(-sin\theta,cos\theta)$ in f'(x) : $asin^2\theta + b = tan\theta$ [ as circle has $tan\theta$ slope at B]

$1,2 $ eqn gives $3asin^2\theta = acos^2\theta$

So, $sin^2\theta = \frac{1}{4}$

But I getting the value of $\theta$ but a answer shows plot of many cubics -> because in my case $ABCD$ is a square.

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  • $\begingroup$ A very interesting issue. I think I have a part of the solution, but not sure... $\endgroup$ – Jean Marie Apr 10 at 21:29
  • $\begingroup$ Can you please post it Sir $\endgroup$ – Nayas Apr 10 at 21:35
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We are looking for a third degree curve $(C)$ with cartesian equation:

$$y=ax(x^2-B)\tag{1}$$

Due to the fact it is an odd function, we can restrict our attention to the $x>0$ part.

The rest will follow by symmetry with respect to the origin.

I am going to show that the solution is:

$$a=\tfrac{8}{\sqrt{3}}, \ \ B=\tfrac{5}{8} \tag{2}$$

A parametric representation of $(C)$ is obtained (a classical method) by intersecting it with the line having the equation $y=tx$, where $t$ is to be interpreted (this is important) as $t:=\tan \theta$:

$$\begin{cases}x&=&\sqrt{\frac{t}{a}+B}\\y&=&t\sqrt{\frac{t}{a}+B}\end{cases}\tag{3}$$

Intersecting now $(C)$ with the unit circle $U$ ($x^2+y^2=1$) gives equation:

$$t^3+t^2aB+t+a(B-1)=0\tag{4}$$

As $\tan \theta = t$ is a root we must have as well $-\tfrac{1}{t}$ for a root the fullfillment of the orthogonality condition (think to $f'(x)$ vs. $-\dfrac{1}{f'(x)}$). Moreover, this root has to be doubled for the tangency condition.

Therefore, equation (4) must have the form:

$$(t-t_0)(t+\tfrac{1}{t_0})^2=0\tag{5}$$

which encompasses all the constraints of the issue.

Identifying coefficients in (4) and (5) gives the 3 equations:

$$aB=\tfrac{2}{t_0}-t_0, \ \ \ \ (\tfrac{1}{t_0})^2 - 2 = 1, \ \ \ \ -\tfrac{1}{t_0}=a(B-1)$$

implying $t_0=\tfrac{1}{\sqrt{3}}$ and, at once, (2).

We retrieve of course as well angle $\theta_0=\operatorname{atan}(\tfrac{1}{\sqrt{3}})=\pi/6$.

The following figure displays different curves giving a rectangle, with, in red, the solution curve with coefficients given by (2) giving a square:

enter image description here

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  • $\begingroup$ Thank you very much Prof Jean Marie. $\endgroup$ – Nayas Apr 11 at 6:45
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    $\begingroup$ Thanks for drawing the family of such cubic . You may see my question I have added a small part from the three answers $\endgroup$ – Nayas Apr 11 at 6:54
  • $\begingroup$ I have erased the complicated former "first solution", only keeping the former "second" solution, so straightforward compared to the first one. $\endgroup$ – Jean Marie Apr 13 at 8:07
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Let $$f(x)=ax^3+bx. \quad \quad (1)$$ Let's find out the values of $a$ and $b$ for a specific value of $R$.

Let the points $ A=(R \cos \theta, R \sin \theta)$ and $B=(-R \sin \theta, R \cos \theta)$ two intersection points of the circle $\Lambda (O, R)$ with $f(x)$, so that B is a tangent point.

Substituting the coordinates of $A$ in $f(x)$, we get: $$\tan \theta =aR^2\cos^2\theta+b. \quad \quad (2)$$

Substituting the coordinates of $A$ and $B$ in $f(x)$, we get with some algebra: $$a =\frac{4}{R^2 \sin4\theta}. \quad \quad (3)$$

Substituting the $x$ of $B$ in $f'(x)$, which is equal to $\tan \theta$, we get: $$\tan \theta =3aR^2 \sin^2\theta +b. \quad \quad (4)$$

Choosing $R =2$ and substituting in $(2)$ and $(4)$, we get: $$\theta = 30°.$$

The values of $a$ and $b$ are:

$$a =\frac{2}{\sqrt 3}$$

and

$$b =-\frac{5\sqrt 3}{3}.$$

Plot: CircleAndPoly

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  • $\begingroup$ So we can't find theta if the coefficient of polynomial is not given? $\endgroup$ – Nayas Apr 11 at 3:33
  • $\begingroup$ Thank you very much RicardoCruz I got theta as R was 1. $\endgroup$ – Nayas Apr 11 at 6:43
  • $\begingroup$ I have attempted to plot the curve with these values of $a$ and $b$ and, unless I have don an error, I don't find it a solution. $\endgroup$ – Jean Marie Apr 11 at 7:26
  • $\begingroup$ In fact, $b$ is good and $a$ should be $8/\sqrt{3}$ $\endgroup$ – Jean Marie Apr 11 at 8:30
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    $\begingroup$ @JeanMarie, I've inserted a graph of the solution for $R = 2$. $\endgroup$ – RicardoCruz Apr 11 at 11:12
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HINT

The cubic must clearly be of type $$ y = bx\left( {x^{\,2} - a^{\,2} } \right) $$

In polar coordinates $$ r\sin \theta = br\cos \theta \left( {r^{\,2} \cos ^{\,2} \theta - a^{\,2} } \right) $$ i.e. $$ 0 = r\left( {br^{\,2} \cos ^{\,3} \theta - \left( {a^{\,2} b\cos \theta + \sin \theta } \right)} \right) $$ and excluding the origin $$ 0 = br^{\,2} \cos ^{\,3} \theta - c\cos \left( {\theta + \beta } \right) $$ where either $b$ and $c$ can be taken as positive.

So $$ r = \sqrt {{{c\cos \left( {\theta + \beta } \right)} \over {b\cos ^{\,3} \theta }}} $$

Then $D$ is a local max for $r$, and you shall impose to find the same $r_{max}$ at $90^{\circ}$ thereafter.

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