User:Neil Parker/Ptolemy

Equilateral triangle


Ptolemy's Theorem yields as a corollary a pretty theorem regarding an equilateral triangle inscribed in a circle.

Given An equilateral triangle inscribed on a circle and a point on the circle.

The distance from the point to the most distant vertex of the triangle is the sum of the distances from the point to the two nearer vertices.

Proof: Follows immediately from Ptolemy's theorem:



qs=ps+rs\Rightarrow q=p+r. $$

Square
Any square can be inscribed in a circle whose center is the barycenter of the square. If the common length of its four sides is equal to $$a$$ then the length of the diagonal is equal to $$a\sqrt{2}$$ according to the Pythagorean theorem and the relation obviously holds.

Rectangle
More generally, if the quadrilateral is a rectangle with sides a and b and diagonal d then Ptolemy's theorem reduces to the Pythagorean theorem. In this case the center of the circle coincides with the point of intersection of the diagonals. The product of the diagonals is then d2, the right hand side of Ptolemy's relation is the sum a2 + b2.

Copernicus - who used Ptolemy's theorem extensively in his trigonometrical work - refers to this result as a 'Porism' or self evident corollary:


 * Furthermore it is clear (manifestum est) that when the chord subtending an arc has been given, that chord too can be found which subtends the rest of the semicircle.

De Revolutionibus Orbium Coelestium: Page 37. See last two lines of this page.

Nor for that matter does Copernicus refer to "Ptolemy's Theorem" but labels it more simply as "Theorema Secundum".

Pentagon
A more interesting example is the relation between the length a of the side and the (common) length b of the 5 chords in a regular pentagon. In this case the relation reads b2 = a2 + ab which yields the golden ratio
 * $$\varphi = {b \over a} = {{1+\sqrt{5}}\over 2}.$$

Side of decagon
If now diameter AF is drawn bisecting DC so that DF and CF are sides c of an inscribed decagon, Ptolemy's Theorem can again be applied – this time to cyclic quadrilateral ADFC with diameter d as one of its diagonals:


 * $$ad=2bc\;$$


 * $$\Rightarrow ad=2\varphi ac$$ where $$\varphi$$ is the golden ratio.
 * $$\Rightarrow c=\frac{d}{2\varphi}.$$

whence the side of the inscribed decagon is obtained in terms of the circle diameter. Pythagoras' Theorem applied to right triangle AFD then yields "b" in terms of the diameter and "a" the side of the pentagon is thereafter calculated as
 * $$a=\frac {b} {\varphi}=b(\varphi-1).\ $$

As Copernicus (following Ptolemy) wrote,


 * "The diameter of a circle being given, the sides of the triangle, tetragon, pentagon, hexagon and decagon, which the same circle circumscribes, are also given." – De Revolutionibus Orbium Coelestium: Liber Primus: Theorema Primum

Complement of pentagon chord
The ancient geometers are not done yet, for if the fifth vertex of the pentagon is marked as E and FE and BF are joined (with FE=BF=z), then cyclic quadrilateral EFBA will be formed with diagonals length d (diameter) and b. Applying the 'Almagest' theorem yet again:


 * $$db=2az\;$$


 * $$\Rightarrow \varphi ad=2az$$ where $$\varphi$$ is the golden ratio.
 * $$\Rightarrow z=\frac{\varphi d}{2}.$$

z subtends 540 at the circumference, c (the side of the decagon) subtends 180 at the circumference and thereby is established:


 * $$\sin(54^\circ)=\frac {\varphi}{2}$$ and $$\sin(18^\circ)=\frac{1}{2\varphi}=\frac {\varphi-1}{2}.$$

And since angle ABF is subtended by the diameter and is therefore right, the side of the pentagon is calculated by an even simpler route:


 * $$\sin(36^\circ)=\frac {\sqrt{3-\varphi}}{2}$$ and $$\sin(72^\circ)=\frac{\sqrt{2+\varphi}}{2}.$$

Based on his circle of diameter 200000 units, Copernicus provides accurate numerical values for the four pentagon related chords corresponding to these angles:


 * "Since the side of the decagon which subtends 360 has been shown to have 61803 parts, whereof the diameter has 200000 parts – the chord which subtends the remaining 1440 of the semicircle has 190211 parts. And in the case of side of the pentagon, which is equal to 117557 parts of the diameter and subtends an arc of 720, a straight line of 161803 parts is given, and it subtends the remaining 1080 of the circle".

