ITERATIONS ON CONVEX QUADRILATERALS

Tamela Underwood

Southern Illinois University at Carbondale

Mangho Ahuja

Southeast Missouri State University

1. Introduction. The object of this paper is to study the effect of the repeated applications of a particular process P, when it is performed on an arbitrary (convex) quadrilateral. The process is described below.

\underline{Process P}. Given a quadrilateral ABCD, we construct squares on the sides AB, BC, CD, and DA [Fig. 1]. All four squares are constructed on the outside of ABCD. Let P_{1}, Q_{1}, R_{1}, and S_{1} denote the centers of the squares on the sides AB, BC, CD, and DA, respectively. By joining the centers of the squares a new quadrilateral P_{1}Q_{1}R_{1}S_{1} is obtained. The process of obtaining quadrilateral P_{1}Q_{1}R_{1}S_{1} from quadrilateral ABCD is defined as the process P.

We will denote P_{1}Q_{1}R_{1}S_{1} by P[ABCD] and also by Π_{1}. In general P_{n}Q_{n}R_{n}S_{n} and Π_{n} will denote the quadrilateral obtained by applying the process n times. In Proposition 1 we will prove that the quadrilateral P_{1}Q_{1}R_{1}S_{1} has the following properties:

(i) P_{1}R_{1} = Q_{1}S_{1}, i.e. the diagonals are equal, and

(ii) P_{1}R_{1} is perpendicular to Q_{1}S_{1}, i.e. the diagonals are perpendicular. We note that properties (i) and (ii) are not sufficient to make P_{1}Q_{1}R_{1}S_{1} a square. For our purpose we may define a square as follows. A quadrilateral P QRS is a square if it has the following three properties:

(i) P R = QS,

(ii) P R is perpendicular to QS,

(iii) the diagonals P R and QS bisect each other.

We have seen that just one application of process P transforms an arbitrary quadrilateral into one which satisfies two of the three properties for a square. One wonders what effect repeated applications of P would have on ABCD. Since Π_{1} satisfies (i) and (ii), it is obvious that every quadrilateral Π_{n} will also satisfy (i) and (ii). Let M_{n} and N_{n} denote the midpoints of the diagonals P_{n}R_{n} and Q_{n}S_{n} respectively, and let s_{n} denote the distance M_{n}N_{n}. Since every Π_{n} satisfies (i) and (ii), Π_{n} will be a square if and only if the diagonals bisect each other, i.e. if and only if s_{n} = 0.

We also note that with each application of process P the quadrilateral increases in size. Let d_{n} denote the length of the diagonals (both diagonals have the same length) of Π_{n}, and let s′_{n} = \frac{s_{n}}{d_{n}}. To measure the effect of P on Π_{n}, we should compare quadrilaterals of the same size. This means that we should not compare s_{n} with s_{n+1}, but we should compare s′_{n} with s′_{n+1}. We would like to know if the quadrilaterals Π_{n} obtained by n applications of P would ‘approach a square’ i.e. whether s′_{n} goes to 0 as n tends to infinity. We will explore this question and also study other effects of P.

2. We now give an analytic proof of Proposition 1. For a proof using rotations of triangles, see [1].

\underline{Proposition 1}. Let ABCD be an arbitrary polygon and P_{1}Q_{1}R_{1}S_{1} be the polygon obtained by applying process P. Then the diagonals P_{1}R_{1} and Q_{1}S_{1} are equal and perpendicular.

\underline{Proof}. Choosing the axes as shown in Fig. 2, let the coordinates of the vertices be given by A = (0, 0), B = (a, b), C = (c, d), and D = (e, 0). Without loss of generality assume b > d. Also c > a (convexity). We construct square BCB′C′ on the outside of BC and denote its center by Q_{1}. Let CN and C′N′ be perpendiculars on the vertical line through B(a, b). The right triangles BCN and C′BN′ are congruent (BC = BC′, also angles CBN and C′BN′ are complementary, making angles BCN and C′BN′ equal). Then CN = BN′ = c − a, and BN = C′N′ = b − d. This gives C′ = (a + b − d, −a + b + c), and the midpoint of CC′,

Q_{1} = (\frac{a + b + c − d}{2}, \frac{−a + b + c + d}{2}).

