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Difference between revisions of "1975 USAMO Problems/Problem 4"

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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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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.
 
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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{{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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