<s>[INST] 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.
 [/INST] 
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} } $
</s>