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Did some testing with a CAS and found this cute fact:

Consider three points $a$, $b$, $c$ in the complex plane, and $d$ the orthocenter of the triangle $a b c$. Then: $$\det \begin{pmatrix} 1 & a & a^2 & \overline a \\ 1& b & b^2 & \overline b \\ 1& c & c^2 & \overline c\\ 1& d & d^2 & \overline d \end{pmatrix}=0$$ The above as an equation in $d$ has $4$ roots: $a$, $b$, $c$ and the orthocenter of $abc$.

Provided as reference. Perhaps a known fact, haven't seen it. References and/or feedback would be appreciated.

Note:

The formula for the center of the circle through $abc$ seems relevant. One can obtain from it a formula for the orthocenter.

$\bf{Added:}$ The $\Re$ and $\Im$ of the above equation give rectangular hyperbolas ( see also this), which pass through the orthocentric system $abcd$.

$\bf{Added:}$

The solution of @Jean Marie: inspired me to get another one a bit more algebraic. The main idea is still this: since the equation is invariant under complex affine transformations $z \mapsto \alpha z + \beta$, we may assume that $|a|=|b|=|c|=1$ ( the circuscribed circle is the unit circle). Recall that we always have the vector equality ( Sylvester theorem)

$$\vec{OH}= \vec{OA}+\vec{OB} + \vec{OC}$$

where $O$ is the center of the circuscribed circle of the $\Delta ABC$, and $H$ is the orthocenter.

In our case, since $o = 0$, we get for $d$ the orthocenter ( its complex coordinate)

$$d = a+b+c$$

Recall that $\bar a = \frac{1}{a}$, $\bar b = \frac{1}{b}$ and $\bar c= \frac{1}{c}$, so $\bar d = \frac{1}{a} + \frac{1}{b}+ \frac{1}{c}$

It is enough to show that

Now we need to show that

$$\det \begin{pmatrix} \frac{1}{p} &1& p& p^2\\ \frac{1}{q} &1& q& q^2\\ \frac{1}{r} &1& r& r^2 \\ t &1& s& s^2 \end{pmatrix}=0$$

if $s= p+q+r$, and $t= \frac{1}{p} + \frac{1}{q} + \frac{1}{r}$. Now there is a vector perpendicular to the first $3$ rows of the above matrix, it is $(-p q r, p q + q r + p r, -(p+q+r), 1)$. We only need to check that it is also orthogonal to the last row, that is

$$\left(-p q r, p q + q r + p r, -(p+q+r), 1\right)\cdot\left( (\frac{1}{p} + \frac{1}{q} + \frac{1}{r}), 1, (p+q+r), (p+q+r)^2\right) = 0$$

The last two products cancel out obviously, and so do the first two ones.

orangeskid
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2 Answers2

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I think worth of presenting the following approach, though its result is only partialy established (confirmed by numerical computations).

Let us assume WLOG that $a,b,c$ belong to the unit circle.

Your result is equivalent to find a non-zero vector in the kernel of the following matrix (I have transposed it because it was more convenient for me, and exchanged two rows, also WLOG) :

$$M=\begin{pmatrix} 1 & 1 & 1 & 1\\ a& b & c & d \\ \overline a & \overline b & \overline c & \overline d\\ a^2& b^2 & c^2 & d^2 \end{pmatrix}$$

(or, said differently, an eigenvector associated to eigenvalue $0$).

This eigenvector is :

$$V=\pmatrix{\tan(A)\\ \tan(B) \\ \tan(C) \\ -(\tan(A)+\tan(B)+\tan(C))}$$

The fact that $MV=0$ doesn't really looks like the rabbit out of the hat. Indeed $(\tan(A),\tan(B),\tan(C))$ are, normalized by dividing by their sum, the barycentric coordinates of the orthocenter, explaining the zero products of the 3 first lines of matrix $M$ with $V$.

The surprize comes from the zero product of the 4th line of the matrix with vector $V$ ; I have verified on different numerical examples that it was OK also for this fourth line, but I haven't found a theoretical reason till now...

Here is the Matlab program I have built :

 clear all;close all;hold on;axis equal;
 t=0:0.01:2*pi;
 p=[0,2*pi/3,5*pi/4,2*pi]; % polar angles
 v=exp(i*p(1:3)); % vertices
 a=v(1);b=v(2);c=v(3);
 A=(p(3)-p(2))/2;
 B=(p(4)-p(3))/2;
 C=(p(2)-p(1))/2;
 an=[A,B,C]; % list of angles
 TL=tan(an); % their tangents
 s=sum(TL);
 d=exp(i*p(1:3))*TL'/s; % orthocenter
 %tri=exp(i*p);plot(tri),hold on;
 %plot(exp(i*t));hold on;
 %plot(d,'or');hold on;
 [TL,-s]*[1 a a^2 conj(a);
          1 b b^2 conj(b);
          1 c c^2 conj(c);
          1 d d^2 conj(d)], % result always equal to (0,0,0,0)
Jean Marie
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    I am asking myself if the answer isn't in lemma 2 of this document saying that "the incenter is the orthocenter of opposite arcs midpoints" – Jean Marie Jan 14 '24 at 00:43
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    now I see that the equation is invariant under complex affine maps ( as it should) so indeed we may assume $a$, $b$, $c$ of norm $1$. – orangeskid Jan 14 '24 at 01:04
  • Wrote another answer based on your idea to consider $a$, $b$, $c$ of modulus $1$. Thanks ! – orangeskid Jan 15 '24 at 16:09
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The equation is invariant under affine transformations $z\mapsto \alpha z + \beta$. So we may assume that we have an orthocentric system formed by the centers of incircle and excircles of a triangle inscribed in a circle of radius $1$. Let the complex coordinates of the vertices be $a^2$, $b^2$, $c^2$ ( $|a|=|b|=|c|=1$). Then (see )the have the orthocentric system formed by

$$-a b - b c - c a\\ a b - b c + c a\\ a b + b c - c a\\ - a b + b c + c a$$

and we must show that

$$\det \left ( \begin{matrix} 1 & a b + b c + c a & (a b + b c + c a)^2 & \frac{1}{a b} + \frac{1}{b c} + \frac{1}{c a} \\ 1 & \ldots & \ldots & \ldots \end{matrix} \right) = 0$$

Now it is enough to see that each row $0$ dot product with the row $$(a^2 b^2 + b^2 c^2 + c^2 a^2, 0, -1, 2 a^2 b^2 c^2)$$

(so column $3$ is a linear combination of $1$ and $4$).

orangeskid
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