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.


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} } $
