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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 \overline{AB} and \overline{CD} are equal. Lines AB and CD intersect at M. Form four circumcircles: (E)=(ACM), (F)=(ADM), (G)=(BDM), (H)=(BCM). Prove that the circumcenters E, F, G, H form a rhombus, with \angle EFG = \angle AMC.

rhombusincircles

Note that if this construction works, it must do so independently of translations, rotations and dilations. We may then assume that M is the origin, that the segments have length one, A=(2,0), B=(1,0), and that for some parameters a>0, \theta \in (0, \pi), it is C=(a+1) (\cos \theta, \sin\theta), D=a (\cos\theta, \sin\theta). 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 EFGH 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 \triangle EFG 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 \pi (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 \angle EFG = \angle AMC with SymPy is not that easy either. This is the second challenge.

How would the reader resolve this situation?


Instant SymPy Starter

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