Ptolemaios' theorem

In mathematics, Ptolemaios' theorem is a relation in Euclidean geometry between the four sides and two diagonals or chords of a quadrilateral inscribed in circle. The theorem is named after the Greek astronomer and mathematician Claudios Ptolemaios (better known as Ptolemy).

If the quadrilateral is given by its four vertices P₁, P₂, P₃ and P₄, then the relation is

$\overline{P_1P_3}\cdot \overline{P_2P_4}=\overline{P_1P_2}\cdot \overline{P_3P_4}+\overline{P_1P_4}\cdot \overline{P_2P_3}$

Here $\overline{P_iP_{i+1}}=\overline{P_{i+1}P_{i}}$ for $i=1,\ldots,4$ denote the four sides of the quadrilateral (with indices taken modulo 4), the two diagonals are then $\overline{P_1P_3}$ and $\overline{P_2P_4}$ .

This relation may be verbally expressed as follows:

"If a quadrilateral is inscribed in a circle then the sum of the products of its two pairs of opposite sides is the product of its diagonals".

Examples

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 $\sqrt{2}a$ according to the Pythagorean theorem and the relation obviously holds.
More generally, if the quadrilateral is a rectangle with sides a and b and diagonal c then Ptolemaios' 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 c², the right hand side of Ptolemaios' relation is the sum a^{2^+ b².}
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 b^{2^{= a^{2^{+ ab which yields}}}}

$b={{(1+\sqrt{5})}\over 2}a$

The Ptolemaios' theorem, applied repeatedly, allows to compute the length of all diagonals for a polygon inscribed in a circle with vertices P₁, ..., P_n, if the sides are given together with all the length values of the "next to sides" chords connecting two vertices P_i and P_i+2 (with indices taken modulo n).

Proof(s)

"Trigonometric" proof of Ptolemaios' theorem

It suffices to prove the theorem for the standard unit circle (the statement of the theorem is invariant under re-scaling and translation). Introducing polar coordinates one may represent the four vertices $\,P_1, \ldots,P_4\,$ in the form

$P_i=(\,\cos\alpha_i,\sin\alpha_i \,)$ where $\alpha_i \in \,[\,0,2\pi).\,$

After a possible renumbering of the P_i one can also assume that the four vertices appear in natural counterclockwise order which means that $\,\alpha_1 < \alpha_2 < \alpha_3 <\alpha_4\,$ .

A basic result from trigonometry states that for two points $x=(\,\cos\alpha,\sin\alpha \,)$ and $y=(\,\cos\beta,\sin\beta\,)$ on the unit circle written in polar coordinates their Euclidean distance ||x − y|| is given as

$||x-y||=2\sin\left({{|\alpha-\beta|}\over 2}\right).$

If $(P_i,P_j),\;\; i<j$ is an (ordered) pair of vertices of the given quadrilateral this formula implies

$\overline{P_i P_j}=2\sin\left({\alpha_j\over 2}-{\alpha_i \over 2}\right).$

Ptolemaios' relation

$\overline{P_1P_3}\cdot \overline{P_2P_4}=\overline{P_1P_2}\cdot \overline{P_3P_4}+\overline{P_1P_4}\cdot \overline{P_2P_3}$

then follows from the quadratic addition relation

$\sin(\theta_3-\theta_1)\cdot \sin(\theta_4-\theta_2)=\sin(\theta_2-\theta_1)\cdot \sin(\theta_4-\theta_3)+\sin(\theta_4-\theta_1)\cdot \sin(\theta_3-\theta_2)$

satisfied by the sine-function which in turn can be deduced from the trigonometric identity (which is the products-to-sum identity for the sine)

$\sin\alpha\sin\beta={1\over2} \left( \cos(\alpha-\beta)-\cos(\alpha+\beta)\right)$

applied to each of the three products of sines (the resulting six terms cancel out in pairs).

