Constructing "Double" circumscribed pentagons.

Question

I want to construct the following thing : A non regular pentagon $ABCDE$ circumscribed about a circle such that if $$A_1 = BC\cap ED, B_1=CD\cap AE...$$ then $A_1B_1C_1D_1E_1$ is also circumscribed about a circle.

I tried to do that in Geogebra for a bit but didn't succeed.

Answer 1

Here is a process that allows you to select two points on a circle, make some decisions about which roots of certain polynomials to pick, and results in a configuration meeting your requirements. I'm using projective geometry, homogeneous coordinates, lots of polynomials, Sage as my computer algebra system of choice, and Cinderella for the final visualization.

1. Assume, without loss of generality, that the inscribed circle of the first pentagon is the unit circle.
2. Start with five generic points on the unit circle, defined by five variables $a,b,c,d,e$ using the tangent half-angle formula to write their homogeneous coordinates as $A_0=[1-a^2:2a:1+a^2]$ and likewise for $B_0, C_0, D_0, E_0$. These are the points where the first pentagon will touch the unit circle.
3. Without loss of generality you can choose $e=\infty$ i.e. $E_0=[-1:0:1]$ to get rid of one variable and resolve the gap in the parametrization at the same time.
4. Now define $a_0$ as the tangent in point $A_0$ and so on for $b_0,c_0,d_0,e_0$. The homogeneous coordinates of a tangent are just the coordinates of the point with the last coordinate negated. (This is just a special case of multiplying the matrix of a conic with the coordinates of a point to get the coordinates of the corresponding polar line, since the polar line of an incident point is the tangent in that point.)
5. Label $A:=a_0\cap b_0, B=b_0\cap c_0, \ldots$ as the corners of the first pentagon. Intersecting lines is just taking the cross product of the homogeneous coordinates of the lines, so $A=a_0\times b_0,\ldots$.
6. Also compute $A_1=BC\cap ED=c_0\times e_0, B_1=CD\cap AE=d_0\times a_0, \ldots$ as the corners of the second pentagon.
7. Now connect these, which is again cross products. $a_1=A_1\times B_1, b_1=B_1\times C_1,\ldots$.
8. Contruct the dual matrix of the conic that is tangent to thse five lines. You can get that from \begin{align*} M_1&=(a_1\times b_1)\cdot(c_1\times d_1)^T+(c_1\times d_1)\cdot(a_1\times b_1)^T\\ M_2&=(a_1\times c_1)\cdot(b_1\times d_1)^T+(b_1\times d_1)\cdot(a_1\times c_1)^T\\ M_3&=\left(e_1^T\cdot M_2\cdot e_1\right)M_1-\left(e_1^T\cdot M_1\cdot e_1\right)M_2 \end{align*} $M_1$ and $M_2$ are degenerate conics, and $M_3$ is a suitable linear combination of these to satisfy all five tangentialities. A line $l$ is tangent to this inscribed conic if it satisfies $l^T\cdot M_3\cdot l$.
9. Compute $M=\operatorname{adj}(M_3)$, the adjugate matrix of $M_3$. This is the primal matrix of the same conic: a point $p$ lies on this conic if $p^T\cdot M\cdot p=0$.
10. This matrix represents a circle if both $M_{1,1}=M_{2,2}$ and $M_{1,2}=0$. The former means that $x^2$ and $y^2$ will have the same coefficient, while the latter means there is no mixed $xy$ term. Note that $M_{1,2}=M_{2,1}$ by construction, so no need to check both. (The conditions I stated are required but not sufficient for getting a circle. To make them sufficient we would also check $M_{1,1}\neq 0$ and ensure that we have any real points on the conict at all. I'll ignore this.)
11. Having done all of the above in a computer algebra system, you get two polynomials $M_{1,1}-M_{2,2}$ and $M_{1,2}$ which become zero when the two conditions are met. You can factor them and drop any factors like $a-b$ which essentially just encode a non-degeneracy condition requiring points to be distinct. There is one big factor left in each of them after that.
12. You can combine these two big factors using resultants to eliminate one more variable, say $d$. Now you have a polynomial condition in $a,b,c$ which has to be zero. Since the polynomial has degree $5$ in $c$ but only degree $4$ in $b$, I would say you can pick $a$ and $c$ as you like, then solve for $b$, finding up to $4$ real solutions.
13. You can then use the big factor from $M_{1,2}$ to solve for $d$. This is quadratic, so you get up to $2$ solutions for each $b$ you feed in.
14. Check that the solution you found matches the $M_{1,1}-M_{2,2}$ condition as well, since eliminating variables can introduce spurious solutions.
15. Now feed all of this to a dynamic geometry system of your choice to get an image.

