Ptolemy's Theorem via Vector Inversion

#Ptolemy'sTheorem Geometry

Problem Statement

Ptolemy's Inequality.

For any four distinct points $A,B,C,D$ in the plane,

$$ AC\cdot BD\leq AB\cdot CD+AD\cdot BC. $$

Equality holds if and only if the four points are concyclic in the order $A,B,C,D$, or are collinear with the corresponding ordering.

Motivation: The desired inequality involves products of distances, whereas the ordinary triangle inequality involves sums of vector magnitudes. We therefore seek a transformation that converts products of distances into ordinary distances, up to a common multiplicative factor.

The key observation is that the transformation

$$ T(\mathbf{x})=\frac{\mathbf{x}}{|\mathbf{x}|^2}, \qquad \mathbf{x}\neq\mathbf{0}, $$

satisfies the claim

$$ \boxed{ \left| \frac{\mathbf{x}}{|\mathbf{x}|^2} - \frac{\mathbf{y}}{|\mathbf{y}|^2} \right| = \frac{|\mathbf{x}-\mathbf{y}|} {|\mathbf{x}||\mathbf{y}|}. } $$

Indeed, using the inner product,

$$ \left| \frac{\mathbf{x}}{|\mathbf{x}|^2} - \frac{\mathbf{y}}{|\mathbf{y}|^2} \right|^2 $$

$$= \frac{1}{|\mathbf{x}|^2} +\frac{1}{|\mathbf{y}|^2} -\frac{2\mathbf{x}\cdot\mathbf{y}} {|\mathbf{x}|^2|\mathbf{y}|^2}$$

$$ = \frac{|\mathbf{x}-\mathbf{y}|^2} {|\mathbf{x}|^2|\mathbf{y}|^2}, $$

which establishes the claim above.

Proof. Take $A$ as the origin and put

$$ \mathbf{b}=\overrightarrow{AB},\qquad \mathbf{c}=\overrightarrow{AC},\qquad \mathbf{d}=\overrightarrow{AD}. $$

By the triangle inequality,

$$ |T(\mathbf{b})-T(\mathbf{d})| \le |T(\mathbf{b})-T(\mathbf{c})| + |T(\mathbf{c})-T(\mathbf{d})|. $$

Applying the claim, we obtain

$$ \frac{BD}{AB \cdot AD} \le \frac{BC}{AB \cdot AC} + \frac{CD}{AC \cdot AD}. $$

Multiplying by $AB\cdot AC\cdot AD$ gives

$$ \boxed{ AC\cdot BD\leq AD\cdot BC+AB\cdot CD. } $$

The Equality Case: Equality holds in the triangle inequality precisely when the inverted points

$$ B'=T(\mathbf{b}),\qquad C'=T(\mathbf{c}),\qquad D'=T(\mathbf{d}) $$

are collinear, with $C'$ lying on the segment $B'D'$. To understand this condition, observe that a line not passing through the origin has an equation

$$ \mathbf{n}\cdot\mathbf{x}=k, \qquad k\neq 0. $$

Under inversion, $\mathbf{x}$ is replaced by $\mathbf{x}/|\mathbf{x}|^2$. Thus its inverse image satisfies

$$ \mathbf{n}\cdot\mathbf{x} =k|\mathbf{x}|^2, $$

or, equivalently,

$$ \left| \mathbf{x}-\frac{\mathbf{n}}{2k} \right|^2 = \frac{|\mathbf{n}|^2}{4k^2}. $$

This is a circle passing through the origin $A$.

Consequently, for four non-collinear points, equality holds precisely when $A,B,C,D$ lie on a circle in the cyclic order $A,B,C,D$. The order ensures that $C'$ lies between $B'$ and $D'$. Therefore, for a cyclic quadrilateral $ABCD$,

$$ \boxed{ AC\cdot BD=AB\cdot CD+AD\cdot BC. } $$

This is Ptolemy's theorem.

Remark. The proof reveals that Ptolemy's theorem is essentially the triangle inequality viewed through inversion. Moreover, the argument for Ptolemy's inequality works in any real inner-product space. If you are a high school student, then you can think of inner product as dot product for now.

Bibliography & References

[1]
Ptolemy'sTheorem