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Difference between revisions of "1979 USAMO Problems/Problem 4"

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==Problem==
 
==Problem==
  
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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.
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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} + \frac{1}{PR}</math> is as large as possible.
  
−
==Solution==
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== Hint ==
 +
 
 +
There are two ways to solve this problem. The first is more subtle, and the second is just bashing.
 +
 
 +
==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) ==
 +
<asy>
 +
pair O = (0,0), A = (14,28), Q = (20,40), B = (16,0), R = (25,0), P = (23,16);
 +
dot(O); dot(A); dot(Q); dot(B); dot(R); dot(P);
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label("O", O, S);
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label("A", A, W);
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label("Q", Q, W);
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label("B", B, S);
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label("R", R, S);
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label("P", P, E);
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draw(O--R--Q--O); draw(O--P);
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label("r", O--P, N);
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</asy>
 +
 +
Let <math>r = OP, x = \angle OPR, a = \angle POR,</math> and <math>b = \angle POQ.</math> Then <math>\angle ORP = \pi - x - a</math> and <math>\angle OQP = x - b.</math> Using the Law of Sines on <math>\triangle OPR</math> gives
 +
<cmath>PR = \sin a * \frac{r}{\sin(\pi - x - a)} = \sin a * \frac{r}{\sin(x + a)},</cmath>
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and using the Law of Sines on <math>\triangle OPQ</math> gives
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<cmath>PQ = \sin b * \frac{r}{\sin(x - b)}.</cmath>
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Note that <math>r, a,</math> and <math>b</math> are given constants.
 +
Hence,
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<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> 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==
 
{{USAMO box|year=1979|num-b=3|num-a=5}}
 
{{USAMO box|year=1979|num-b=3|num-a=5}}
 +
{{MAA Notice}}
  
 
[[Category:Olympiad Geometry Problems]]
 
[[Category:Olympiad Geometry Problems]]

Latest revision as of 12:24, 11 March 2018

Problem

$P$ lies between the rays $OA$ and $OB$. Find $Q$ on $OA$ and $R$ on $OB$ collinear with $P$ so that $\frac{1}{PQ} + \frac{1}{PR}$ is as large as possible.

Hint

There are two ways to solve this problem. The first is more subtle, and the second is just bashing.

Solution (inversions)

Perform the inversion with center $P$ and radius $\overline{PO}.$ Lines $OA,OB$ go to the circles $(O_1),(O_2)$ passing through $P,O$ and the line $QR$ cuts $(O_1),(O_2)$ again at the inverses $Q',R'$ of $Q,R.$ Hence

$\frac{1}{PQ}+\frac{1}{PR}=\frac{PQ'+PR'}{PO^2}=\frac{Q'R'}{PO^2}$

Thus, it suffices to find the line through $P$ that maximizes the length of the segment $\overline{Q'R'}.$ If $M,N$ are the midpoints of $PQ',PR',$ i.e. the projections of $O_1,O_2$ onto $QR,$ then from the right trapezoid $O_1O_2NM,$ we deduce that $O_1O_2 \ge MN = \frac{_1}{^2}Q'R'.$ Consequently, $2 \cdot O_1O_2$ is the greatest possible length of $Q'R',$ which obviously occurs when $O_1O_2NM$ is a rectangle. Hence, $Q,R$ are the intersections of $OA,OB$ with the perpendicular to $PO$ at $P.$

Solution (trig bash)

[asy] pair O = (0,0), A = (14,28), Q = (20,40), B = (16,0), R = (25,0), P = (23,16); dot(O); dot(A); dot(Q); dot(B); dot(R); dot(P); label("O", O, S); label("A", A, W); label("Q", Q, W); label("B", B, S); label("R", R, S); label("P", P, E); draw(O--R--Q--O); draw(O--P); label("r", O--P, N); [/asy]

Let $r = OP, x = \angle OPR, a = \angle POR,$ and $b = \angle POQ.$ Then $\angle ORP = \pi - x - a$ and $\angle OQP = x - b.$ Using the Law of Sines on $\triangle OPR$ gives \[PR = \sin a * \frac{r}{\sin(\pi - x - a)} = \sin a * \frac{r}{\sin(x + a)},\] and using the Law of Sines on $\triangle OPQ$ gives \[PQ = \sin b * \frac{r}{\sin(x - b)}.\] Note that $r, a,$ and $b$ are given constants. Hence, \[\frac{1}{PR} + \frac{1}{PQ} = \frac{1}{r} (\frac{\sin(x + a)}{\sin a} + \frac{\sin(x - b)}{\sin b}) = \frac{\sin(x+a)\sin b + \sin(x-b)\sin a}{r \sin a \sin b} = \frac{\sin x\cos a\sin b + \sin a\cos x\sin b + \sin x\cos b\sin a - \cos x\sin b\sin a}{r \sin a \sin b} = \frac{\sin x(\sin b \cos a + \sin a \cos b)}{r \sin a \sin b}\]

Clearly, this quantity is maximized when $\sin x = 1.$ Because $x$ must be less than $\pi$, $\frac{1}{PQ} + \frac{1}{PR}$ is as large as possible when $x = \frac{\pi}{2},$ or when line $QR$ is perpendicular to line $PO$.

See Also

1979 USAMO (Problems • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions

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