Difference between revisions of "1979 USAMO Problems/Problem 4"
Mrdavid445 (talk | contribs) (Created page with "==Problem== <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...") |
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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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Revision as of 23:43, 11 April 2012
Problem
lies between the rays and . Find on and on collinear with 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 and radius Lines go to the circles passing through and the line cuts again at the inverses of Hence
Thus, it suffices to find the line through that maximizes the length of the segment If are the midpoints of i.e. the projections of onto then from the right trapezoid we deduce that Consequently, is the greatest possible length of which obviously occurs when is a rectangle. Hence, are the intersections of with the perpendicular to at
1979 USAMO (Problems • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |