Prove $(-a,-\frac{a}{2})$ is orthocenter for triangle formed by given lines

Question

I need to prove that $(-a,-\frac{a}{2})$ is the orthocentre of the triangle formed by the lines

$$y = m_ix+\frac{a}{m_i}$$

with $i = 1,2,3$; $m_1,m_2,m_3$ being the roots of the equation $$x^3-3x^2+2=0$$

My attempt:

The roots of the equation are $x=1,1+\sqrt3,1-\sqrt3$ So I found the slopes of the lines perpendicular to that and tried to solve for the vertices of the triangle and then solve for the orthocenter using slopes. But that seems way too long and lengthy.

I noticed that the given lines are the equations for tangents to a parabola of the form $y^2=4ax$.

Is there any way I could use that to my advantage?
I think I'm missing something here.

Answer 1

As shown below, you do not need to find $m_i$ explicitly; instead, only the Vieta's relationships below for $x^3-3x^2+2=0$,

$$m_1m_2m_3=-2,\>\>\>\>\>m_1m_2+m_2m_3+m_3m_1=0\tag 1$$

The vertexes of the triangle, which are the intersections of the sides $y = m_ix+\frac{a}{m_i}$, can be computed directly,

$$(x_1,y_1) = a\left(\frac 1{m_2m_3},\frac 1{m_2}+\frac 1{m_3}\right),\>\>\>\>\>\> (x_2,y_2) = a\left(\frac 1{m_3m_1},\frac 1{m_3}+\frac 1{m_1}\right)\tag 2$$

The corresponding altitude lines are,

$$y-y_1= -\frac 1{m_1}(x-x_1),\>\>\>\>\>\>y-y_2= -\frac 1{m_2}(x-x_2)$$

whose intersection is the orthocenter. Its $x$-coordinate is solved by taking the difference, i.e.

$$y_2-y_1= \left(\frac1{m_2}-\frac1{m_1}\right)x + \left( \frac {x_1}{m_1} -\frac {x_2}{m_2} \right) $$

From (2), we have $y_2-y_1 = -a(\frac1{m_2}-\frac1{m_1})$ and $\frac {x_1}{m_1} =\frac {x_2}{m_2}$. Plug them into above equation to obtain $x= -a$ immediately. The $y$-coordinate is, in turn,

$$y=a\frac {1- m_1m_2-m_2m_3-m_3m_1}{m_1m_2m_3} =-\frac a2$$

where (1) is used in the last step. Thus, the orthocenter is $(-a,-\frac a2)$.

Answer 2

$\require{begingroup} \begingroup$

$\def\i{\mathbf{i}}$ Let denote the lines \begin{align} L_1:\quad y&=m_1\,x+\frac a{m_1} ,\\ L_2:\quad y&=m_2\,x+\frac a{m_2} ,\\ L_3:\quad y&=m_3\,x+\frac a{m_3} , \end{align}

where $m_1=1$, $m_2=1+\sqrt3$, $m_3=1-\sqrt3$.

Using complex coordinates $A_a,B_a,C_a$

\begin{align} A_a = L_2\cap L_3&= -\frac a2-a\i ,\\ B_a = L_3\cap L_1&= -\frac a2\,(\sqrt3+1)-\frac a2\,(\sqrt3-1)\,\i ,\\ C_a = L_1\cap L_2&= \frac a2\,(\sqrt3-1)+\frac a2\,(\sqrt3+1)\,\i . \end{align}

Instead of $\triangle A_aB_aC_a$ we can consider $\triangle ABC$, scaled by $1/a$ for simplicity:

\begin{align} A &= -\frac 12-\i ,\\ B &= -\frac 12\,(\sqrt3+1)-\frac 12\,(\sqrt3-1)\,\i ,\\ C &= \frac 12\,(\sqrt3-1)+\frac 12\,(\sqrt3+1)\,\i . \end{align}

