TITLE: Tracing a circle by a sliding triangle
QUESTION [7 upvotes]: An isosceles triangle with a unit length base is sliding on two lines which make an angle of $60^\circ$ between them.  The third vertex traces a circle centered at the intersection of the two lines.  What is the altitude of the triangle, and what is the radius of the circle ?

What I have tried:

I found the coordinates of points $X$ and $Y$ as shown above on the two lines in terms of the angle $\theta$, then found the coordinates of the tip of the triangle (the third vertex) as a function of $\theta$, and finally found the altitude $h$ that results in the distance of this tip from the origin being constant.  And that constant is the radius of the circle.

REPLY [2 votes]: The solution by @dfnu is indeed simple.
But something even simpler can be given (avoiding in particular two cases).
Let $\varphi$ be the angle between the two "guiding" axes $A_1$ (horizontal) and $A_2$ intersecting in $O$, where $\varphi$ can be any value from $0$ to $\pi/2$, generalizing the case $\varphi=\pi/3$ given in the question.
Let us work backwards. Let us consider a circle with center $O$ ; we can assume WLOG that its radius is $1$.
With complex notation $e^{i\alpha}$ for the current point $C$ of the circle, let us consider the triangle with vertices $A,B,C$ defined in this way:
$$\begin{cases}A&=&e^{i\alpha}+e^{-i\alpha}&=&2\cos(\alpha),\\
B&=&e^{i\alpha}+e^{-i\alpha}e^{2i\varphi}&=&2 \cos(\alpha-\varphi)e^{i\varphi},\\
C&=&e^{i\alpha}&&\end{cases}\tag{1}$$
(use Euler formulas for converting between the two forms for $A$ and $B$).
$A$ belongs clearly to the horizontal axis $A_1$ and $B$ to axis $A_2$. Furthermore $CA=CB$ because $|A-C|=|B-C| \ \iff \ |e^{-i \alpha}|=|e^{i (-\alpha+2 \varphi)}|=1$ proving that $ABC$ is an isosceles triangle.
Some particular cases are worth to be considered: $\varphi=\pi/6,0,\pi/2$.

Fig. 1: $\varphi=\pi/6$ gives equilateral triangles.

Fig. 2: The degenerated case $\varphi=\pi/2$ (orthogonal axes $A_1$ and $A_2$) with flat triangles as well. The envelope of these lines is an astroid. This is the famous problem of the ladder falling down ; see the nice animation here.

Fig. 3: Another degenerated case $\varphi=0$ where axes $A_1$ and $A_2$ are superimposed gives flat triangles. One can observe that it is in fact a part of the previous figure (the envelope is the middle part of the astroid).
Remark: I discovered (thanks to @brainjam) that this is known as the van Schooten's mechanism. A connected presentation can be found as well here.