# Cubic equation with circle intersection to form a square 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.

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.

• A very interesting issue. I think I have a part of the solution, but not sure... – Jean Marie Apr 10 at 21:29
• Can you please post it Sir – Nayas Apr 10 at 21:35

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: • Thank you very much Prof Jean Marie. – Nayas Apr 11 at 6:45
• Thanks for drawing the family of such cubic . You may see my question I have added a small part from the three answers – Nayas Apr 11 at 6:54
• I have erased the complicated former "first solution", only keeping the former "second" solution, so straightforward compared to the first one. – Jean Marie Apr 13 at 8:07

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

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

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

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.