### Archive

Archive for July, 2013

## Some results related to the Feuerbach Point

Given a triangle $\triangle ABC,$ the circle that goes through the midpoints of each side, $D, E, F,$ is called the Feuerbach circle. It has very surprising properties:

• It also goes through the feet of the heights, points $G, H, I.$
• If $Oc$ denotes the orthocenter of the triangle, then the Feuerbach circle also goes through the midpoints of the segments $OcA, OcB, OcC.$ For this reason, the Feuerbach circle is also called the nine-point circle.
• The center of the Feuerbach circle is the midpoint between the orthocenter and circumcenter of the triangle.
• The area of the circumcircle is precisely four times the area of the Feuerbach circle.

Most of these results are easily shown with sympy without the need to resort to Gröbner bases or Ritt-Wu techniques. As usual, we realize that the properties are independent of rotation, translation or dilation, and so we may assume that the vertices of the triangle are $A=(0,0), B=(1,0)$ and $C=(r,s)$ for some positive parameters $r,s>0.$ To prove the last statement, for instance we may issue the following:

>>> import sympy
>>> from sympy import *
>>> A=Point(0,0)
>>> B=Point(1,0)
>>> r,s=var('r,s')
>>> C=Point(r,s)
>>> D=Segment(A,B).midpoint
>>> E=Segment(B,C).midpoint
>>> F=Segment(A,C).midpoint
>>> simplify(Triangle(A,B,C).circumcircle.area/Triangle(D,E,F).circumcircle.area)
4


But probably the most amazing property of the nine-point circle, is the fact that it is tangent to the incircle of the triangle. With exception of the case of equilateral triangles, both circles intersect only at one point: the so-called Feuerbach point.

## An Automatic Geometric Proof

We are familiar with that result that states that, on any given triangle, the circumcenter, centroid and orthocenter are always collinear. I would like to illustrate how to use Gröbner bases theory to prove that the incenter also belongs in the previous line, provided the triangle is isosceles.

We start, as usual, indicating that this property is independent of shifts, rotations or dilations, and therefore we may assume that the isosceles triangle has one vertex at $A=(0,0)$, another vertex at $B=(1,0)$ and the third vertex at $C=(1/2, s)$ for some value $s \neq 0.$ In that case, we will need to work on the polynomial ring $R=\mathbb{R}[s,x_1,x_2,x_3,y_1,y_2,y_3,z],$ since we need the parameter $s$ free, the variables $x_1$ and $y_1$ are used to input the conditions for the circumcenter of the triangle, the variables $x_2$ and $y_2$ for centroid, and the variables $x_3$ and $y_3$ for the incenter (note that we do not need to use the orthocenter in this case).

We may obtain all six conditions by using sympy, as follows:

>>> import sympy
>>> from sympy import *
>>> A=Point(0,0)
>>> B=Point(1,0)
>>> s=symbols("s",real=True,positive=True)
>>> C=Point(1/2.,s)
>>> T=Triangle(A,B,C)
>>> T.circumcenter
Point(1/2, (4*s**2 - 1)/(8*s))
>>> T.centroid
Point(1/2, s/3)
>>> T.incenter
Point(1/2, s/(sqrt(4*s**2 + 1) + 1))


This translates into the following polynomials

$h_1=2x_1-1, h_2=8sy_1-4s^2+1$ (for circumcenter)
$h_3=2x_2-1, h_4=3y_2-s$ (for centroid)
$h_5=2x_3-1, h_6=(4sy_3+1)^2-4s^2-1$ (for incenter)

The hypothesis polynomial comes simply from asking whether the slope of the line through two of those centers is the same as the slope of the line through another choice of two centers; we could use then, for example, $g=(x_2-x_1)(y_3-y_1)-(x_3-x_1)(y_2-y_1).$ It only remains to compute the Gröbner basis of the ideal $I=(h_1, \dotsc, h_6, 1-zg) \subset \mathbb{R}[s,x_1,x_2,x_3,y_1,y_2,y_3,z].$ Let us use SageMath for this task:

sage: R.<s,x1,x2,x3,y1,y2,y3,z>=PolynomialRing(QQ,8,order='lex')
sage: h=[2*x1-1,8*r*y1-4*r**2+1,2*x2-1,3*y2-r,2*x3-1,(4*r*y3+1)**2-4*r**2-1]
sage: g=(x2-x1)*(y3-y1)-(x3-x1)*(y2-y1)
sage: I=R.ideal(1-z*g,*h)
sage: I.groebner_basis()
[1]


This proves the result.