Back to MathWords and Other Words

The area of a quadrilateral can be found by an extension of Heron's formula, that is often credited to the Indian Mathematician, Brahamagupta.

If
the lengths of the four sides are given as a, b, c and d; and the semiperimeter
(half the perimeter),

s = (a+b+c+d)/2, of the polygon, then the
area of the cyclic quadrangle is given by the formula

It is often possible to make several quadrangles with the same length sides, but of all the possible quadrangles with the given sides, the inscribed quadrilateral has the largest area.

The length of the two diagonals
of a cyclic quadrilateral are related to the four sides in **Ptolemy's
Theorem** which states (using *m* and *n* for the diagonals
lengths) mn=ac+bd. In words, the product of the diagonals is equal to the
sum of the products of the oppsite sides.

Using the same four sides we can
reorder them to produce another cyclic quadrangle with the same side lengths. By
drawing a diagonal (in either direction) we split the quadrangle into two
triangles. Now if we reflect that triangle in the perpendicular bisector
of the diagonal, we get a new quadrangle with a new permutation, or ordering,
of the sides. It is easy to see that by choosing different diagonals
to draw we could continue to alter the quadrangle to put the four lengths
in any order we want. Since there are four sides in a cyclic permutation,
this means there can be three distinct quadrilaterals drawn with
the same four side lengths, and suprisingly, at least it surprised me, is that there are only three possible lenghts for the diagonals. If we use a, b, c,
and d for the lengths of the sides, then the three lengths are given by

A simple and elegant proof of Ptolemy's theorem understandable at the highschool level is given at the Cut-the-Knot web site.

Ptolemy's Thm can be generalized to an inequality for all quadrilaterals. If the quadrilateral is not cyclic, then the product of the diagonals is always less than the sum of the products of the opposite sides.

Ptolemy of Alexandria lived around 150 BC, and was the author of what is perhaps the most significant trigonometry treatise of all antiquity. The collection of thirteen volumes was officially called __Mathematical Synthesis__ but because it was frequently compared to other collections as "** megiste"**, the greater, it is better known today as the

Professor Eisso J. Atzema of the Department of Mathematics & Statistics at the University of Maine sent the following historical note about the often confused names of Ptolemy's Thm and Menelaus' Thm to the Historia Matematica discussion list.

The theorem that you refer to is Ptolemy's Theorem, not Menelaus'. This mix-up by your colleague is rather ironic as until the 1830s or so, what we now call Menelaus' theorem (the dual of Ceva's theorem) often went by the name of Ptolemy's theorem as well since both our current Ptolemy's theorem and our current Menelaus' theorem had Ptolemy's Almagest as their only known primary source. More than likely, Ptolemy himself discovered the theorem called after him. As to Menelaus' theorem, however, he probably took it without acknowledgement from the Spherics of Menelaus of Alexandria, an astronomer of about a generation earlier. The existence of Menelaus' Spherics was known throughout post-Ptolemian Classical Antiquity and after, but rather unfortunate for Menelaus, knowledge of its actual contents had been lost to the Western world by the time the Renaissance came around. It was only sometime in the 1830s when Michel Chasles unearthed an Arabic translation of the parts of the Spherics, that the nature and content of this work could be reconstructed. In his Apercu of 1837, Chasles gives Menelaus due credit and the latter's name seems to have been attached to the theorem ever since.

Ptolemy himself actually never proved his theorem in the version that you state{the inequality}. He only proved equality for the case of cyclic quadrilaterals.

Euler found a relationship between the diagonals and the sides of any convex quadrilateral. If the length of the segment between the two midpoints of the diagonals is called x, and the sides of the quadrilateral are a,b,c,d and the diagonals are m and n, then it is true that a^{2} + b^{2} +c^{2} +d^{2} = m^{2} +n^{2} + 4 x ^{2} .

