I don't really get why you say it's so easy, but maybe it's just my headache that don't let me see such trivial things .

vinoth_90_2004 wrote: Let $ A_1$, $ A_2$, $ A_3$ and $ A_4$ be four circles such that the circles $ A_1$ and $ A_3$ are tangent at a point $ P$, and the circles $ A_2$ and $ A_4$ are also tangent at the same point $ P$. Suppose that the circles $ A_1$ and $ A_2$ meet at a point $ T_1$, the circles $ A_2$ and $ A_3$ meet at a point $ T_2$, the circles $ A_3$ and $ A_4$ meet at a point $ T_3$, and the circles $ A_4$ and $ A_1$ meet at a point $ T_4$, such that all these four points $ T_1$, $ T_2$, $ T_3$, $ T_4$ are distinct from $ P$. Prove: $ \frac {\overline{T_1T_2}\cdot\overline{T_2T_3}}{\overline{T_1T_4}\cdot\overline{T_3T_4}} = \frac {\overline{PT_2}^2}{\overline{PT_4}^2}$ (where $ \overline{ab}$, of course, means the distance between points $ a$ and $ b$). Good old inversion.. . Consider the inversion of pole $ P$ and any power we want and it should yield a pretty nice and neat solution. Let $ F'$ be the image the figure $ F$ by the inversion (no matter what $ F$ is: circle, line, point etc.). We have $ \frac {T_{1}T_{2}}{T_{1}'T_{2}'} = \frac {PT_{2}}{PT_{1}'}$ and the analogous relations. From all of that we find $ \frac {T_{1}T_{2}\cdot T_{3}T_{2}}{T_{1}T_{4}\cdot T_{3}T_{4}} = \frac {T_{1}'T_{2}'\cdot T_{3}'T_{2}'}{T_{1}'T_{4}'\cdot T_{3}'T_{4}'}\cdot \frac {PT_{2}^{2}}{PT_{4}^{2}} = \frac {PT_{2}^{2}}{PT_{4}^{2}}$ because $ A_{1}'\parallel A_{3}',\ A_{2}'\parallel A_{4}'$, so $ T_{1}'T_{2}'T_{3}'T_{4}'$ is a parallelogram, so $ T_{1}'T_{2}' = T_{3}'T_{4}',\ T_{3}'T_{2}' = T_{1}'T_{4}'$.

please is there another solution without inversion ??

grobber wrote: Let $ F'$ be the image the figure $ F$ by the inversion (no matter what $ F$ is: circle, line, point etc.). We have $ \frac {T_{1}T_{2}}{T_{1}'T_{2}'} = \frac {PT_{2}}{PT_{1}'}$ and the analogous relations. Can somebody explain me how to get this: $\frac{T_1 T_2}{T_1 ' T_2 '} =\frac{PT_2}{PT_1 '}$

Dr N0 wrote: grobber wrote: Let $ F'$ be the image the figure $ F$ by the inversion (no matter what $ F$ is: circle, line, point etc.). We have $ \frac {T_{1}T_{2}}{T_{1}'T_{2}'} = \frac {PT_{2}}{PT_{1}'}$ and the analogous relations. Can somebody explain me how to get this: $\frac{T_1 T_2}{T_1 ' T_2 '} =\frac{PT_2}{PT_1 '}$ We have $\frac{T_1T_2}{T_1'T_2'}=\frac{PT_1\cdot PT_2}{r^2}$ From definition, $\frac{PT_1}{r^2}=\frac{1}{PT_1'}$ and substituting there gives the result. Just invert it with respect to $P$ and you get a parallelogram. Apply properties of parallelogram to conclude the proof.

Thank you. I know that $\frac{PT_1}{r^2}=\frac{1}{PT_1 '}$, but I still don't get why $\frac{T_1 T_2}{T_1 ' T_2 '}=\frac{PT_1 PT_2}{r^2}$ I substituted $\frac{PT_1}{r^2}=\frac{1}{PT_1 '}$ into $\frac{T_1 T_2}{T_1 ' T_2 '}=\frac{PT_1 PT_2}{r^2}$ but it didn't give me anything that I could recognize

Dr N0 wrote: Thank you. I know that $\frac{PT_1}{r^2}=\frac{1}{PT_1 '}$, but I still don't get why $\frac{T_1 T_2}{T_1 ' T_2 '}=\frac{PT_1 PT_2}{r^2}$ I substituted $\frac{PT_1}{r^2}=\frac{1}{PT_1 '}$ into $\frac{T_1 T_2}{T_1 ' T_2 '}=\frac{PT_1 PT_2}{r^2}$ but it didn't give me anything that I could recognize It is very obvious. $\frac{T_1 T_2}{T_1 ' T_2 '}=\frac{PT_1 PT_2}{r^2}=\left(\frac{PT_1}{r^2}\right)\cdot PT_2=\left(\frac{1}{PT_1'}\right)\cdot PT_2$ and so, we get the expresssion. I use the fact that If two points $A, B$ are inverted with centre $O$, and they map to $A', B'$ respectively, we have $A'B'=\frac{AB\cdot r^2}{OA\cdot OB}$

Oh, thank you, I got it now. What I was not realizing is that triangles $OAB$ and $OA'B'$ are similar. That's how you got $A'B'=\frac{AB\cdot r^2}{OA\cdot OB}$