Golden ratio aficionados will instantly recognize the digits 161803 and 61803 as corresponding to $$\varphi$$ and its reciprocal. The chords are of considerable historical importance because, along with the sides of the triangle and tetragon (square), they enable the generation of a table of half chords (effectively sine values) which in turn underpins many of the key astronomical measurements and calculations effected by Copernicus in the development of his helio-centric model:


 * "Because the proofs which we shall use in almost the entire work deal with straight lines and arcs, with plane and spherical triangles and because Euclid's Elements, although they clear up much of this, do not have what is here most required, namely, how to find the sides from angles and the angles from the sides ... there has accordingly been found a method whereby the lines subtending any arc may be known."

Euclid 13:10
Next our attention is drawn to point G midway between points B and C on the circumference. CG, FG and DG are joined forming cyclic quadrilateral DFCG in which three sides belong to the regular decagon (length c) and the third DG is of length z. Diagonals DC and FG are both of length a (side of the pentagon) Then:


 * $$c^2+cz=a^2\;$$


 * $$\Rightarrow c^2+\frac{d}{2 \varphi}\frac{\varphi d}{2}=a^2.$$


 * $$\Rightarrow c^2+\frac{d^2}{4}=a^2$$


 * $$\Rightarrow c^2+r^2=a^2$$

where r is the radius of the circle and also the side of the inscribed hexagon.

Whence with relative ease is proved Proposition 10 in Book XIII of Euclid's Elements: The square on the side of the pentagon equals the sum of the squares on the sides of the hexagon and the decagon inscribed in the same circle.

In modern trigonometric notation this corresponds to the identity:


 * $$\sin^2(18^\circ)+\sin^2(30^\circ)=\sin^2(36^\circ)\;$$ as may be verified on a scientific calculator.

The Pythagorean nature of this relationship makes possible the construction of a regular pentagon as demonstrated here.

Theorema Tertium
A well documented classical application of the "Second Theorem" as illustrated in the diagram is the determination of chord BC subtending 12 degrees of arc. Referring to the diagram:


 * $$\overline{BC}=\frac{\overline{BD}\cdot\overline{AC}-\overline{AB}\cdot\overline{CD}}{\overline{AD}}$$


 * $$=\frac{\sqrt{3}ra-rz}{2r}$$


 * $$=\frac{a\sqrt{3}-z}{2}.$$


 * "When, for example, the sides of the pentagon and hexagon are given from the above, by this computation a line is given subtending 120 which is the difference between the arcs - and it is equal to 20905 parts of the diameter." De Revolutionibus Orbium Coelestium: Liber Primus: Theorema Tertium

Based on his circle of diameter 200000 units and already established chords of pentagon, hexagon and triangle the calculation effected by Copernicus would have been:


 * $$\frac{173205\times117557-100000\times161803}{200000}\approx20906.$$

A small rounding error is evident in the result but the corresponding entry (in the Copernican table of half chords) of 10453 units against 6 degrees is correct as may readily be verified on a calculator (sin 6).

In modern trigonometric notation, the above calculation corresponds to the following application of a compound angle formula:


 * $$\sin(6^\circ)=\sin(36^\circ)\cos(30^\circ)-\cos(36^\circ)\sin(30^\circ).$$

Theorema Quintum
The previous diagram demonstrated a general technique for calculating the chord subtending the difference between two arcs. The following diagram neatly reverses this procedure to obtain the chord subtending the sum of arcs: ie determination of chord AC given chords AB and BC.

Compared with the previous we note that diameter BE has been swung across from point B to point E. EC and ED are joined. Since AEDB is a rectangle DE=AB. Thus in cyclic quadrilateral BEDC, sides BE, BC and ED are known along with diagonals CE and BD by application of the "Porism" (Pythagoras Thm). Then:


 * $$\overline{CD}=\frac{\overline{EC}\cdot\overline{BD}-\overline{ED}\cdot\overline{BC}}{\overline{BE}}.$$

In the specific example illustrated in the diagram, calculation of chord CD in cyclic quadrilateral BEDC corresponds to the following application of a compound angle formula:


 * $$\cos(30+6)=\sin(60-6)=\sin(60^\circ)\cos(6^\circ)-\cos(60^\circ)\sin(6^\circ).$$

The required chord AC (in this example corresponding to sin(30+6)) is then calculated by application of the "Porism".

De Revolutionibus Orbium Coelestium: Liber Primus: Theorema Quintum

Other Polygons
When applied repeatedly, Ptolemy's theorem allows one to compute the lengths of all diagonals for polygons inscribed in a circle with vertices P1, ..., Pn, if the sides are given together with all the length values of the "next to sides" chords connecting two vertices Pi and Pi+2 (with indices taken modulo n).