 # Consider a conic section. There are 2 rectangles such that all of the 8 vertices of the 2 rectangles lie on the conic section. Further assume that the 2 rectangles have different orientation (ie. a side of one rectangle is not parallel to any side of the other rectangle). What are all the possible conic section for that to be the case? kjukks1234531 2022-09-14 Answered
Orientation of rectangle on conic section
Consider a conic section. There are 2 rectangles such that all of the 8 vertices of the 2 rectangles lie on the conic section. Further assume that the 2 rectangles have different orientation (ie. a side of one rectangle is not parallel to any side of the other rectangle). What are all the possible conic section for that to be the case?
Intuitively, the only possibility is the circle. It appears that the problem could potentially be solved by write down all the conic section equation and solving some complicated set of equations, but I would rather not do that. I am hoping for a nice and element geometrical approach to the problem, or one that base on group theory regarding isometry in Euclidean plane/space.
You can still ask an expert for help

• Live experts 24/7
• Questions are typically answered in as fast as 30 minutes
• Personalized clear answers

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it Haylee Krause
Step 1
A rectangle has two lines of mirror symmetry that are perpendicular to each other and pass through its center. If a rectangle is to have all 4 vertices lie on a conic section, then it must be centered at the center of the conic section (a parabola's center is at infinity), since otherwise the breaking of symmetry would prevent some vertices from lying on the conic section. Now you also ask that 2 rectangles with non-parallel sides simultaneously satisfy this condition, which says that the conic section must have at least 4 distinct lines of mirror symmetry. Only a circle satisfies this constraint.
Let me clarify the center of symmetry assertion that you asked for in the comment. A generic non-degenerate conic section can be written in matrix form as ${\left[\begin{array}{c}x\\ y\end{array}\right]}^{T}\left[\begin{array}{cc}P& Q/2\\ Q/2& R\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]+{\left[\begin{array}{c}D\\ E\end{array}\right]}^{T}\left[\begin{array}{c}x\\ y\end{array}\right]+F={r}^{T}Ar+{b}^{T}r+c=0\phantom{\rule{2em}{0ex}}\left(0\right)$
For degenerate cases, A is identically zero, and we ignore this for now. This form can be re-centered by considering a change of coordinates: ${r}^{\prime T}{A}^{\prime }{r}^{\prime }+{c}^{\prime }=0\phantom{\rule{2em}{0ex}}\left(1\right)$
where $r={r}^{\prime }-{r}_{0}$ and the center ${r}_{0}$ is defined by $2{A}^{T}{r}_{0}=b$. Note that in the case of a parabola, $detA=0$, so it is not possible to define a center, but otherwise in the cases of a circle, ellipse, or hyperbola, ${r}_{0}$ exists.
Step 2
Now, in Eq. (1), it's fairly obvious that if r′ is a solution, then so is −r′, so the equation is centered at the origin, and the original uncentered equation (Eq. (0)) has a point of inversion symmetry about ${r}_{0}$.
To demonstrate mirror symmetry, we simply note that A always has an eigenvalue decomposition since it is a real symmetric matrix, so $A=U\mathrm{\Lambda }{U}^{T}$ where $\mathrm{\Lambda }=\mathrm{diag}\left({\lambda }_{1},{\lambda }_{2}\right)$ is diagonal and U is orthogonal. Let us change coordinates again to $\rho ={U}^{T}{r}^{\prime }=\left[\xi ,\eta {\right]}^{T}$, so
${\rho }^{T}\left[\begin{array}{cc}{\lambda }_{1}& \\ & {\lambda }_{2}\end{array}\right]\rho +{c}^{\prime }={\lambda }_{1}{\xi }^{2}+{\lambda }_{2}{\eta }^{2}+{c}^{\prime }=0\phantom{\rule{2em}{0ex}}\left(2\right)$
From this is it is obvious that the Eq. (2) has two mirror symmetries along the $\xi$ and $\eta$ axes. Furthermore, since U is orthogonal, this means that translating back, Eq. (0) has two orthogonal lines of mirror symmetry going through ${r}_{0}$

We have step-by-step solutions for your answer! Valentina Holland
Step 1
It is safe to assume the conic is an ellipse (explanation below).
There are many affine transformations that convert the ellipse to a circle, and these send parallelograms (which in this problem happen to be rectangles) inscribed in the ellipse to rectangles inscribed in the circle. Declaring one of the right angles in the ellipse-rectangles as the center and x,y axes of a coordinate system, one of those affine transformations keeps the coordinate axes in place; take any transformation that changes the ellipse to a circle, and follow it with a translation and rotation to fix the position of the coordinate system. Such a coordinate-friendly transformation can only be $\left(x,y\right)\to \left(ax,by\right)$ for nonzero a and b, which multiplies y/x by the constant $c=\frac{b}{a}$.
Step 2
This transformation has the unusual property of also preserving orthogonality of a second pair of lines, the sides at any corner of the other rectangle, that are not parallel to the coordinate axes. It will therefore preserve the orthogonality of pairs of lines in those directions through 0. Thus for some positive real m, the perpendicular lines $y=mx$ and $y=-\frac{x}{m}$, are sent to $y=cmx$ and $y=-\frac{c}{m}x$ which are still perpendicular, $\left(cm\right)\left(-c/m\right)=-1$, so that $c=±1$ or $a=±b$. Thus the chosen affine transformation from ellipse to circle to was scaling by a factor of |a|, maybe with a reflection, and the ellipse was already a circle.
Validity of reduction to ellipse: The assertion to be proved has the form that n polynomial equations in the parameters of the problem imply one more polynomial equation, and the restriction to real coefficients with positive determinant is a fully general (called "Zariski dense") subset for the purpose of proving such identities.

We have step-by-step solutions for your answer!