Difference between revisions of "1979 USAMO Problems/Problem 4"
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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> | ||
| − | ==See | + | ==See also== |
{{USAMO box|year=1979|num-b=3|num-a=5}} | {{USAMO box|year=1979|num-b=3|num-a=5}} | ||
Revision as of 23:45, 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 
See also
| 1979 USAMO (Problems • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 1 • 2 • 3 • 4 • 5 | ||
| All USAMO Problems and Solutions | ||
