Difference between revisions of "1975 USAMO Problems/Problem 4"
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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}} | {{USAMO box|year=1975|num-b=3|num-a=5}} | ||
+ | {{MAA Notice}} | ||
[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] | ||
[[Category:Olympiad Inequality Problems]] | [[Category:Olympiad Inequality Problems]] |
Latest revision as of 15:13, 13 August 2014
Problem
Two given circles intersect in two points and
. Show how to construct a segment
passing through
and terminating on the two circles such that
is a maximum.
Solution
A maximum cannot be attained if
intersects segment
because a larger value can be attained by making one of
or
diametrically opposite
, which (as is easily checked) increases the value of both
and
. Thus, assume
does not intersect
.
Let and
be the centers of the small and big circles, respectively, and
and
be their respective radii.
Let and
be the feet of
and
to
, and
and
We have:
is maximum when the product
is a maximum.
We have
But and is fixed, so is
.
So its maximum depends on which occurs when
. To draw the line
:
Draw a circle with center and radius
to cut the radius
at
. Draw the line parallel to
passing through
. This line meets the small and big circles at
and
, 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.