Difference between revisions of "1979 USAMO Problems/Problem 4"
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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. | <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 (inversions) == |
Perform the inversion with center <math>P</math> and radius <math>\overline{PO}.</math> Lines <math>OA,OB</math> go to the circles <math>(O_1),(O_2)</math> passing through <math>P,O</math> and the line <math>QR</math> cuts <math>(O_1),(O_2)</math> again at the inverses <math>Q',R'</math> of <math>Q,R.</math> Hence | Perform the inversion with center <math>P</math> and radius <math>\overline{PO}.</math> Lines <math>OA,OB</math> go to the circles <math>(O_1),(O_2)</math> passing through <math>P,O</math> and the line <math>QR</math> cuts <math>(O_1),(O_2)</math> again at the inverses <math>Q',R'</math> of <math>Q,R.</math> Hence | ||
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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> | ||
+ | |||
+ | ==Solution (trig bash) == | ||
+ | PLEASE PROVIDE A FIGURE. | ||
+ | |||
+ | Let <math>r = OP, x = <OPR, a = <POR,</math> and <math>b = <POQ.</math> Then <math><ORP = \pi - x - a</math> and <math><OQP = x - b.</math> Using the Law of Sines on triangle OPR gives | ||
+ | <cmath>PR = \sin a * \frac{r}{\sin(\pi - x - a)} = \sin a * \frac{r}{\sin(x + a)},</cmath> | ||
+ | and using the Law of Sines on triangle OPQ gives | ||
+ | <cmath>PQ = \sin b * \frac{r}{\sin(x - b)}.</cmath> | ||
+ | Note that <math>r, a,</math> and <math>b</math> are given constants. | ||
+ | Hence, | ||
+ | <cmath>\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}</cmath> | ||
+ | |||
+ | Clearly, this quantity is maximized when <math>\sin x = 1.</math> Because <math>x</math> clearly 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 10:47, 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
clearly 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.