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Difference between revisions of "1975 USAMO Problems/Problem 4"
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</asy>
 
</asy>
  
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Solution with graph posted at
 
  
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http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.USA1975Problem4
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==Solution==
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A maximum <math>AP \cdot PB</math> cannot be attained if <math>AB</math> intersects segment <math>O_1O_2</math> because a larger value can be attained by making one of <math>A</math> or <math>B</math> diametrically opposite <math>P</math>, which (as is easily checked) increases the value of both <math>AP</math> and <math>PB</math>. Thus, assume <math>AB</math> does not intersect <math>O_1O_2</math>.
  
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and here:
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Let <math>E</math> and <math>F</math> be the centers of the small and big circles, respectively, and <math>r</math> and <math>R</math> be their respective radii.
  
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Let E and F be the centers of the small and big circles, respectively, and r and R be their respective radii.
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Let <math>M</math> and <math>N</math> be the feet of <math>E</math> and <math>F</math> to <math>AB</math>, and <math>\alpha = \angle APE</math> and <math>\epsilon = \angle BPF</math>
  
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Let M and N be the feet of E and F to AB, and α = ∠APE and ε = ∠BPF
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We have:
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<cmath>AP \times PB = 2r \cos{\alpha} \times 2R \cos{\epsilon} = 4 rR \cos{\alpha} \cos{\epsilon}</cmath>
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<math>AP\times PB</math> is maximum when the product <math>\cos{\alpha} \cos{\epsilon}</math> is a maximum.
  
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We have:
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We have <math>\cos{\alpha} \cos{\epsilon}= \frac{1}{2} [\cos(\alpha +\epsilon) + \cos(\alpha -\epsilon)]</math>
  
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AP × PB = 2r cosα × 2R cosε = 4 rR cosα cosε
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But <math>\alpha +\epsilon = 180^{\circ} - \angle EPF</math> and is fixed, so is <math>\cos(\alpha +\epsilon)</math>.
  
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AP × PB is maximum when the product cosα cosε is a maximum.
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So its maximum depends on <math>cos(\alpha -\epsilon)</math> which occurs when <math>\alpha=\epsilon</math>. To draw the line <math>AB</math>:
  
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We have cosα cosε = ½ [cos(α + ε) + cos(α - ε)]
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Draw a circle with center <math>P</math> and radius <math>PE</math> to cut the radius <math>PF</math> at <math>H</math>. Draw the line parallel to <math>EH</math> passing through <math>P</math>. This line meets the small and big circles at <math>A</math> and <math>B</math>, respectively.
  
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But α + ε = 180° - ∠EPF and is fixed, so is cos(α + ε)
 
  
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So its maximum depends on cos(α - ε) which occurs when α = ε. To draw the line AB:
 
  
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Draw a circle with center P and radius PE to cut the radius PF at H. Draw the line parallel to EH passing through P. This line meets the small and big circles at A and B, respectively.
 
  
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by Vo Duc Dien
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{{alternate solutions}}
  
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==See also==
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==See Also==
 +
[http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.USA1975Problem4 Solution with graph at Cut the Knot]
  
 
{{USAMO box|year=1975|num-b=3|num-a=5}}
 
{{USAMO box|year=1975|num-b=3|num-a=5}}
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{{MAA Notice}}
  
 
[[Category:Olympiad Geometry Problems]]
 
[[Category:Olympiad Geometry Problems]]
 
[[Category:Olympiad Inequality Problems]]
 
[[Category:Olympiad Inequality Problems]]

Latest revision as of 14:13, 13 August 2014

Problem

Two given circles intersect in two points $P$ and $Q$. Show how to construct a segment $AB$ passing through $P$ and terminating on the two circles such that $AP\cdot PB$ is a maximum.

[asy] size(150); defaultpen(fontsize(7)); pair A=(0,0), B=(10,0), P=(4,0), Q=(3.7,-2.5); draw(A--B); draw(circumcircle(A,P,Q)); draw(circumcircle(B,P,Q)); label("A",A,(-1,1));label("P",P,(0,1.5));label("B",B,(1,1));label("Q",Q,(-0.5,-1.5)); [/asy]


Solution

A maximum $AP \cdot PB$ cannot be attained if $AB$ intersects segment $O_1O_2$ because a larger value can be attained by making one of $A$ or $B$ diametrically opposite $P$, which (as is easily checked) increases the value of both $AP$ and $PB$. Thus, assume $AB$ does not intersect $O_1O_2$.

Let $E$ and $F$ be the centers of the small and big circles, respectively, and $r$ and $R$ be their respective radii.

Let $M$ and $N$ be the feet of $E$ and $F$ to $AB$, and $\alpha = \angle APE$ and $\epsilon = \angle BPF$

We have:

\[AP \times PB = 2r \cos{\alpha} \times 2R \cos{\epsilon} = 4 rR \cos{\alpha} \cos{\epsilon}\]

$AP\times PB$ is maximum when the product $\cos{\alpha} \cos{\epsilon}$ is a maximum.

We have $\cos{\alpha} \cos{\epsilon}= \frac{1}{2} [\cos(\alpha +\epsilon) + \cos(\alpha -\epsilon)]$

But $\alpha +\epsilon = 180^{\circ} - \angle EPF$ and is fixed, so is $\cos(\alpha +\epsilon)$.

So its maximum depends on $cos(\alpha -\epsilon)$ which occurs when $\alpha=\epsilon$. To draw the line $AB$:

Draw a circle with center $P$ and radius $PE$ to cut the radius $PF$ at $H$. Draw the line parallel to $EH$ passing through $P$. This line meets the small and big circles at $A$ and $B$, respectively.



Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

See Also

Solution with graph at Cut the Knot

1975 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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