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The incircle of triangle $A_1 A_2 A_3$ touches the sides $A_2 A_3$, $A_3 A_1$, and $A_1 A_2$ at $S_1$, $S_2$, and $S_3$, respectively. Let $O_1$, $O_2$, and $O_3$ be the incenters of triangles $A_1 S_2 S_3$, $A_2 S_3 S_1$, and $A_3 S_1 S_2$, respectively. Prove that $O_1 S_1$, $O_2 S_2$, $O_3 S_3$ are concurrent.

I have tried drawing a big diagram, as accurate as possible, as a start, and saw that $O_1$, $O_2$, and $O_3$ seem to touch the incircle of triangle $A_1A_2A_3$. However, I am not sure of how to prove this, or how this helps with proving the concurrency. Can I have a little hint as to how to prove this and how it helps answer the question? Much appreciated.

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  • $\begingroup$ Showing that, say, $O_3$ lies on the incircle can be done with a little angle chasing. Using the fact that $\triangle S_1 S_2 A_3$, etc, are isosceles, and that $\overline{O_3S_1}$ and $\overline{O_3S_2}$ are angle bisectors, verify that $\angle S_1 S_3 S_2 = (\angle A_1 + \angle A_2 )/2$ and that $\angle S_1 O_3 S_2$ is its supplement; this guarantees that $\square S_1 S_3 S_2 O_3$ is cyclic. $\endgroup$
    – Blue
    Jan 20, 2015 at 3:00
  • $\begingroup$ For the final proof, note that $O_3$ is on the perpendicular bisector of $\overline{S_1S_2}$ (why?), which, because all the $O_i$s and $S_i$s lie on the incircle, says something interesting about $\angle S_1 S_3 O_3$ and $\angle S_2 S_3 O_3$. $\endgroup$
    – Blue
    Jan 20, 2015 at 3:07
  • $\begingroup$ @Blue I like your last part. For the first part, I think it might be better not to think about $A_3$ too much. $\endgroup$
    – user208259
    Jan 20, 2015 at 3:09
  • $\begingroup$ @user208259: There's not a great deal of thinking to be done, about $A_3$ or anything else. The angle chase is over pretty quickly. $\endgroup$
    – Blue
    Jan 20, 2015 at 3:14
  • $\begingroup$ @Blue If you draw the bisector of $\angle S_2 S_1 A_3$, it subtends (with the ray $S_1 A_3$) an arc on the incircle that must be half as large as that subtended by $\angle S_2 S_1 A_3$. Hence it meets the circle at the midpoint of the arc $S_1 S_2$. Then do the same for $\angle S_1 S_2 A_3$. $\endgroup$
    – user208259
    Jan 20, 2015 at 3:19

1 Answer 1

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You're correct about $O_1$, etc.

Here's a hint to prove it.

Take two points $S_2$ and $S_3$ on a circle. Draw the tangent line at $S_2$, and take a bisector between the tangent line and $S_2S_3$. Prove that the bisector meets the circle midway along the arc from $S_2$ to $S_3$. (You'll want to use the version of the inscribed angle theorem that works for tangent lines.)

Once you have that, the problem is simplified somewhat because you can forget about $A_1$, $A_2$ and $A_3$ and just focus on the points $S_i$ and $O_i$ on the circle, with the points $O_i$ placed on the circle at the midpoints. In this case, say $S_2 O_2$ and $S_3 O_3$ meet at $P_1$, and define $P_2$ analogously. Prove that $P_1O_3 = S_2 O_3 = S_1 O_3 = P_2 O_3$.

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  • $\begingroup$ An earlier version of my answer had an error in it. $\endgroup$
    – user208259
    Jan 20, 2015 at 2:58

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