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Difference between revisions of "1977 Canadian MO Problems/Problem 2"

 
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Let <math>\displaystyle O</math> be the center of a circle and <math>\displaystyle A</math> be a fixed interior point of the circle different from <math>\displaystyle O.</math> Determine all points <math>\displaystyle P</math> on the circumference of the circle such that the angle <math> \displaystyle OPA</math> is a maximum.  
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Let <math>O</math> be the center of a circle and <math>A</math> be a fixed interior point of the circle different from <math>O.</math> Determine all points <math>P</math> on the circumference of the circle such that the angle <math>OPA</math> is a maximum.  
  
 
[[Image:CanadianMO-1977-2.jpg]]
 
[[Image:CanadianMO-1977-2.jpg]]
  
 
== Solution ==
 
== Solution ==
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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>.
  
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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>.
  
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== See Also ==
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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>.
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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>
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==Solution 2==
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Let <math>XY</math> be the chord perpendicular to <math>OA</math> through <math>A</math>. We claim that the choices of <math>P</math> that maximize <math>\angle{OPA}</math> are <math>X</math> and <math>Y</math>.
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Let <math>\omega</math> be the circumcircle of <math>OPA</math>. Then clearly <math>\angle{OPA}</math> is maximized if the radius of <math>\omega</math> is minimized, which occurs if <math>\omega</math> is internally tangent to circle <math>O</math>. Hence <math>OP</math> is a diameter of <math>\omega</math> (whose center <math>OP</math> is collinear with), and so <math>\angle{OAP} = 90^\circ</math>. It follows that <math>P = X</math> or <math>Y</math>, as desired.
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{{Old CanadaMO box|num-b=1|num-a=3|year=1977}}
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[[Category:Olympiad Geometry Problems]]

Latest revision as of 22:28, 21 September 2014

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.

CanadianMO-1977-2.jpg

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 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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