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

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 minimized 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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1977 Canadian MO Problems/Problem 2

Solution

Solution 2