Додаток: Узагальненна теорема Птолемея

The generalized version of Ptolemy's theorem, also known, as Casey's theorem states the following.

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 circle that lie inside $O$ and tangent to it. Denote by $t_{ij}$ the length of the exterior common tangent of the circles $O_i,O_j$. Then, $$t_{12} \cdot t_{34} + t_{41} \cdot t_{23} = t_{13} \cdot t_{24}.$$

Think about why this is a generalized case of Ptolemy's theorem. When is Ptolemy's Theorem a special case of this?

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

Proof, taken from Wikipedia. Use this diagram.

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, from the Pythagorean Theorem, that $$t_{ij}^2 = \left( O_i O_j \right)^2 - \left( R_i - R_j \right)^2.$$ We will try to express this length in terms of the points $K_i,K_j$. By the law of cosines on triangle $O_iOO_j$, we have $$\left( O_i O_j \right)^2 = \left( O O_i \right)^2 + \left( O O_j \right)^2 - 2 \left( O O_i \right) \cdot \left( O O_j \right) \cdot \cos \angle O_i O O_j.$$Since the circles $O$ and $O_i$ are tangent to each other, we find that $OO_i = R - R_i$ and $\angle O_i O O_j = \angle K_i O K_j$. Let $C$ be a point on the circle $O$. According to the law of sines on triangles $K_i C K_j$, we have $$K_i K_j = 2R \cdot \sin \angle K_i C K_j = 2R \cdot \sin \frac {\angle K_i O K_j}{2}.$$ Therefore, $$\cos \angle K_i O K_j = 1 - 2 \sin^2 \frac {\angle K_i O K_j}{2} = 1 - 2 \cdot \left( \frac {K_i K_j}{2R} \right)^2 = 1 - \frac {\left( K_i K_j \right)^2}{2R^2}.$$ Substituting and simplifying, you should get $$\left( O_i O_j \right)^2 = \left( \left( R - R_i \right) - \left( R - R_j \right) \right)^2 + \left( R - R_i \right) \left( R - R_j \right) \cdot \frac {\left( K_i K_j \right)^2}{R^2}.$$And eventually, after much algebra, we find that $$t_{ij} = \sqrt {\left( O_i O_j \right)^2 - \left( R_i - R_j \right)^2} = \frac {\sqrt {R - R_i} \cdot \sqrt {R - R_j} \cdot \left( K_i K_j \right)}{R}$$ and in fact, the left-hand-side in the original theorem becomes $$\begin {align*} t_{12} t_{34} + t_{14} t_{23} &= \frac {1}{R^2} \cdot \sqrt {R-R_1} \cdot \sqrt {R - R_2} \cdot \sqrt {R - R_3} \cdot \sqrt {R - R_4} \cdot \left( \left( K_1 K_3 \right) \cdot \left( K_2 K_4 \right) \right) \\&= t_{13} t_{24}, \end {align*}$$as desired. $\blacksquare$

Exercise for the reader. Verify the steps we skipped and make sure you understand where everything comes from!