AoPSWiki

Want to learn how to tackle those tough MATHCOUNTS and AMC counting and probability problems? Check out Art of Problem Solving's Introduction to Counting & Probability by David Patrick.

Personal tools

Search

Toolbox

1977 Canadian MO Problems/Problem 2

From AoPSWiki

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 $B$ . $\Box$

1977 Canadian MO (Problems)
Preceded by Problem 1	1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10	Followed by Problem 3

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/1977_Canadian_MO_Problems/Problem_2"

Category: Olympiad Geometry Problems

Try our innovative online adaptive learning system, Alcumus.
Over 1100 problems and 60+ video lessons. FREE!

© Copyright 2008 AoPS Incorporated. All Rights Reserved. • Foundation • Privacy • Contact Us