Law of cosines, Ptolemy's, angle chasing on an isosceles triangle inscribed in a circle

From HMMT:

Triangle $$\triangle PQR$$, with $$PQ=PR=5$$ and $$QR=6$$, is inscribed in circle $$\omega$$. Compute the radius of the circle with center on $$QR$$ which is tangent to both $$\omega$$ and $$PQ$$.

I haven't made much progress. I've set $$QS=x$$ and $$SR=y$$ to try for Law of Cosines, since we know $$\cos\angle QSR$$, but that really hasn't lead anywhere. With Ptolemy's, I 've found that $$PS=\displaystyle\frac{5(x+y)}{6}$$, but unfortunately $$PS$$ isn't colinear with anything useful (like the line connecting the centers). I also haven't really been able to use the tangent properties.

Hints beyond what I've done or any useful insights would be greatly appreciated!

Given that $$|PQ|=|PR|=5,\ |QR|=6$$, the area, the height and the circumradius of $$\triangle PQR$$ are $$S=12$$, $$|PF|=4$$ and $$R_0=\tfrac{25}8$$, respectively. Let $$\angle PQR=\alpha$$, $$\angle FOE=\phi$$.

Assuming that the center of the circle $$O_t\in QR$$, we must have $$|DQ_t|=|EQ_t|=r$$.

\begin{align} \sin\alpha&=\frac{|PF|}{|PQ|} =\frac45 ,\\ |OF|&=|PF|-R_0=\tfrac78 \tag{1}\label{1} . \end{align}

We have two conditions for $$r$$, $$\phi$$:

\begin{align} |QO_t|+|FO_t|&=\tfrac12\,|QR| \tag{2}\label{2} ,\\ \frac r{\sin\alpha} + (R_0-r)\sin\phi &=3 \tag{3}\label{3} ,\\ (R_0-r)\cos\phi&=|OF| \tag{4}\label{4} . \end{align}

Excluding $$\phi$$ from \eqref{3},\eqref{4} and using known values, we get

\begin{align} r&=\frac{20}9 . \end{align}

I've done this using coordinate geometry. Let the sides of the triangle be along the sides as shown in the figure. The angle between $$PQ$$ and $$QR$$ is $$53°$$ (approximately) because of the $$(3,4,5)$$ triangle formed by the sides $$PQ,QB,BP$$ where $$B$$ is midpoint of $$QR$$. So, the equation line along $$PQ$$ is $$3y=4x$$.

Now, let the centre and radius of the unknown circle be $$C(h, 0)$$ and $$r$$ respectively. We get two constraints from these,

1. Perpendicular distance from $$C$$ to the line $$PQ$$ is $$r$$.

2. $$\displaystyle AC=\frac{25}{8}-r$$.

You've two equations and two variable and may proceed now

• How do you get that the angle between $PQ$ and $QR$ is $53^{\circ}$?
– mpnm
Jul 20, 2020 at 3:09
• The $(3,4,5)$ Pythagorean triplet has well know trigonometric inverses(they occur a lot in elementary Physics exercises); however, explicitly assuming the angle does not seem fair and I think @SarGe should offer an explanation. Jul 20, 2020 at 4:02
• @David Dong, I've edited the solution. Let me know if there is still any confusion. Jul 20, 2020 at 5:20