Suppose we have acute triangle $ABC$ and a single arbitrary point $D$ that lies on $AB$. We then repeat the following process:
- Find point $E$ on $BC$ such that $DE$ and $BC$ are perpendicular.
- Find point $F$ on $CA$ such that $EF$ and $CA$ are perpendicular.
- Find point $G$ on $AB$ such that $FG$ and $AB$ are perpendicular.
- Ad infinitum...
That is, we repeatedly draw lines from our current point that are perpendicular to the next edge (always going either clockwise or counterclockwise), giving our new point on that edge.
My questions are:
This process seems to always converge to a stable fixed cycle of three points forming a smaller similar triangle inside the original one. Is this true?
If yes, is there a name for this smaller inner triangle, or another construction, or any other interesting known facts?
Yes, it is true.
First note that such a smaller triangle exists.
In fact, if we consider the line passing through $A$ and perpendicular to $AB$, the line passing through $B$ and perpendicular to $BC$, the line passing through $C$ and perpendicular to $CA$, then these three lines form a larger triangle, which is easily seen to be similar to the original triangle $ABC$.
Now "shrinking" the big triangle to $ABC$, and the image of the triangle $ABC$ will be the small triangle that we look for.
To prove convergence:
We denote by $XYZ$ the small triangle constructed above, with $X$ on $AB$, $Y$ on $BC$, $Z$ on $CA$.
Let $S$ be the union of the three line segments $AB \cup BC \cup CA$. Define a map $\phi$ from $S$ to itself, which maps any point $x \in AB$ to the point $y\in BC$ such that $xy$ is perpendicular to $BC$, and similarly defined for points on $BC$ and $CA$. By convention, we view the vertex $A$ as belonging to $AB$ but not $BC$, and similarly for $B$ and $C$.
The properties of $\phi$ that we will use are:
We now start with a point $X_0$ on $AB$, and define $Y_0 = \phi(X_0)$, $Z_0 = \phi(Y_0)$, $X_1 = \phi(Z_0)$, $Y_1 = \phi(X_1)$, ..., i.e. $X_i = \phi^{3i}(X_0) \in AB$, $Y_i = \phi^{3i + 1}(X_0) \in BC$, $Z_i = \phi^{3i + 2}(X_0) \in CA$.
The aim is to show that the sequence of points $(X_n)_{n \geq 0}$ converges to the point $X$, and similarly for $Y$ and $Z$.
Without loss of generality, we assume that $X_0 \neq X$ lies between $A$ and $X$, written as $X_0\in AX$ for short. Then it is easy to see: $Y_0\in CY$, $Z_0\in CZ$, $X_1\in BX$, $Y_1 \in BY$, $Z_1 \in AZ$, $X_2 \in AX$, etc.
In short, we have $X_{2n} \in AX$ and $X_{2n + 1} \in BX$ for all $n \geq 0$, and similarly for $Y$ and $Z$. We also have $|X_nX| > |Y_nY| > |Z_nZ| > |X_{n + 1}X|$ for all $n \geq 0$, by the injectivity and the second property of $\phi$.
It follows that the sequence of points $(X_{2n})_{n \geq 0}$ monotonically approachs $X$ from the side of $A$. Since every bounded monotone sequence of real numbers converge, we see that the sequence $(X_{2n})_{n \geq 0}$ converges to a point $X' \in AB$.
I claim that $X'$ must be the point $X$.
Suppose $X' \neq X$, so that $|X'X| > 0$. Since $X'$ is the limit $\lim\limits_{n \rightarrow \infty}X_{2n}$, by continuity of $\phi$, we have $\phi^6(X') = \phi^6(\lim\limits_{n\rightarrow\infty}X_{2n}) =\lim\limits_{n \rightarrow\infty}\phi^6(X_{2n}) = \lim\limits_{n\rightarrow\infty} X_{2n + 2} = X'$.
But by the second property of $\phi$, we have $|X'X| > |\phi(X')Y| > |\phi^2(X')Z| > |\phi^3(X')X| > |\phi^4(X')Y| > |\phi^5(X')Z| > |\phi^6(X')X| = |X'X|$, a contradiction.
Thus we have shown that $\lim\limits_{n\rightarrow \infty}X_{2n} = X$. By continuity of $\phi$, we have $X = \phi^3(X) = \phi^3(\lim\limits_{n \rightarrow\infty}X_{2n}) = \lim\limits_{n\rightarrow\infty}\phi^3(X_{2n}) = \lim\limits_{n\rightarrow\infty}X_{2n + 1}$.
Combining the two limits, we obtain $\lim\limits_{n\rightarrow\infty}X_n = X$.
Applying $\phi$ twice, we get the corresponding limits for $Y$ and $Z$.
This in particular shows that the small triangle $XYZ$ with the property $\phi(X) = Y$, $\phi(Y) = Z$, $\phi(Z) = X$ is unique, as it must be the limit of any starting point $X_0$.
Unfortunately I don't know the answer to your second question. But I guess there is a name and previous study of this triangle, which others may know.