In a quadrilateral,
the orthongonal or perpendicular segment from the midpoint of one side
to the opposite side is called a Maltitude. The word is apparently
short for "Median Altitude" or something similar. I have not yet
been able to date its origin, but it has been suggested it was used as
early as the mid-nineteenth century. Three maltitudes are drawn in
the figure above. In a cyclic quadrilateral the maltitudes are concurrent
(intersect in a single point). The point (L in the figure) is called
the **Anti-Center **of the quadrilateral. The midpoint of the line segment
between the center of the circle, and the anti-center is the Centroid,
or center of Gravity of the four vertices. The centroid of the cyclic quadrilateral is also the common point of intersection of the lines joining midpoints of opposite edges. The line joining the midpoints of the diagonals also passes through the geocenter of the cyclic quadrilateral.

The **anti-center** is a busy point as other lines and curves coverge there also. If a line like the maltitude is constructed on the two diagonals of a cyclic quadrilateral (that is from the midpoint of each diagonal and perpendicular to the other diagonal) then these two lines also intersect at the anti-center. This means that if a triangle is formed whose vertices are the two midpoints of the diagonals, and the intersection of the diagonals, then the anti-center of the quadrilateral is the orhtocenter of that triangle (see figure at right). .

Four triangles [ABC, ABD, ACD, BCD] can be formed using the vertices of the cyclic quadrilateral. If the nine-point circles of each of these is constructed, the circles will all pass through the anti-center. All of these circles are congruent, with radii equal to 1/2 the circumradius of the cyclic quadrilateral. The centers of the four circles N1, N2, N3, and N4, also lie on a circle (in green) that is congruent to each of these, with a center at the anti-center, L. Since the construction of a Simson line uses a triangle and a fourth point on the cirucmcircle of the triangle, the Simson line for each of the triangles with respect to the fourth point passes through the anti-center also.

I recently (April 2003) learned from an outstanding young German geometer named Darij Grinberg that the anti-center is also referred to as **Mathot's point**. Darij kindly added the following note on the origin of this term:

The term Mathot point comes from the solution of Problem 2276 in Crux Mathematicorum 8/24 pages 514-515, where the editor writes: "It seems that this result is from Lemoine, 1869, Nouvells Annals de Mathématiques (pp. 174 and 317). M is sometimes called the Mathot point, after Jules Mathot, Mathesis, 1901, p. 25. In [1] M is called the anticentre of the cyclic quadrangle ABCD."

The reference [1] is Nathan Altshiller Court, College Geometry. Barnes and Noble, 1965 (theorem 263, p. 132).

Here are some additional relationships involving the anti-center. If the four vertices are considered three at a time and the orthocenters found, the four orthocenters lie on a circle that is congruent to the original circumcircle. The center of this "orthic cyclic quadrilateral" is the reflection of the circumcenter of the original quadrilateral in the anti-center. The diagonals of the orthic quadrilateral intersect in a point that is the reflection of the intersection of the diagonals of the original quadrilateral about the anti-center. The anti-center of the orthic quadrilateral is the same as the anti-center of the original quadrilateral, and so the orthocenters of the triangles formed by the orthic quadrilateral are the vertices of the original cyclic quadrilateral.

Another way to describe this whole picture is that the orhtic quadrilateral is the image under a 180 degree rotation of the orginal quadrilateral about the anti-center, taking each point to the orthocenter of the triangle formed by the three other points.

Thanks also to Gordon Brown for reminding me that the vectors from the circumcenter, O, to the four vertices, A, B, C, and D, add up to twice the vector from O to the Anti-center, E; OA+OB+OC+OD = 2 OE = OQ where Q is the circumcenter of the orhtic quadrilateral.

I have also posted an interactive javascript of the orthic quadrilateral generated by a cyclic quadrilateral.

A nice proof of the congruence between the original quadrilateral and the orhtic quadrilateral is at a web site by John Scholes. I find the proof especially nice since it uses the centroids of the four individual triangles as an intermediate step, and shows that the four centroids form a cyclic quadrilateral that is similar to the original and 1/3 the size. This property of the centroids forming a similar quadrilateral is not dependent on the original quadrilateral being cyclic. If the four centroids of any quadrilateral are found, they will form a quadrilateral similar to the original dilated by 1/3.

The relationship between the original and orthic quadrilaterals appeared as a problem in the 1984 Balkan Mathematics Olympiad. I have no earlier history of the relationship.