Tracing a circle by a sliding triangle 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.
 A: A very fast synthetic approach
As in your notation, let $\phi$ and $\pi -\phi$ be the angles between the straight lines intersecting in $O$.
We want to construct an isosceles triangle with sides $BC \cong AC$, and vertex $C$ on the circle centered in $O$ having radius $\overline{BC}$.
It is sufficient then to analyse the two cases shown below.
Case 1
Let $\measuredangle AOC = \alpha$. Since $\triangle AOC$ is isosceles we have
$\measuredangle ACO = \pi -2\alpha$. Similarly, $\triangle BOC$ is isosceles, and $\measuredangle BOC = \pi - \phi -\alpha$, so that $\measuredangle BCO = \pi - 2(\pi-\phi-\alpha) = 2\phi + 2\alpha - \pi$. Therefore $\measuredangle ACB = 2\phi$.
Case 2
Let now $\measuredangle A'OC' = \beta$. Since $\triangle A'OC'$ is isosceles we have
$\measuredangle A'C'O = \pi -2\beta$. Similarly, $\triangle B'OC'$ is isosceles, and $\measuredangle B'OC' = \phi -\beta$, so that $\measuredangle B'C'O = \pi - 2\phi +2 \beta$. Thus $\measuredangle A'C'B' = 2\pi - (\pi -2\beta) - (\pi - 2\phi +2\beta) = 2\phi$.
Your thesis follows immediately, since the previous results do not depend on the choice of either $\alpha$ or $\beta$.

A: 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.
A: 
Let one line be horizontal and the other making an angle of $\phi$ with it.
I'll take the origin of the coordinate system at their intersection.  Placing the triangle at a general orientation with its base making an angle of $\theta$ with the horizontal as shown above, we can write the following equation from the law of sines
$ \dfrac{y}{\sin \theta } = \dfrac{1}{\sin \phi} $
Hence, $ y = \dfrac{\sin \theta}{\sin \phi} $
Since the unit vector pointing from the origin to $Y$ is $(-\cos \phi, -\sin \phi) $, then point $Y$ has the following coordinates
$Y = (- \sin \theta \cot \phi, - \sin \theta ) $
Now, the unit vector along $\vec{YX}$ is $(-cos \theta, \sin \theta)$, therefore,
$X = Y + (1) (-\cos \theta, \sin \theta) = (- \sin \theta \cot \phi - \cos \theta, 0 ) $
The midpoint of the base has coordinates
$ M = \frac{1}{2} (X + Y) = Y + \frac{1}{2} ( - \cos \theta, \sin \theta ) $
Hence,
$M = (- \sin \theta \cot \phi - \frac{1}{2}\cos \theta, - \frac{1}{2} \sin \theta ) $
Finally the tip of the triangle $T$ is a distance $h$ in the direction $(\sin \theta, \cos \theta) $, thus its coordinates are
$T = ( \sin \theta (h - \cot \phi) - \frac{1}{2} \cos \theta, -\frac{1}{2} \sin \theta + h \cos \theta ) $
If $T$ traces a circle centered at the origin, then the sum of squares of its $x$ and $y$ coordinates must be constant, hence
$ T_x^2 + T_y^2 = \sin^2 \theta ( (h - \cot \phi)^2 + \frac{1}{4} ) + \cos^2 \theta ( \frac{1}{4} + h^2 ) - \sin \theta \cos \theta ( 2 h - \cot \phi ) $
For this to be constant, we must have
$ (h - \cot \phi)^2 = h^2 + \frac{1}{4} $
and
$ 2 h - \cot \phi = 0 $
We can see that the two equations are identical and have the solution
$ h = \frac{1}{2} \cot \phi $
And the radius of the circle is $\sqrt{ \frac{1}{4} + h^2 } = \dfrac{1}{2 \sin \phi } $
For $\phi = \dfrac{\pi}{3} $, we get
$ h = \dfrac{1}{2 \sqrt{3}} $ and $R = \dfrac{1}{\sqrt{3} } $
