Difference between revisions of "1979 USAMO Problems/Problem 4"
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= \frac{\sin x(\sin b \cos a + \sin a \cos b)}{r \sin a \sin b}</cmath> | = \frac{\sin x(\sin b \cos a + \sin a \cos b)}{r \sin a \sin b}</cmath> | ||
− | Clearly, this quantity is maximized when <math>\sin x = 1.</math> Because <math>x</math> | + | Clearly, this quantity is maximized when <math>\sin x = 1.</math> Because <math>x</math> must be less than <math>\pi</math>, <math>\frac{1}{PQ} + \frac{1}{PR}</math> is as large as possible when <math>x = \frac{\pi}{2},</math> or when line <math>QR</math> is perpendicular to line <math>PO</math>. |
==See Also== | ==See Also== |
Revision as of 15:11, 19 April 2014
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 (inversions)
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
Solution (trig bash)
PLEASE PROVIDE A FIGURE.
Let and
Then
and
Using the Law of Sines on triangle OPR gives
and using the Law of Sines on triangle OPQ gives
Note that
and
are given constants.
Hence,
Clearly, this quantity is maximized when Because
must be less than
,
is as large as possible when
or when line
is perpendicular to line
.
See Also
1979 USAMO (Problems • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.