Shortlist JBMO, 2007

Geometry Difficulty 6.3 National Olympiad Prove it JBMO

Problem:
Let $S$ be a point inside $\text{p O q}$ , and let $k$ be a circle which contains $S$ and touches the legs $O p$ and $O q$ in points $P$ and $Q$ respectively. Straight line $s$ parallel to $O p$ from $S$ intersects $O q$ in a point $R$ . Let $T$ be the point of intersection of the ray $P S$ and circumscribed circle of $\triangle S Q R$ and $T \neq S$ . Prove that $O T \| S Q$ and $O T$ is a tangent of the circumscribed circle of $\triangle S Q R$ .

Solution

Solution:
Let $\text{O P S= 1}$ and $\text{O Q S= 2}$ . We have that $\text{O P S= P Q S= 1}$ and $\text{O Q S= Q P S= 2}$ (tangents to circle $k$ ).

Because $R S \| O P$ we have $\text{O P S= R S T= 1}$ and $\text{R Q T= R S T= 1}$ (cyclic quadrilateral $R S Q T$ ). So, we have as follows $\text{O P T= 1 = R Q T= O Q T}$ , which implies that the quadrilateral $O P Q T$ is cyclic. From that we directly obtain $\text{Q O T= Q P T= 2 = O Q S}$ , so $O T \| S Q$ .

From the cyclic quadrilateral $O P Q T$ by easy calculation we get
$\text{O T R= O T P- R T S= O Q P- R Q S= ( 1 + 2 )- 2 = 1 = R Q T}$
Thus, $O T$ is a tangent to the circumscribed circle of $\triangle S Q R$ .

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.