## Advanced Problem #18

Another fun problem, as requested by Qinfeng Li in MA598R: Measure Theory

Let so that too.

Show that and

I got this problem by picking some strictly positive value and breaking the integral as follows:

Let us examine now the factors and above:

We have thus proven that with At this point, all you have to do is pick (provided the denominator is not zero) and you are done.

## Some results related to the Feuerbach Point

Given a triangle the circle that goes through the midpoints of each side, is called the Feuerbach circle. It has very surprising properties:

- It also goes through the feet of the heights, points
- If denotes the orthocenter of the triangle, then the Feuerbach circle also goes through the midpoints of the segments 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 and for some positive parameters 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 , another vertex at and the third vertex at for some value In that case, we will need to work on the polynomial ring since we need the parameter free, the variables and are used to input the conditions for the circumcenter of the triangle, the variables and for centroid, and the variables and 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

(for centroid)

(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, It only remains to compute the Gröbner basis of the ideal 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.

## Sympy should suffice

I have just received a copy of Instant SymPy Starter, by Ronan Lamy—a no-nonsense guide to the main properties of `SymPy`, the Python library for symbolic mathematics. This short monograph packs everything you should need, with neat examples included, in about 50 pages. Well-worth its money.

To celebrate, I would like to pose a few coding challenges on the use of this library, based on a fun geometric puzzle from cut-the-knot: **Rhombus in Circles**

Segments and are equal. Lines and intersect at Form four circumcircles: Prove that the circumcenters form a rhombus, with

Note that if this construction works, it must do so independently of translations, rotations and dilations. We may then assume that is the origin, that the segments have length one, and that for some parameters it is We let `SymPy` take care of the computation of circumcenters:

import sympy from sympy import * # Point definitions M=Point(0,0) A=Point(2,0) B=Point(1,0) a,theta=symbols('a,theta',real=True,positive=True) C=Point((a+1)*cos(theta),(a+1)*sin(theta)) D=Point(a*cos(theta),a*sin(theta)) #Circumcenters E=Triangle(A,C,M).circumcenter F=Triangle(A,D,M).circumcenter G=Triangle(B,D,M).circumcenter H=Triangle(B,C,M).circumcenter

Finding that the alternate angles are equal in the quadrilateral is pretty straightforward:

In [11]: P=Polygon(E,F,G,H) In [12]: P.angles[E]==P.angles[G] Out[12]: True In [13]: P.angles[F]==P.angles[H] Out[13]: True

To prove it a rhombus, the two sides that coincide on each angle must be equal. This presents us with the first challenge: Note for example that if we naively ask SymPy whether the triangle is equilateral, we get a `False` statement:

In [14]: Triangle(E,F,G).is_equilateral() Out[14]: False In [15]: F.distance(E) Out[15]: Abs((a/2 - cos(theta))/sin(theta) - (a - 2*cos(theta) + 1)/(2*sin(theta))) In [16]: F.distance(G) Out[16]: sqrt(((a/2 - cos(theta))/sin(theta) - (a - cos(theta))/(2*sin(theta)))**2 + 1/4)

Part of the reason is that we have not indicated anywhere that the parameter `theta` is to be strictly bounded above by (we did indicate that it must be strictly positive). The other reason is that `SymPy` does not handle identities well, unless the expressions to be evaluated are perfectly simplified. For example, if we *trust* the routines of simplification of trigonometric expressions alone, we will not be able to resolve this problem with this technique:

In [17]: trigsimp(F.distance(E)-F.distance(G),deep=True)==0 Out[17]: False

Finding that with `SymPy` is not that easy either. This is the second challenge.

How would the reader resolve this situation?

## A nice application of Fatou’s Lemma

Let me show you an exciting technique to prove some convergence statements using exclusively functional inequalities and Fatou’s Lemma. The following are two classic problems solved this way. Enjoy!

Exercise 1Let be a measurable space and suppose is a sequence of measurable functions in that converge almost everywhere to a function and such that the sequence of norms converges to . Prove that the sequence of integrals converges to the integral for every measurable set .

*Proof:* Note first that

Set then (which are non-negative functions) and apply Fatou’s Lemma to that sequence. We have then

which implies

It must then be . But this proves the statement, since

Exercise 2Let be a finite measure space and let . Suppose that is a sequence of measurable functions in whose norms are uniformly bounded in and which converge almost everywhere to a function . Prove that the sequence converges to for all where is the conjugate exponent of .

*Proof:* The proof is very similar to the previous problem. We start by noticing that under the hypotheses of the problem,

If we prove that , we are done.

We will achieve this by using the convexity of , since in that case it is

hence,

Set then (which are non-negative functions) and apply Fatou’s Lemma as before.