TITLE: When you inscribe a triangle in a circle in a triangle in a circle in a triangle ...
QUESTION [36 upvotes]: Here's a cute question I came up with.

Start with a circle, and then choose three points $a$, $b$, and $c$ on the circle, and proceed as follows:

Draw the triangle inside the circle with vertices $a$, $b$, and $c$
Draw the inscribed circle of that triangle, which is tangent to each of the three sides of triangle, and now label these three points of tangency as $a$, $b$, and $c$ (so we're updating which points we're calling points $a$, $b$, and $c$).
Repeat from Step 1.

This construction gives us a sequence of inscribed circles and triangles that telescope down to a point, a limit point, which is determined only by the initial choice of the points $a$, $b$, and $c$. Which points in the interior of the circle are limit points of this construction?

Also, if anyone has ideas for more interesting variations of this question, I'd like to hear them.

REPLY [18 votes]: This answer has very much the same flavor as Micah's answer, since it relies on there being a continuous map from the points $(a,b,c)$ to the limit point, and on the Intermediate Value Theorem.

Take any point $X$ (red-violet and blinking for some reason) in the interior of your circle and let $O$ (green) be the center of your circle. Draw the ray $\overrightarrow{OX}$, and let $a$ (violet) be the point where this ray intersects the circle. Draw the other two points $b$ and $c$ (both violet) such that $\angle aOb = \angle aOc$ and each are less than or equal to $2\pi/3$. By the symmetry of this construction, the limit point must lie on the segment $\overline{Oa}$. 
Now the claim is that this function from $(0,2\pi/3] \to \overline{Oa}$ where $\angle aOb$ maps to the limit point is continuous. In order to rigorously show that it's continuous (without just trusting the picture) we'd have to do some calculations like in Thomas Andrews's answer. But if you trust that it is continuous, since the limit point is the center of the original circle when $\angle aOb = 2\pi/3$, and since the limit point approaches $a$ as $\angle aOb \to 0$, by the Intermediate Value Theorem the limit point must be $X$ at some point in between. The point $X$ was chosen arbitrarily in the interior of the circle, so we're good: every point in the interior of the circle is the limit point of this construction for some choice of $a$, $b$, and $c$.