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Difference between revisions of "1979 USAMO Problems/Problem 4"

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==Problem==
 
==Problem==
  
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<math>P</math> lies between the rays <math>OA</math> and <math>OB</math>. Find <math>Q</math> on <math>OA</math> and <math>R</math> on <math>OB</math> collinear with <math>P</math> so that <math>\frac{1}{PQ}\plus{} \frac{1}{PR}</math> is as large as possible.[/
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<math>P</math> lies between the rays <math>OA</math> and <math>OB</math>. Find <math>Q</math> on <math>OA</math> and <math>R</math> on <math>OB</math> collinear with <math>P</math> so that <math>\frac{1}{PQ}\plus{} \frac{1}{PR}</math> is as large as possible.
  
 
==Solution==
 
==Solution==
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Thus, it suffices to find the line through <math>P</math> that maximizes the length of the  segment <math>\overline{Q'R'}.</math> If <math>M,N</math> are the midpoints of <math>PQ',PR',</math> i.e. the projections of <math>O_1,O_2</math> onto <math>QR,</math> then from the right trapezoid <math>O_1O_2NM,</math> we deduce that <math>O_1O_2 \ge MN = \frac{_1}{^2}Q'R'.</math> Consequently, <math>2 \cdot O_1O_2</math> is the greatest possible length of <math>Q'R',</math> which obviously occurs when <math>O_1O_2NM</math> is a rectangle. Hence, <math>Q,R</math> are the intersections of <math>OA,OB</math> with the perpendicular to <math>PO</math> at <math>P.</math>
 
Thus, it suffices to find the line through <math>P</math> that maximizes the length of the  segment <math>\overline{Q'R'}.</math> If <math>M,N</math> are the midpoints of <math>PQ',PR',</math> i.e. the projections of <math>O_1,O_2</math> onto <math>QR,</math> then from the right trapezoid <math>O_1O_2NM,</math> we deduce that <math>O_1O_2 \ge MN = \frac{_1}{^2}Q'R'.</math> Consequently, <math>2 \cdot O_1O_2</math> is the greatest possible length of <math>Q'R',</math> which obviously occurs when <math>O_1O_2NM</math> is a rectangle. Hence, <math>Q,R</math> are the intersections of <math>OA,OB</math> with the perpendicular to <math>PO</math> at <math>P.</math>
  
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==See also==
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==See Also==
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{{USAMO box|year=1979|num-b=3|num-a=5}}
  
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{{USAMO box|year=1979|num-b=3|num-a=5}}
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[[Category:Olympiad Geometry Problems]]

Revision as of 09:59, 5 October 2012

Problem

$P$ lies between the rays $OA$ and $OB$. Find $Q$ on $OA$ and $R$ on $OB$ collinear with $P$ so that $\frac{1}{PQ}\plus{} \frac{1}{PR}$ (Error compiling LaTeX. Unknown error_msg) is as large as possible.

Solution

Perform the inversion with center $P$ and radius $\overline{PO}.$ Lines $OA,OB$ go to the circles $(O_1),(O_2)$ passing through $P,O$ and the line $QR$ cuts $(O_1),(O_2)$ again at the inverses $Q',R'$ of $Q,R.$ Hence

$\frac{1}{PQ}+\frac{1}{PR}=\frac{PQ'+PR'}{PO^2}=\frac{Q'R'}{PO^2}$

Thus, it suffices to find the line through $P$ that maximizes the length of the segment $\overline{Q'R'}.$ If $M,N$ are the midpoints of $PQ',PR',$ i.e. the projections of $O_1,O_2$ onto $QR,$ then from the right trapezoid $O_1O_2NM,$ we deduce that $O_1O_2 \ge MN = \frac{_1}{^2}Q'R'.$ Consequently, $2 \cdot O_1O_2$ is the greatest possible length of $Q'R',$ which obviously occurs when $O_1O_2NM$ is a rectangle. Hence, $Q,R$ are the intersections of $OA,OB$ with the perpendicular to $PO$ at $P.$

See Also

1979 USAMO (Problems • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions
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