Given that the intersection point of two lines, passing through points $z_1,z_2$ and $z_3,z_4$ is found as

\begin{align} z&= f_{\times}(z_1,z_2,z_3,z_4)= \frac{ (z_1-z_2)\,(\overline{z_3}\,z_4-\overline{z_4}\,z_3) -(z_3-z_4)\,(\overline{z_1}\,z_2-\overline{z_2}\,z_1) }{ (z_1-z_2)\,(\overline{z_3}-\overline{z_4}) -(z_3-z_4)\,(\overline{z_1}-\overline{z_2}) } , \end{align} where $\overline{z}$ is the complex conjugate of $z$,

the orthocenter $H$ of $\triangle ABC$ is pound as

\begin{align} H&=f_{\times}(A,A+\i\cdot(B-C),B,B+\i\cdot(C-A)) \\ &=-1-\tfrac12\,\i , \end{align}

Hence, the orthocenter of the original $\triangle A_aB_aC_a$

\begin{align} H_a&=a\cdot H=-a-\frac a2\,\i \end{align}

as desired.

Here is corresponding Maxima session:

Maxima 5.38.1 http://maxima.sourceforge.net
using Lisp GNU Common Lisp (GCL) GCL 2.6.12
Distributed under the GNU Public License. See the file COPYING.
Dedicated to the memory of William Schelter.
The function bug_report() provides bug reporting information.
(%i1) _(z):=conjugate(z)$

(%i2) xpoint(z1,z2,z3,z4):=
((z1-z2)*(_(z3)*z4-_(z4)*z3)-(z3-z4)*(_(z1)*z2-_(z2)*z1))
/((z1-z2)*(_(z3)-_(z4))-(z3-z4)*(_(z1)-_(z2)))$

(%i3) declare([A,B,C,H],complex)$

(%i4) A:-1/2-%i;
                                           1
(%o4)                             (- %i) - -
                                           2
(%i5) B:-1/2*(1+sqrt(3))-1/2*(-1+sqrt(3))*%i;
                         (sqrt(3) - 1) %i    sqrt(3) + 1
(%o5)                 (- ----------------) - -----------
                                2                 2
(%i6) C:1/2*(-1+sqrt(3))+1/2*(1+sqrt(3))*%i;
                        (sqrt(3) + 1) %i   sqrt(3) - 1
(%o6)                   ---------------- + -----------
                               2                2
(%i7) H:expand(xpoint(A,A+%i*(B-C),B,B+(C-A)*%i));
                                     %i
(%o7)                             (- --) - 1
                                     2

$\endgroup$

Answer 3

(see figure below where we have taken $a=1$.)

Yes, you can take advantage of your remark : in this case, you know that the orthocenter $H$ is on the directrix of parabola $y^2=4ax$ : see for that a recent answer of mine here containing references. The directrix of the parabola with equation $y^2=4ax=2px$ has equation $x=-\tfrac{p}{2}=-a$ ; therefore the abscissa $x_H$ of $H$ is equal to $-a$.

Now, what is the ordinate $y_H$ of H ?

In order to find it, it suffices to find the equation of any altitude : its intersection with straight line $x=-a$ will give in particular $y_H$. I have done the same computations as @g.kov to find vertex : $(x_0,y_0)=(-a/2,-a)$ (intersection $L_2 \cap L_3$) ; as the opposite side $L_1$ has equation $y=x+a$, with slope $+1$, the altitude dropped on this side (colored in magenta) will have slope $s=-1$, therefore will have the following equation :

$$y-y_0=s(x-x_0) \ \ \iff \ \ y+a=-(x+a/2)$$

Taking $x=-a$ in this equation gives, as expected, $y_H=-a/2$.

A side remark : the tangency points (contact points of the tangents with equation $y = m_ix+\frac{a}{m_i}$ with the parabola) are $\left(\dfrac{a}{m_i^2},\dfrac{2a}{m_i}\right)$.