Problem — Triple central angle

Text

In a circle with centre $O$ and radius $r$, consider a chord $MN$. Extend the chord beyond $N$ by a segment $NP$ congruent to the radius. Then join $P$ to the centre $O$ and draw the line $PO$; let $Q$ be the point where it meets the circle on the opposite side of the centre from $P$. Prove that the central angle $\widehat{QOM}$ is three times the angle $\widehat{OPN}$.

Solution

Set $\widehat{OPN} = \alpha$. Triangle $ONP$ is isosceles, because $\overline{ON} = \overline{NP} = r$; hence its base angles are congruent: $\widehat{NOP} = \widehat{NPO} = \alpha$. The third angle is then $\widehat{ONP} = 180^\circ - 2\alpha$. Since $M$, $N$, $P$ are collinear, the adjacent angle is $\widehat{ONM} = 180^\circ - \widehat{ONP} = 2\alpha$. Triangle $OMN$ is also isosceles, since $\overline{OM} = \overline{ON} = r$; therefore $\widehat{OMN} = \widehat{ONM} = 2\alpha$. Now $\widehat{QOM}$ is the exterior angle of triangle $OMP$ at the vertex $O$ (the points $P$, $O$, $Q$ are collinear), so it equals the sum of the two non-adjacent interior angles: $\widehat{QOM} = \widehat{OPM} + \widehat{OMP} = \alpha + 2\alpha = 3\alpha.$ Therefore $\widehat{QOM} = 3\,\widehat{OPN}$. $\blacksquare$