Art of Problem Solving
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus
During AMC 10A/12A testing, the AoPS Wiki is in read-only mode and no edits can be made.

1977 Canadian MO Problems/Problem 2: Difference between revisions

← Older editNewer edit →
Temperal (talk | contribs)
3,154 edits
box
H34N1 (talk | contribs)
86 edits
No edit summary
Line 4: Line 4:


== Solution ==
== Solution ==
{{solution}}
If <math>AB</math> is the chord perpendicular to <math>OX</math> through point <math>P</math>, then extend <math>AO</math> to meet the circle at point <math>C</math>. It is now evident that <math>O</math> is the midpoint of <math>AC</math>, <math>X</math> is the midpoint of <math>AB</math>, and hence <math>OX=\dfrac{BC}{2}</math>.
 
Similarly, let <math>P</math> be a point on arc <math>AB</math>. Extend <math>PO</math> to meet the circle at point <math>R</math>. Extend <math>PX</math> to meet the circle a second time at <math>Q</math>.
 
We now plot <math>S</math> on <math>XQ</math> such that <math>XS=XP</math>. Then, <math>OX=\dfrac{RS}{2}</math>. Since <math>\angle RQS=90</math>, <math>RS>RQ</math>. Hence, <math>RQ<\dfrac{OX}{2}</math>, and therefore, <math>\angle OPX=\angle OAX=\angle RPQ</math>.
 
Ergo, the points <math>P</math> such that <math>\angle OPA</math> is maximized are none other than points <math>A</math> and <math>B</math>. <math>\Box</math>


{{Old CanadaMO box|num-b=1|num-a=3|year=1977}}
{{Old CanadaMO box|num-b=1|num-a=3|year=1977}}

Revision as of 18:55, 11 April 2008

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.

Error creating thumbnail: Unable to save thumbnail to destination

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 • Followed by
Problem 3
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1977_Canadian_MO_Problems/Problem_2&oldid=24669"