1977 Canadian MO Problems/Problem 2

Let $O$ be the center of a circle and $A$ be a fixed interior point of the circle different from $O.$ Determine all points $P$ on the circumference of the circle such that the angle $OPA$ is a maximum.

Solution

If $AB$ is the chord perpendicular to $OX$ through point $P$ , then extend $AO$ to meet the circle at point $C$ . It is now evident that $O$ is the midpoint of $AC$ , $X$ is the midpoint of $AB$ , and hence $OX=\dfrac{BC}{2}$ .

Similarly, let $P$ be a point on arc $AB$ . Extend $PO$ to meet the circle at point $R$ . Extend $PX$ to meet the circle a second time at $Q$ .

We now plot $S$ on $XQ$ such that $XS=XP$ . Then, $OX=\dfrac{RS}{2}$ . Since $\angle RQS=90$ , $RS>RQ$ . Hence, $RQ<\dfrac{OX}{2}$ , and therefore, $\angle OPX=\angle OAX=\angle RPQ$ .