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.



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, $\angle OPX=\angle OAX=\angle RPQ$.

Ergo, the points $P$ such that $\angle OPA$ is maximized are none other than points $A$ and $B$. $\Box$

Solution 2

Let $XY$ be the chord perpendicular to $OA$ through $A$. We claim that the choices of $P$ that maximize $\angle{OPA}$ are $X$ and $Y$.

Let $\omega$ be the circumcircle of $OPA$. Then clearly $\angle{OPA}$ is maximized if the radius of $\omega$ is minimized, which occurs if $\omega$ is internally tangent to circle $O$. Hence $OP$ is a diameter of $\omega$ (whose center $OP$ is collinear with), and so $\angle{OAP} = 90^\circ$. It follows that $P = X$ or $Y$, as desired.

1977 Canadian MO (Problems)
Preceded by
Problem 1
1 2 3 4 5 6 7 8 Followed by
Problem 3
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