Casey's theorem

Template:Short description In mathematics, Casey's theorem, also known as the generalized Ptolemy's theorem, is a theorem in Euclidean geometry named after the Irish mathematician John Casey.

Formulation of the theorem

File:Casey new1a.svg

$t_{12}\cdot t_{34}+t_{14}\cdot t_{23}-t_{13}\cdot t_{24}=0$

Note that in the degenerate case, where all four circles reduce to points, this is exactly Ptolemy's theorem.

$t_{12}\cdot t_{34}+t_{14}\cdot t_{23}=t_{13}\cdot t_{24}.$

Proof

The following proof is attributable to Zacharias. Denote the radius of circle $O_i$ by $R_i$ and its tangency point with the circle $O$ by $K_i$. We will use the notation $O,O_i$ for the centers of the circles. Note that from Pythagorean theorem,

$t_{ij}^2=\stackrel{2}{\stackrel{‾}{O_i{O}_j}}-(R_i-R_j{)}^2.$

$\stackrel{2}{\stackrel{‾}{O_i{O}_j}}=\stackrel{2}{\stackrel{‾}{O{O}_i}}+\stackrel{2}{\stackrel{‾}{O{O}_j}}-2\overline{O{O}_i}\cdot \overline{O{O}_j}\cdot \cos \mathrm{\angle }{O}_{i}O{O}_j$

Since the circles $O,O_i$ tangent to each other:

$\overline{O{O}_i}=R-R_i,\mathrm{\angle }{O}_{i}O{O}_j=\mathrm{\angle }{K}_{i}O{K}_j$

Let $C$ be a point on the circle $O$. According to the law of sines in triangle $K_{i}C{K}_j$:

$\overline{K_i{K}_j}=2R\cdot \sin \mathrm{\angle }{K}_{i}C{K}_j=2R\cdot \sin \frac{\mathrm{\angle }{K}_{i}O{K}_j}{2}$

Therefore,

$\cos \mathrm{\angle }{K}_{i}O{K}_j=1-2{\sin }^{}\frac{\mathrm{\angle }{K}_{i}O{K}_j}{2}=1-2\cdot {\left(\frac{\overline{K_i{K}_j}}{2R}\right)}^2=1-\frac{\stackrel{2}{\stackrel{‾}{K_i{K}_j}}}{2R^2}$

and substituting these in the formula above:

$\stackrel{2}{\stackrel{‾}{O_i{O}_j}}=(R-R_i{)}^2+(R-R_j{)}^2-2(R-R_i)(R-R_j)(1-\frac{\stackrel{2}{\stackrel{‾}{K_i{K}_j}}}{2R^2})$

$\stackrel{2}{\stackrel{‾}{O_i{O}_j}}=(R-R_i{)}^2+(R-R_j{)}^2-2(R-R_i)(R-R_j)+(R-R_i)(R-R_j)\cdot \frac{\stackrel{2}{\stackrel{‾}{K_i{K}_j}}}{R^2}$

$\stackrel{2}{\stackrel{‾}{O_i{O}_j}}=((R-R_i)-(R-R_j){)}^2+(R-R_i)(R-R_j)\cdot \frac{\stackrel{2}{\stackrel{‾}{K_i{K}_j}}}{R^2}$

And finally, the length we seek is

$t_{ij}=\sqrt{\stackrel{2}{\stackrel{‾}{O_i{O}_j}}-(R_i-R_j{)}^2}=\frac{\sqrt{R-R_i}\cdot \sqrt{R-R_j}\cdot \overline{K_i{K}_j}}{R}$

We can now evaluate the left hand side, with the help of the original Ptolemy's theorem applied to the inscribed quadrilateral $K_1{K}_2{K}_3{K}_4$:

$\begin{array}{rl}& t_{12}{t}_{34}+t_{14}{t}_{23}\\ =& \frac{1}{R^2}\cdot \sqrt{R-R_1}\sqrt{R-R_2}\sqrt{R-R_3}\sqrt{R-R_4}(\overline{K_1{K}_2}\cdot \overline{K_3{K}_4}+\overline{K_1{K}_4}\cdot \overline{K_2{K}_3})\\ =& \frac{1}{R^2}\cdot \sqrt{R-R_1}\sqrt{R-R_2}\sqrt{R-R_3}\sqrt{R-R_4}(\overline{K_1{K}_3}\cdot \overline{K_2{K}_4})\\ =& t_{13}{t}_{24}\end{array}$

Further generalizations

It can be seen that the four circles need not lie inside the big circle. In fact, they may be tangent to it from the outside as well. In that case, the following change should be made:

If $O_i,O_j$ are both tangent from the same side of $O$ (both in or both out), $t_{ij}$ is the length of the exterior common tangent.

If $O_i,O_j$ are tangent from different sides of $O$ (one in and one out), $t_{ij}$ is the length of the interior common tangent.

The converse of Casey's theorem is also true. That is, if equality holds, the circles are tangent to a common circle.

Applications

Casey's theorem and its converse can be used to prove a variety of statements in Euclidean geometry. For example, the shortest known proofTemplate:Rp of Feuerbach's theorem uses the converse theorem.

External links

• Shailesh Shirali: "'On a generalized Ptolemy Theorem'". In: Crux Mathematicorum, Vol. 22, No. 2, pp. 49-53