Grégoire Nicollier
March 29, 2016

The following problem has been treated in and

Erect right-hand squares on the sides of a planar quadrilateral Q. For each vertex of Q, take the midpoint of the extremities of the two square sides issued from this vertex. The segments joining the pairs of opposite midpoints are perpendicular and congruent. When Q is a parallelogram, the midpoints form a square.

We give here a generic proof of this kind of problem. The method applies to every linear circulant transformation of a planar n-gon (see for example the reference below, Sections 3-7). The proofs are then always very short. Napoleons's theorem, Van Aubel's theorem, the Petr-Douglas-Neumann theorem have for example such one-line proofs. It is also easy to discover "new" theorems in this way. We illustrate the method with the above problem.

We consider a quadrilateral Q of the complex plane as the point Q=(z_0,\,z_1,\,z_2,\,z_3) of \mathbf{C}^4 formed by the vertices in order. The dot product of two quadrilaterals Q and Q' (in this order) is

The regular quadrilaterals

F_0=(1,\,1,\,1,\,1),\quad F_1=(1,\,i,\,-1,\,-i) \text{ (positively oriented square)},\quad F_2=(1,\,-1,\,1,-\,1),\quad F_3=(1,\,-i,\,-1,\,i)=\overline{F_1} \text{ (negatively oriented square)}

form the orthonormal Fourier basis of \mathbf{C}^4. In the Fourier basis, Q=(z_0,\,z_1,\,z_2,\,z_3) is thus given by

Q=\sum_{k=0}^3 \hat z_kF_k\quad\text{with}\quad \hat z_k=\frac14Q\cdot\overline{F_k},\quad k=0,\,1,\,2,\,3.
The quadrilateral \widehat Q=(\hat z_0,\,\hat z_1,\,\hat z_2,\,\hat z_3) is the spectrum or discrete Fourier transform of Q. Note that all Fourier basis polygons but F_0 are centered at the origin.

The Fourier coefficient \hat z_0 is the vertex centroid of Q. One has \hat z_2=0 if and only if z_0-z_1=z_3-z_2, that is, if and only if Q is a parallelogram. Each planar quadrilateral is hence the sum of a parallelogram P and a multiple of F_2, which translates the pairs of opposite vertices of P by opposite vectors.

Let T(Q) be the above quadrilateral M formed by the midpoints. The transformation T is linear. One sees graphically at once that

T(F_0)=F_0,\quad T(F_1)=2F_1,\quad T(F_2)=F_2,\quad \text{and}\quad T(F_3)=(0,\,0,\,0,\,0):

the transformation T is a filter that deletes the F_3-part of Q. Thus T(Q)-\hat z_0F_0 is the sum of the square 2\hat z_1F_1 and \hat z_2F_2: T(Q) has diagonals 4\hat z_1 and 4i\hat z_1. When Q is a parallelogram, one has \hat z_2=0 and T(Q)=\hat z_0F_0+2\hat z_1F_1 is a square.

Note that the Fourier basis vectors are eigenvectors of T: this is not by chance! The Fourier basis vectors are in fact eigenvectors of any circulant linear transformation of a polygon.

 G. Nicollier, Some Theorems on Polygons With One-line Spectral Proofs, Forum Geom. 15 (2015) 267-273.