$P$ and $Q$ are points on the semicircular arc with diameter $AB$. $R$ is a point on the radius $OB$, such that $O P R \angle = O Q R \angle = 10^{\circ}$. If $P O A \angle = 40^{\circ}$, what is the measure of $Q O B \angle$?
I. Solution. Rotate the triangle $O R Q$ around $O$ so that the image of point $Q$ is $P$. Let the image of $R$ be $R^{\prime}$ (Figure 1). Since $\angle O P R = \angle O Q R$, $R^{\prime}$ lies on the ray $P R$. Due to the rotation, $O R = O R^{\prime}$, so the triangle $O R R^{\prime}$ is isosceles. We know that $\angle P O B = 180^{\circ} - \angle P O A = 140^{\circ}$, thus $\angle O R P = 180^{\circ} - (\angle P O R + \angle O P R) = 30^{\circ}$. In the triangle $O R R^{\prime}$, $\angle R O R^{\prime} = 180^{\circ} - 2 \cdot \angle O R P = 120^{\circ}$. Due to the rotation,

\[
\angle Q O B = \angle P O R^{\prime} = \angle P O B - \angle R^{\prime} O R = 140^{\circ} - 120^{\circ} = 20^{\circ}.
\]

![](https://cdn.mathpix.com/cropped/2024_05_02_4c720f7d3cf22c013b6cg-1.jpg?height=323&width=548&top_left_y=462&top_left_x=775)

Figure 1

János Barát (Szeged, Radnóti M. Gymnasium, 2nd grade)

II. Solution. Since $\angle O P R = \angle O Q R$ and points $P$ and $Q$ lie on the same side of the line $O R$, $P$ and $Q$ lie on one of the $10^{\circ}$-degree circular arcs of segment $O R$, so the quadrilateral $O P Q R$ is a cyclic quadrilateral (Figure 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_4c720f7d3cf22c013b6cg-1.jpg?height=377&width=569&top_left_y=1091&top_left_x=759)

Thus, $\angle P Q R = 180^{\circ} - \angle P O R = \angle P O A = 40^{\circ}$, from which $\angle P Q O = \angle P Q R - \angle O Q R = 40^{\circ} - 10^{\circ} = 30^{\circ}$. Since $P$ and $Q$ are points on a semicircle with center $O$, $O P = O Q$, so the triangle $O P Q$ is isosceles. Therefore, $\angle Q P O = \angle P Q O = 30^{\circ}$. By the inscribed angle theorem, $\angle Q O B = \angle Q O R = \angle Q P R = \angle Q P O - \angle R P O = 30^{\circ} - 10^{\circ} = 20^{\circ}$.