Concluding remark (explaining the naming "addition relation"):

If one introduces the difference angles $\,\delta_i=\theta_{i+1}-\theta_{i}\,$ for $\,i=1, \ldots,3\,$ then the relation

$\sin(\theta_3-\theta_1)\cdot \sin(\theta_4-\theta_2)=\sin(\theta_2-\theta_1)\cdot \sin(\theta_4-\theta_3)+\sin(\theta_4-\theta_1)\cdot \sin(\theta_3-\theta_2)$

turns into

$\sin(\delta_1 +\delta_2 )\cdot \sin(\delta_2 +\delta_3 )=\sin(\delta_1)\cdot \sin(\delta_3)+\sin(\delta_1+\delta_2+\delta_3)\cdot \sin(\delta_2). \,$

Solving for $\sin(\delta_1+\delta_2+\delta_3\,)$ , this relation may be interpreted as a "triple" addition relation expressing the sine of a triple angle sum $\,\delta_1+\delta_2+\delta_3\,$ as a rational expression in the sine values $\sin(\delta_i+\delta_j\,)$ and $\sin \,\delta_i\,$ . Written out explicitly:

$\sin(\delta_1+\delta_2+\delta_3)={{\sin(\delta_1 +\delta_2 )\cdot \sin(\delta_2 +\delta_3 )-\sin(\delta_1)\sin(\delta_3)}\over {\sin\delta_2}}$

"Algebraic" proof of Ptolemaios' theorem

An alternative proof of Ptolemaios' Theorem can be given using complex number calculus and projective analytic geometry, introducing complex coordinates for the vertices of the quadrilateral. Again it suffices to prove the theorem for the standard unit circle $S^1=\{z \in \mathbb{C}, \; z\overline{z}=1\}$ .

Ptolemaios' Relation

$\overline{P_1P_3}\cdot \overline{P_2P_4}=\overline{P_1P_2}\cdot \overline{P_3P_4}+\overline{P_1P_4}\cdot \overline{P_2P_3}$

can be reformulated as

${{\overline{P_1P_3}\cdot \overline{P_2P_4}}\over{\overline{P_1P_4}\cdot \overline{P_2P_3}}} =1+{{\overline{P_1P_2}\cdot \overline{P_3P_4}}\over{\overline{P_1P_4}\cdot \overline{P_2P_3}}} \ .$

Written in this form Ptolemaios' theorem is in fact a "disguised" form of the relation

$\,\mbox{cr}(z_1,z_2,z_3,z_4)=1- \mbox{cr}(z_1,z_3,z_2,z_4)$

valid for the cross-ratio $\,\mbox{cr}(z_1,z_2,z_3,z_4)={{(z_1-z_3)(z_2-z_4)}\over{(z_1-z_4)(z_2-z_3)}}$ of any four (pairwise different) complex numbers $z_1,\ldots,z_4$ .

To make this connection explicit one represents the four vertices $P_1, \ldots, P_4$ as four complex numbers $z_1, \ldots, z_4$ of norm one, arranged in (counterclockwise) cyclic order on the unit circle. For two complex numbers $x, y$ on the unit circle their squared distance equals

$|x-y|^2=(x-y)\cdot(\overline{x}-\overline{y})=(x-y)\cdot({1\over x}-{1\over y})=-{(x-y)^2\over{xy}} \ .$

Therefore for any quadruple of (pairwise different) complex numbers $(z 1, z 2, z 3, z 4)$ on the unit circle the square of the "length cross-ratio"

${{|z_1-z_3|\cdot |z_2-z_4|}\over{|z_1-z_4|\cdot |z_2-z_3|}}$

is equal to the square $\,\mbox{cr}^2(z_1,z_2,z_3,z_4)$ of the ordinary ("complex points" ) cross-ratio $\,{{(z_1-z_3)(z_2-z_4)}\over{(z_1-z_4)(z_2-z_3)}}$ . Taking square roots one first deduces

${{|z_1-z_3|\cdot |z_2-z_4|}\over{|z_1-z_4|\cdot |z_2-z_3|}}=\epsilon {{(z_1-z_3)(z_2-z_4)}\over{(z_1-z_4)(z_2-z_3)}}=\epsilon\, \mbox{cr}( z_1,z_2,z_3,z_4)$

for any quadruple $\,(z_1,\ldots,z_4)$ of points on the unit circle. The sign factor $\epsilon \in \{ -1,1\}$ depends on the relative position of the four points $\,z_1, \ldots, z_4$ on the unit circle and can be determined using the invariance of the cross-ratio under a linear fractional transformation $z \mapsto {{az+b}\over{cz+d}}$ . Assume that the quadruple $\,(z_1, \ldots,z_4)$ on the unit circle is arranged in natural (counterclockwise) cyclic order. Then

