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:

  1. Draw the triangle inside the circle with vertices $a$, $b$, and $c$
  2. 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 $c$).
  3. 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.

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    $\begingroup$ I assume all $a,b,c$ are distinct? I'd be shocked if its not the interior of the disk bounded by the circle. $\endgroup$ Commented Mar 1, 2018 at 20:13
  • $\begingroup$ @PeteCaradonna The way I described the construction, yeah, you need to assume $a \neq b \neq c$. But I think that there's a natural choice for the limit point in this degenerate case. $\endgroup$ Commented Mar 1, 2018 at 20:18
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    $\begingroup$ Next question: start with the triangle, draw the inscribed circle, draw the next triangle, etc. :-) Which of the triangle centers is the limit of this process? $\endgroup$ Commented Mar 2, 2018 at 6:04

3 Answers 3


Without loss of generality assume that the circle has radius $1$, and that the triple $(a,b,c)$ is positively oriented (i.e., they go counterclockwise around the circle). As the problem is circularly symmetric, it suffices to show that we can choose $a,b,c$ so that the limit point has any specified distance from its center.

For any $\epsilon > 0$, we can choose $a, b, c$ such that the triangle $abc$ does not intersect the circle of radius $1-\epsilon$. Since the limit point must lie inside the triangle $abc$, it follows that the limit point can come arbitrarily close to the boundary.

On the other hand, if the triangle $abc$ is equilateral, the limit point is clearly the center of the circle (as it is fixed under a rotation by $\frac{2 \pi}{3}$ about the center).

So, if we can show that the map from $(a,b,c)$ to the limit point is continuous, it will follow that the limit point can have any distance less than $1$ from the center. The set of all possible distances will be a continuous image in $\Bbb{R}$ of the connected set of all possible oriented triples $(a,b,c)$, so it must be an interval; we have showed that this interval both contains $0$ and comes arbitrarily close to $1$.

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    $\begingroup$ Yeah, this is more-or-less what I had in mind. I've got a different geometric set-up, but so long as the function that takes $(a,b,c)$ to the limit point continuous, you can say a lot by evoking the Intermediate Value Theorem. $\endgroup$ Commented Mar 1, 2018 at 20:28
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    $\begingroup$ No, you should be invoking it :) $\endgroup$
    – Narasimham
    Commented Mar 1, 2018 at 21:31
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    $\begingroup$ I like the equilateral road you're going down. Loosely to me too it seemed like this process forces the triangles to become more equilateral over time. $\endgroup$ Commented Mar 1, 2018 at 22:11
  • $\begingroup$ @PeteCaradonna, my first thought on reading the question was that the triangles would likely each be similar, but I can see that's not the case. However, it still seems possible at a glance they may be alternately similar (e.g. a "skinny tall" isosceles, then a "wide short" isosceles, etc.). I don't see why the rough alternation should have to converge. (But I haven't worked it out precisely; these are just rough observations.) $\endgroup$
    – Wildcard
    Commented Mar 2, 2018 at 1:02
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    $\begingroup$ @Wildcard: It's not that hard to see that the angles in the $(n+1)$st triangle are the pairwise averages of the angles in the $n$th triangle, so they do converge to equilateral. (But I agree that you have to actually draw a diagram to see this; it's not intuitively clear to me.) $\endgroup$
    – Micah
    Commented Mar 2, 2018 at 1:29

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.

Diagram of the construction

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 $c$.

  • $\begingroup$ Very pretty. It is not necessary for the proof, but if you extended beyond an equilateral triangle to a very narrow isosceles triangle, the limit point would reach arbitrarily close to the bottom of the circle $\endgroup$
    – Henry
    Commented Mar 2, 2018 at 16:27
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    $\begingroup$ @Henry constructing it that way would prove that all possible $X$ that lie on the diameter of the original circle are a limit point, but I think it sufficient just to prove it just for a radius of the circle (since every interior point $X$ lies on the radius $\overrightarrow{OX}$). $\endgroup$ Commented Mar 2, 2018 at 16:31
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    $\begingroup$ I completely agree - which is why I said "It is not necessary for the proof". It is just that I had been mildly surprised when the animation stopped at an equilateral triangle as my idea for a solution had been exactly the opposite of your animation: start with a narrow isosceles triangle and widen to equilateral $\endgroup$
    – Henry
    Commented Mar 2, 2018 at 16:38
  • $\begingroup$ @Henry Ah! I've got you. I'm sorry I misread your comment. Your idea is actually the first thing I thought of too. :) $\endgroup$ Commented Mar 2, 2018 at 16:44

Not an answer, but too long for a comment.

We can get an explicit formula for this operation if we use vector operations.

Letting $A=|\vec b-\vec c|,B=|\vec a-\vec c|, C=|\vec a-\vec b|$, define $$\alpha=\frac{1}{2}(B+C-A),\\\beta=\frac{1}{2}(A+C-B),\\ \gamma=\frac{1}{2}(A+B-C).$$

Then the new points $\vec {a'},\vec {b'},\vec {c'}$ are $$\vec {a'}=\frac{\gamma \vec b+\beta \vec c}{\gamma+\beta}\\\vec {b'}=\frac{\gamma \vec a+\alpha \vec c}{\gamma+\alpha}\\\vec {c'}=\frac{\beta \vec a+\alpha \vec b}{\beta+\alpha}$$

Worth noting that $\gamma+\beta=A, \gamma+\alpha=B,\gamma+\beta=C.$

We can thus write this as:

$$\vec {a'}=\frac{1}{2}\left(\left(1+\frac{B-C}{A}\right)\vec b+\left(1+\frac{C-B}{A}\right)\vec c\right)$$

or $$\vec {a'}=\frac{\vec b+\vec c}{2}+\frac{|\vec a-\vec c|-|\vec a-\vec b|}{2}\frac{\vec b-\vec c}{|\vec b-\vec c|}$$

So, if $$f(\vec u,\vec v,\vec w)=\frac{\vec v+\vec w}{2}+\frac{|\vec u-\vec w|-|\vec u-\vec v|}{2}\frac{\vec v-\vec w}{|\vec v-\vec w|}$$

We can define $f$ so that it is continuous by defining $f(\vec u,\vec v,\vec v)=\vec v.$

Then our transformation is:

$$M:(\vec a,\vec b,\vec c)\mapsto (f(\vec a,\vec b,\vec c),f(\vec b,\vec c,\vec a),f(\vec c.\vec a,\vec b))$$

  • $\begingroup$ I do not know why on certain sites I have completely different sets of common typing errors. On this site, and no other, I frequent write "write" as "right" and "two" and "too" as "to" and I never do that on any other site. $\endgroup$
    – fleablood
    Commented Mar 1, 2018 at 20:37
  • $\begingroup$ Part of it might be mental contexts. As I was drawing pictures, I had a lot of right angles, so that word was in my head. $\endgroup$ Commented Mar 1, 2018 at 20:43

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