1979 USAMO Problems/Problem 4: Difference between revisions
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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 14:11, 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
![\[PR = \sin a * \frac{r}{\sin(\pi - x - a)} = \sin a * \frac{r}{\sin(x + a)},\]](http://latex.artofproblemsolving.com/7/1/0/71058985a39467316ac4c1b374f4bd049cfe10a4.png)
and using the Law of Sines on triangle OPQ gives
![\[PQ = \sin b * \frac{r}{\sin(x - b)}.\]](http://latex.artofproblemsolving.com/1/c/6/1c6201da87f5fc3ad883710472a1f0cc9614cf1a.png)
Note that 
and 
are given constants.
Hence,
![\[\frac{1}{PR} + \frac{1}{PQ} = \frac{1}{r} (\frac{\sin(x + a)}{\sin a} + \frac{\sin(x - b)}{\sin b}) = \frac{\sin(x+a)\sin b + \sin(x-b)\sin a}{r \sin a \sin b} = \frac{\sin x\cos a\sin b + \sin a\cos x\sin b + \sin x\cos b\sin a - \cos x\sin b\sin a}{r \sin a \sin b} = \frac{\sin x(\sin b \cos a + \sin a \cos b)}{r \sin a \sin b}\]](http://latex.artofproblemsolving.com/5/b/a/5ba92e7ef8c000e3ed97b05ded6d9ca9e9dbf696.png)
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 | ||
These problems are copyrighted © by the Mathematical Association of America. 