$\,\mbox{cr}(z_1,z_2,z_3,z_4)>1 \ .$

This property can be proved using the projective transformation $r:\; z \mapsto i{{(1+z)}\over{(1-z)}}$ (which is the "inverse Cayley transform"). It maps the punctured unit circle $S^{1}\setminus \{z=1\}$ (continuously) to the real line $\mathbb{R}$ (with the upper (resp. lower) arc of the unit circle mapping to the negative (resp. positive) half-line). In polar coordinates the map is given as $\,r(e^{ i \alpha})=-\cot( \alpha /2)$ which shows that it defines a monotone function in the "angle" coordinate $\alpha \in )0,2\pi($ . Therefore the sign of the cross-ratio can be read off from the mutual order of the image points on the real line. After multiplying the $z i$ with a suitable scalar $z'$ of norm 1 one may in addition assume that $z_i \ne 1$ for all $i$ . If the quadruple $(\,z_1, \ldots,z_4)$ on the unit circle (punctured at $z = 1$ ) is arranged in natural (counterclockwise) cyclic order the image quadruple $\,(y_1, \ldots, y_4):=(\,r(z_1),r(z_2),r(z_3),r(z_4)\,)$ satisfies $\, y_1 <y_2 <y_3<y_4$ . The relation

$\mbox{cr}(y_1,y_2,y_3,y_4)-1={{(y_1-y_3)(y_2-y_4)}\over{(y_1-y_4)(y_2-y_3)}} -1={{(y_1-y_2)(y_3-y_4)}\over{(y_1-y_4)(y_2-y_3)}}>0$

then shows that $\,\mbox{cr}(z_1,z_2,z_3,z_4)=\mbox{cr}(y_1,y_2,y_3,y_4)>1$ . On the other hand, if one interchanges the middle pair $(z 2, z 3)$ in a cyclically ordered quadruple then the cross-ratio will become negative because $\,\mbox{cr}(z_1,z_3,z_2,z_4)=1-\mbox{cr}(z_1,z_2,z_3,z_4)<0$ , using the relation of cross-ratio's

${{(z_1-z_3)(z_2-z_4)}\over{(z_1-z_4)(z_2-z_3)}}=1-{{(z_1-z_2)(z_3 -z_4)}\over{(z_1-z_4)(z_3-z_2)}} \ .$

Summarizing the sign discussion one obtains that for a quadruple $(z_1, \ldots, z_4)$ of (pairwise different) points on the unit circle given in (counterclockwise) cyclic order one has

${{|z_1-z_3|\cdot |z_2-z_4|}\over{|z_1-z_4|\cdot |z_2-z_3|}}=+{{(z_1-z_3)(z_2-z_4)}\over{(z_1-z_4)(z_2-z_3)}}$

and

${{|z_1-z_2|\cdot |z_3-z_4|}\over{|z_1-z_4|\cdot |z_3-z_2|}}=-{{(z_1-z_2)(z_3-z_4)}\over{(z_1-z_4)(z_3-z_2)}} \ .$

Ptolemaios' relation

${{\overline{P_1P_3}\cdot \overline{P_2P_4}}\over{\overline{P_1P_4}\cdot \overline{P_2P_3}}} =1+{{\overline{P_1P_2}\cdot \overline{P_3P_4}}\over{\overline{P_1P_4}\cdot \overline{P_2P_3}}}$

can now be interpreted as the algebraic relation (already used above) between cross-ratio's

${{(z_1-z_3)(z_2-z_4)}\over{(z_1-z_4)(z_2-z_3)}}=1-{{(z_1-z_2)(z_3 -z_4)}\over{(z_1-z_4)(z_3-z_2)}}$

using the representation of the vertices $P_1,\ldots,P_4$ as the points $z_1, \ldots, z_4$ on the unit circle.

Ptolemaios' theorem

Examples

Proof(s)

External links

See also: Ptolemaios' theorem, Analytic geometry, Astronomy, Cayley transform, Circle, Complex number, Cross-ratio, Distance, Euclidean geometry, Linear fractional transformation