[ In similar auxiliary triangles ] [ Angle between tangent and chord $\quad$]

In triangle $P Q R$, point $T$ lies on side $P R$, $\angle Q T R = \angle P Q R$, $P T = 8$, $T R = 1$.

Find a) side $Q R$; b) angle $Q R P$, if the radius of the circumscribed circle of triangle $P Q T$ is $3 \sqrt{3}$.
Triangles $Q T R$ and $P Q R$ are similar, and the line $R Q$ is a tangent to the circumcircle of triangle $P Q T$.

## Solution

Triangles $Q T R$ and $P Q R$ are similar by two angles, so $Q R: P R = T R: Q R$, from which $Q R^{2} = P R \cdot T R = 9$. Since $\angle T Q R = \angle Q P R$, by the converse of the theorem about the angle between a tangent and a chord, the line $R Q$ is a tangent to the circumcircle of triangle $P Q T$. Therefore, $O Q \perp R Q$ (where $O$ is the center of the circle). $\operatorname{tg} \angle O R Q = O Q / Q R = \sqrt{3}$, so $\angle O R Q = \pi / 3$ and $O R = 2 Q R = 6$.

Let $S$ be the foot of the perpendicular dropped from the center of the circle to the chord $P T$. Then $S$ is the midpoint of $P T$. From the right triangle $O S R$, we find that $\cos \angle O R S = R S / O R = 5 / 6$. Therefore, $\angle Q R P = \pi / 3 \pm \arccos 5 / 6$ (points $P$ and $Q$ can lie either on the same side or on opposite sides of the line $O R$).

![](https://cdn.mathpix.com/cropped/2024_05_06_4c124ad32a76c9faf91dg-13.jpg?height=803&width=812&top_left_y=1188&top_left_x=629)

## Answer

a) 3 ; b) $\pi / 3 \pm \arccos 5 / 6$.