Knock Euler’s Line

my imagination
is a piece of board
my sole instrument
is a wooden stick

I strike the board
it answers me
yes—yes
no—no

“A Knocker” by Zbigniew Herbert

Euler’s line theorem states
In every triangle
the intersection of the medians
the intersection of the heights
and the center of the circumscribed circle
are on a straight line

Let’s prove it
with the aid of Omega CAS Explorer

We know

x^2+y^2+d \cdot x + e \cdot y +f =0\quad\quad\quad(1)

is a circle centered at (-{d \over 2}, -{e \over 2}) with radius

r^2={{d^2+e^2-4f} \over 4}\quad\quad\quad(2)

provide (2) is positive

Find d, e, f from triangle’s vertices (-x_1, 0), (x_1, 0), (x_2, y_2):

ceq:x^2+y^2+d*x+e*y+f=0;
eq1:ev(ceq, x=x1, y=0);
eq2:ev(ceq, x=-x1, y=0);
eq3:ev(ceq, x=x2, y=y2);
sol:linsolve([eq1, eq2, eq3], [d,e,f]);

\implies d=0, e=-{{y_2^2-x_2^2+x_1^2} \over y_2}, f=-x_1^2

Evaluate (2):

ev(d^2 + e^2-4*f, sol);

\implies {(-y_2^2-x_2^2+x1^2)^2\over y_2^2} + 4x_1^2

always positive for x_1>0, y2 \neq 0

Find the center of the circumscribed circle x_c, y_c:

xc:ev(-d/2, sol)$

\implies 0

yc:ev(-e/2, sol)$

\implies {{-y_2^2-x_2^2+x_1^2} \over {2y_2}}

Find the intersection of the medians x_m, y_m:

eq1:y*((x1+x2)/2+x1)=y2/2*(x+x1)$
eq2:y*((x2-x1)/2-x1)=y2/2*(x-x1)$
sol:solve([eq1, eq2], [x,y])$
xm:ev(x,sol);

\implies {x_2 \over 3}

ym:ev(y, sol);

\implies {y_2 \over 3}

is(ev(x2*y=y2*x, x=xm, y=ym));

\implies true

Find the intersection of the heights x_h, y_h:

eq1:y2*y=-(x2+x1)*(x-x1)$
eq2:y2*y=-(x2-x1)*(x+x1)$
sol:solve([eq1, eq2], [x,y])$
xh:ev(x, sol);

\implies x_2

yh:ev(y, sol);

\implies -{{x_2^2-x_1^2} \over y_2}

Compute the area of triangle with vertices (x_c, y_c), (x_m, y_m), (x_h, y_h):

m:matrix([xc, yc,1], [xm, ym, 1], [xh, yh,1]);
determinant(m)$
ratsimp(%);

\implies 0

Indeed
(x_c, y_c), (x_m, y_m) and (x_h, y_h) are on a straight line.

Screen Shot 2018-02-02 at 9.59.16 AM.png

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