Let S be the circumcenter of ABC. $A_0$ is the middle of arc BC not containing A, $C_0$ also the middle of arc AB without C.

Let $S_1$ be a circle with center $A_0$, tangent to BC, $S_2$ with center $C_0$ and tangent to AB. Showing that the incenter I of ABC lies on the common extangent to $S_1$ and $S_2$.

I tried some angle chasing, considered common extangent of the circumcircle of ABC and both $S_1$ and $S_2$, tried Power of a Point theorem, polar bijection, but I don't know how to prove that the point I lies on the extangent. I'm really clueless how to proceed. I tried thise things bluntly but they seem to be irrelevant. How is it done?

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    $\begingroup$ No problem is beautiful if it's somebody else's homework that he just needs help with. Do you mind if I consider the title kinda inaccurate? $\endgroup$ Dec 13 '14 at 8:03
  • $\begingroup$ Consider explaining what you have done instead of only mentioning them. That way, other readers can know what's your thought process and can assist you easier. $\endgroup$
    – Andrew T.
    Dec 13 '14 at 8:05
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    $\begingroup$ Yes, this abuse of "beautiful ..." needs to stop. $\endgroup$ Dec 13 '14 at 8:05
  • $\begingroup$ I assume the middle of an arc is meant to be its intersection with its bisector? $\endgroup$ Dec 13 '14 at 8:09
  • $\begingroup$ @jandvorak yes the intersection of the circumcircle and the bisector $\endgroup$ Dec 13 '14 at 9:10

For now, we'll consider just the circle about $A_0$.

Let $M$ be the midpoint of $\overline{BC}$ (and therefore also the foot of the perpendicular from $A_0$). Drop a perpendicular from $A_0$ to $\overline{AB}$ at $X$, and let $P$ be the point where this perpendicular crosses $\bigcirc A_0$. Finally, let $I$ be the incenter of $\triangle ABC$, so that $\overline{IA}$ and $\overline{IC}$ bisect $\angle A$ and $\angle C$.

enter image description here

  • Since $\angle BAA_0$ and $\angle BCA_0$ subtend the same arc $\stackrel{\frown}{A_0B}$ of the circumcircle, they are congruent.

  • By the Exterior Angle Theorem applied to $\angle I$ of $\triangle IAC$, we have $$\angle CIA_0 = \color{blue}{\angle IAC} + \color{red}{\angle ICA} = \color{blue}{\angle BCA_0} + \color{red}{\angle ICB} = \angle ICA_0$$ so that $\triangle A_0CI$ is isosceles, with $\color{violet}{\overline{A_0C}}\cong\color{violet}{\overline{A_0I}}$.

  • As $\triangle AA_0X \sim \triangle CA_0 M$, we have $$\frac{|\overline{A_0X}|}{|\overline{A_0A}|} = \frac{|\color{green}{\overline{A_0M}}|}{|\color{violet}{\overline{A_0C}}|} \quad\to\quad \frac{|\overline{A_0X}|}{|\overline{A_0A}|} = \frac{|\color{green}{\overline{A_0P}}|}{|\color{violet}{\overline{A_0I}}|}$$ which shows that $\triangle A_0AX \sim \triangle A_0IP$, and therefore that $\overline{IP}\perp\overline{A_0P}$.

  • Therefore, the line through $I$ parallel to $\overline{AC}$ is tangent to $\bigcirc A_0$; likewise, this line is tangent to $\bigcirc C_0$ (not shown).

  • $\begingroup$ This is very clear and cool! Struggled on it the whole day. Thanks a lot! :) $\endgroup$ Dec 13 '14 at 15:35

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