Casey's theorem

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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.

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[edit] Formulation of the theorem

Let \,O be a circle of radius \,R. Let \,O_1, O_2, O_3, O_4 be (in that order) four non-intersecting circles that lie inside \,O and tangent to it. Denote by \,\ell_{i,j} the length of the exterior common tangent of the circles \,O_i, O_j. Then:

\,\ell_{1,2}\ell_{3,4}+\ell_{1,4}\ell_{2,3}=\ell_{1,3}\ell_{2,4}

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

[edit] Proof

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,

\,\ell_{i,j}^2=\overline{O_iO_j}^2-(R_i-R_j)^2.

We will try to express this length in terms of the points \,K_i,K_j. By the law of cosines in triangle \,O_iOO_j,

\overline{O_iO_j}^2=\overline{OO_i}^2+\overline{OO_j}^2-2\overline{OO_i}\cdot \overline{OO_j}\cdot \cos\angle O_iOO_j

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

\overline{OO_i} = R - R_i,\, \angle O_iOO_j = \angle K_iOK_j

Let \,C be a point on the circle \,O. According to the law of sines in triangle \,K_iCK_j:

\overline{K_iK_j} = 2R\cdot \sin\angle K_iCK_j = 2R\cdot \sin\frac{\angle K_iOK_j}{2}

Therefore,

\cos\angle K_iOK_j = 1-2\sin^2\frac{\angle K_iOK_j}{2}=1-2\cdot \left(\frac{\overline{K_iK_j}}{2R}\right)^2 = 1 - \frac{\overline{K_iK_j}^2}{2R^2}

and substituting these in the formula above:

\overline{O_iO_j}^2=(R-R_i)^2+(R-R_j)^2-2(R-R_i)(R-R_j)\left(1-\frac{\overline{K_iK_j}^2}{2R^2}\right)
\overline{O_iO_j}^2=(R-R_i)^2+(R-R_j)^2-2(R-R_i)(R-R_j)+(R-R_i)(R-R_j)\cdot \frac{\overline{K_iK_j}^2}{R^2}
\overline{O_iO_j}^2=((R-R_i)-(R-R_j))^2+(R-R_i)(R-R_j)\cdot \frac{\overline{K_iK_j}^2}{R^2}

And finally, the length we seek is

\ell_{i,j}=\sqrt{\overline{O_iO_j}^2-(R_i-R_j)^2}=\frac{\sqrt{R-R_i}\cdot \sqrt{R-R_j}\cdot \overline{K_iK_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_1K_2K_3K_4:

\ell_{1,2}\ell_{3,4}+\ell_{1,4}\ell_{2,3}=\frac{1}{R^2}\cdot \sqrt{R-R_1}\sqrt{R-R_2}\sqrt{R-R_3}\sqrt{R-R_4}\left(\overline{K_1K_2}\cdot \overline{K_3K_4}+\overline{K_1K_4}\cdot \overline{K_2K_3}\right)
=\frac{1}{R^2}\cdot \sqrt{R-R_1}\sqrt{R-R_2}\sqrt{R-R_3}\sqrt{R-R_4}\left(\overline{K_1K_3}\cdot \overline{K_2K_4}\right)=\ell_{1,3}\ell_{2,4}

Q.E.D.

[edit] Further generalizations

It can be seen that the four circles must 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 (in or out), \,\ell_{i,j} is the length of the exterior common tangent.

If \,O_i, O_j are tangent from different sides of \,O (in or out), \,\ell_{i,j} is the length of the interior common tangent.

It is also worth noting that the opposite direction of this statement is also true. That is, if equality holds, the circles are tangent.

[edit] Applications

Casey's theorem can be used to prove a variety of statements in Euclidean geometry.

For example, the shortest known proof of Feuerbach's theorem uses it.

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