Using the same method we find the centers of the other squares,

P_{1} = (\frac{a − b}{2}, \frac{a + b}{2}),

R_{1} = (\frac{c + d + e}{2}, \frac{−c + d + e}{2})

S_{1} = (\frac{e}{2}, \frac{−e}{2}).

The lengths and the slopes of the diagonals are as follows:

P_{1}R_{1} = \frac{1}{2}\sqrt{(a − b − c − d − e)^{2} + (a + b + c − d − e)^{2}} = Q_{1}S_{1}.

The slope of P_{1}R_{1} is

\frac{a + b + c − d − e}{a − b − c − d − e}

and the slope of Q_{1}S_{1} is

\frac{−a + b + c + d + e}{a + b + c − d − e}.

This shows that the diagonals P_{1}R_{1} and Q_{1}S_{1} are equal and perpendicular, which completes the proof.

From Proposition 1, we note that M_{1}, N_{1}, the midpoints of P_{1}R_{1} and Q_{1}S_{1} are given by

M_{1} = \frac{1}{4}(a − b + c + d + e, a + b − c + d + e)

and

N_{1} = \frac{1}{4}(a + b + c − d + e, −a + b + c + d − e),

and the distance s_{1} = M_{1}N_{1} is given by

s_{1}^{2} = (\frac{−b + d}{2})^{2} + (\frac{a − c + e}{2})^{2}.

If the original quadrilateral ABCD was a parallelogram, then b = d, and c − e = a. This would make s_{1} = 0, and P_{1}Q_{1}R_{1}S_{1} would be a square.

We now set the stage to study the process P and the properties of the quadrilateral Π_{n}. Since the diagonals of Π_{1} are perpendicular, they will be chosen as the coordinate axes [Fig. 3]. As mentioned earlier, P_{2}Q_{2}R_{2}S_{2} will denote the quadrilateral obtained by applying P to P_{1}Q_{1}R_{1}S_{1}. Let M_{2}, N_{2} denote the midpoints of the diagonals P_{2}R_{2} and Q_{2}S_{2} respectively, and d_{n} denote the length (both diagonals have the same length) of the diagonals of the quadrilateral Π_{n}.

\underline{Proposition 2}.

(i) The diagonals of Π_{2} make a 45^{∘} angle with the diagonals of Π_{1}, and both sets of diagonals intersect at the same point,

(ii) The line segments M_{1}N_{1}, M_{2}N_{2} are equal and bisect each other, (iii) d_{2} = \sqrt{2}d_{1}.

\underline{Proof}. Let the coordinates of P_{1}, Q_{1}, R_{1}, and S_{1} be given by P_{1} = (0, y_{1}), Q_{1} = (x_{1}, 0), R_{1} = (0, y_{1}−d_{1}), S_{1} = (x_{1}−d_{1}, 0) (see Fig. 3). Then M_{1} = (0, y_{1}− \frac{1}{2}d_{1}), N_{1} = (x_{1}− \frac{1}{2}d_{1}, 0). Using the same method as in Proposition 1, the vertices of Π_{2} are obtained as:

\begin{matrix}P_{2} = (\frac{x_{1} + y_{1}}{2} , \frac{x_{1} + y_{1}}{2} ), Q_{2} = (\frac{x_{1} − y_{1} + d_{1}}{2} , \frac{−x_{1} + y_{1} − d_{1}}{2} ), \\ R_{2} = (\frac{x_{1} + y_{1}}{2} − d_{1}, \frac{x_{1} + y_{1}}{2} − d_{1}), S_{2} = (\frac{x_{1} − y_{1} − d_{1}}{2} , \frac{−x_{1} + y_{1} + d_{1}}{2} ). \end{matrix}

Then

M_{2} = (\frac{x_{1} + y_{1} − d_{1}}{2} , \frac{x_{1} + y_{1} − d_{1}}{2} ), N_{2} = (\frac{x_{1} − y_{1}}{2} , \frac{−x_{1} + y_{1}}{2} ).

