A curious connection between the orthocenter of a pentagon and linear algebra

Question

This question is a follow-on of this one which has received a complete answer by @Intelligenti Pauca.

Playing around with fig. 1 below, I have found a way to place the issue of the orthocenter of a pentagon (not necessarily cyclic) into a linear algebra context.

In this figure :

• points $V_k,H,H_k$ are aligned ;

• Orthogonality properties $\vec{HV_k} \perp \vec{V_{k+2}V_{k+3}}$ are verified, with cyclic numbering convention : $V_6=V_1, V_7=V_2$ etc.) ;

• All angles $a_k$ are present twice with $a_1+a_2+a_3+a_4+a_5=\pi$ ;

• Orthocenter $H$ is taken as the origin of coordinates.

Let us furthermore take the following notations :

$$h_k:=HH_k, \ \ c_k:=\cos(a_k), \ \ d=HV_1\tag{1}$$

We have :

$$\begin{cases}dc_k&=&h_4\\ h_4c_5&=&h_5c_2\\ h_5c_1&=&h_1c_3\\ h_1c_2&=&h_2c_4\\ h_2c_3&=&h_3c_5\\ dc_1&=&h_3\end{cases}\tag{2}$$

Remark : These equations are not independent : the product of the first five gives the sixth one.

The $h_k$s and the $c_k$s in equations (2) can be separated by rewriting them under the matrix form :

$$\underbrace{\pmatrix{0&0&0&0&c_1/c_3\\ c_2/c_4&0&0&0&0\\ 0&c_3/c_5&0&0&0\\ 0&0&c_4/c_1&0&0\\ 0&0&0&c_5/c_2&0}}_C\underbrace{\pmatrix{h_1\\h_2\\h_3\\h_4\\h_5}}_V=\underbrace{\pmatrix{h_1\\h_2\\h_3\\h_4\\h_5}}_V\tag{3}$$

which expresses the fact that vector $V$ is an eigenvector of circulant matrix $C$ associated with eigenvalue $1$.

The characteristic polynomial of $C$ being $-x^5+1$, (whatever the values of the $c_k$s) has indeed $1$ among its eigenvalues, with multiplicity $1$.

What is the interest of this "re-writing" ? Knowing angles $a_k$, it gives an efficient access, using any software (by computing a basis of the kernel of $C-I_5$) to the values of the $h_k$s knowing the $a_k$s.

But, in its present state, it doesn't prove the existence of the orthocenter of the corresponding pentagon because we need to know in prior hand the $a_k$s...

So, my question is : Consider a pentagon (and in particular a cyclic pentagon) given by the coordinates of its vertices in a coordinate system which hasn't the (unknown...) orthocenter as its origin as is the case for Fig. 1, how would it be possible to take profit of relationship (3) to obtain the coordinates of orthocenter $H$ ?

Remark 1 : The pentagon case can be extended to any $(2n+1)$-gons.

Remark 2 : This linear algebra issue is maybe connected to Discrete Fourier Transform.

The figure has been constructed with the following Matlab program :

 a=[0.75, 0.25, 0.5, 0.82,0.82]; % sum of angles = pi
 c=cos(a);
 C=[0,0,0,0,c(1)/c(3);
    c(2)/c(4),0,0,0,0;
    0,c(3)/c(5),0,0,0;
    0,0,c(4)/c(1),0,0;
    0,0,0,c(5)/c(2),0];
 x=1;h=null(C-x*eye(5));h=[h;h(1)]';
 L=[0,a(4)+a(2),a(5)+a(3),a(1)+a(4),a(2)+a(5),a(3)+a(1)];
 Lc=cumsum();
 R=-(h.*exp(i*Lc));
 plot(R,'ob');hold on;axis equal
 plot([0,R(1),0,R(2),0,R(3),0,R(4),0,R(5)],'-g');hold on;
 V=exp(i*Lc(1:5)).*[h(4)/c(4),h(5)/c(5),h(1)/c(1),h(2)/c(2),h(3)/c(3)];
 plot([0,V(1),0,V(2),0,V(3),0,V(4),0,V(5)],'-sg');hold on;