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.
Best Answer
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 $\beta$.