We note that the points P_{2} and R_{2} satisfy the equation y = x, whereas the points Q_{2} and S_{2} satisfy the equation y = −x. In other words, the diagonals of Π_{2} lie along the lines y = ±x. Thus they intersect at (0, 0) and make a 45^{∘} angle with the previous set of diagonals. Using distance formula,

M_{1}N_{1}^{2} = (x_{1} − \frac{1}{2}d_{1})^{2} + (y_{1} − \frac{1}{2}d_{1})^{2},

and

M_{2}N_{2}^{2} = (y_{1} − \frac{1}{2}d_{1})^{2} + (x_{1} − \frac{1}{2}d_{1})^{2}.

Thus, M_{1}N_{1} = M_{2}N_{2}. Also, M_{1}N_{1} and M_{2}N_{2} have the same midpoint

G = (\frac{2x_{1} − d_{1}}{4} , \frac{2y_{1} − d_{1}}{4} ).

Lastly,

d_{2}^{2} = P_{2}R_{2}^{2}

= \frac{1}{4}((x_{1} + y_{1} − (x_{1} + y_{1} − 2d_{1}))^{2} + ((x_{1} + y_{1}) − (x_{1} + y_{1} − 2d_{1}))^{2})

= \frac{1}{4}(4d_{1}^{2} + 4d_{1}^{2}) = 2d_{1}^{2}.

Hence, d_{2} = \sqrt{2}d_{1}. This completes the proof.

3. We come back to the question of whether the quadrilaterals Π_{n} ‘approach a square.’ Here is our main theorem.

\underline{Theorem}. With repeated applications of the process P the quadrilaterals Π_{n} ‘approach a square,’ i.e. s′_{n} approaches 0, as n goes to infinity.

\underline{Proof}. From Proposition 2 we have M_{1}N_{1} = M_{2}N_{2}, i.e. s_{1} = s_{2}. This means that s_{n} = s_{1} for all n. Also, d_{2} = \sqrt{2}d_{1} implies d_{n} = (\sqrt{2})^{n−1}d_{1}. Hence,

s′_{n} = \frac{s_{n}}{d_{n}} = \frac{s_{1}}{(\sqrt{2})^{n−1}d_{1}} .

The quantity s′_{n} goes to 0 as n tends to infinity. In other words, the quadrilaterals Π_{n} approach a square with repeated applications of the process P.

4. We now turn our attention to other effects of the process P. The following observations are based on the information provided by Propositions 1 and 2.

1. The diagonals of every quadrilateral Π_{n} intersect at the same point.

2. The diagonals of every quadrilateral Π_{n} make a 45^{∘} angle with the diagonals of the previous one.

3. The centroid of every quadrilateral Π_{n} is the same point. We know that if the vertices of a quadrilateral are given by (x_{i}, y_{i}) for i = 1 to 4, then its centroid G is given by,

G = (\frac{x_{1} + x_{2} + x_{3} + x_{4}}{4} , \frac{y_{1} + y_{2} + y_{3} + y_{4}}{4} ).

The centroid of a quadrilateral is also the midpoint of the line segment joining the midpoints of the diagonals, i.e. G is the midpoint of M_{1}N_{1}. But from Proposition 2, we learn that M_{1}N_{1} and M_{2}N_{2} have the same midpoint, hence, the quadrilaterals Π_{1} and Π_{2} have the same centroid. It follows that every quadrilateral Π_{n} has the same centroid.

5. To summarize: With each application of the process P, the diagonals rotate by 45^{∘}, and their lengths decrease by a factor of \sqrt{2}. But the diagonals of every quadrilateral intersect at the same point. Moreover while the quadrilaterals increase in size, their growth in each direction is such that the centroid remains the same. Thus, each application of process P leaves two points fixed, the point where the diagonals intersect and the centroid G. Lastly, the quadrilateral Π_{n} never becomes a square, only its “non-squareness” as measured by s′_{n} tends to 0.

\underline{References}

1. I. M. Yaglom, Geometric Transformations, Random House Inc., New York, NY, 1962.

2. T. Underwood, Masters Degree Thesis, Southeast Missouri State University, Cape Girardeau, MO, May 1992.

3. J. R. Smart, Modern Geometries, Brooks Cole Publishing Co., Monterey, CA, 1973.

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