Here is the expression you can use to find $b$ given $a$ and $c$:

$$ (-a^4c^2 - a^3c^3 + a^2c^4 - a^3c - 5a^2c^2 + 3ac^3 - 5ac + 2c^2 - 1)\,b^4 \\ + (a^4c^3 - 2a^3c^4 + a^4c + 2a^3c^2 - 4a^2c^3 - 3ac^4 + c^5 + 8a^2c - 12ac^2 + c^3 + 3a - 4c)\,b^3 \\ + (a^4c^4 - a^4c^2 + 7a^2c^4 - 3ac^5 - 4a^3c + 4a^2c^2 + 6ac^3 - 3c^4 - 3a^2 + 5ac - c^2)\,b^2 \\ + (a^4c^3 - 5a^3c^4 + 3a^2c^5 + a^4c - 10a^2c^3 + 6ac^4 + a^3 - a^2c - 6ac^2 + 3c^3 - c)\,b \\ + (a^4c^4 - a^3c^5 + 3a^3c^3 - 3a^2c^4 + 3a^2c^2 - 3ac^3 + ac - c^2) = 0 $$

And here is the expression to find $d$ given $a,b,c$:

$$ (-ab + bc)\,d^2 + (ab^2 + a^2c - ac^2 + bc^2 - b - c)\,d + (-a^2bc - ab^2c + ab + ac) = 0 $$

The picture at the top is what I get for $a=-1.2, c=0.5, b\approx-0.183, d\approx1.490$. The exact solution there satisfies $18032b^4 + 81385b^3 - 68156b^2 - 21040b - 1088 = 0$ and $78608d^4 + 167620d^3 - 575824d^2 - 21430d + 368293 = 0$. The Galois group of these polynomials is primitive and has order $24$, which is not a power of two. So even though the input points $A_0,C_0,E_0$ have rational coordinates, the resulting $B_0$ and $D_0$ are not constructible. You can't do a compass and straightedge construction of the figure I came up with. That doesn't preclude the possibility of some specific constructible examples, as members of this fairly large family. Perhaps there is one additional condition which would fix $c$ for a given $a$ in such a way that everything becomes constructible. But I don't know how to search for this.

Update 2023-12-27: Reading an answer from Tito Piezas III encouraged me to search for constructible solutions by just trying out some rational choices for $a$ and $c$ and checking whether the resulting quartic polynomial in $b$ might factor. Instructing my computer algebra system to do so, but to also check that the values $a$ through $d$ occurred in strictly ascending or strictly descending order, I found a surprise: There is even a completely rational solution!

$$ a=-1 \qquad b=-\frac{6}{61} \qquad c=\frac{10}{11} \qquad d=\frac{23}{7} $$

Multiplying by the common denominator of the resulting fractions and shfting by the minimal values, I can use this to describe a solution with only non-negative integer coordinates:

\begin{align*} A_0 &= \begin{pmatrix}\phantom{0}18795876608925\\\phantom{0}91890952310300\end{pmatrix} & A &= \begin{pmatrix}\phantom{0}27024917114325\\\phantom{0}91890952310300\end{pmatrix} & A_1 &= \begin{pmatrix}\phantom{00}8771409084165\\112942334112296\end{pmatrix} \\ B_0 &= \begin{pmatrix}\phantom{0}28628232964725\\\phantom{0}99962289617300\end{pmatrix} & B &= \begin{pmatrix}\phantom{0}30789142992885\\110840641330100\end{pmatrix} & B_1 &= \begin{pmatrix}\phantom{00000000000000}0\\\phantom{0}91890952310300\end{pmatrix} \\ C_0 &= \begin{pmatrix}\phantom{0}19748427821685\\111894527778260\end{pmatrix} & C &= \begin{pmatrix}\phantom{0}13799969340885\\112462335178700\end{pmatrix} & C_1 &= \begin{pmatrix}\phantom{00}8771409084165\\\phantom{00000000000000}0\end{pmatrix} \\ D_0 &= \begin{pmatrix}\phantom{0}10471059287325\\107499984788300\end{pmatrix} & D &= \begin{pmatrix}\phantom{00}8771409084165\\104966344733900\end{pmatrix} & D_1 &= \begin{pmatrix}229309694628885\\\phantom{0}91890952310300\end{pmatrix} \\ E_0 &= \begin{pmatrix}\phantom{00}8771409084165\\101915419835060\end{pmatrix} & E &= \begin{pmatrix}\phantom{00}8771409084165\\\phantom{0}91890952310300\end{pmatrix} & E_1 &= \begin{pmatrix}\phantom{0}38393884053525\\149124071396300\end{pmatrix} \\ && O &= \begin{pmatrix}\phantom{0}18795876608925\\101915419835060\end{pmatrix} & P &= \begin{pmatrix}\phantom{0}58598909427825\\\phantom{0}83045833906100\end{pmatrix} \end{align*}

Here $O$ is the center of the incircle of pentagon $ABCDE$ while $P$ is the center of the incircle of pentagon $A_1B_1C_1D_1E_1$. Since an integer grid is trivially easy to construct in principle, this proves the fact that this specific solution can be constructed. At the same time I wouldn't want to count much less construct the quantities above in practice, so there may still be room for some other answer to offer a more practical construction method.