Difference between revisions of "1975 USAMO Problems/Problem 4"
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+ | ==Solution== | ||
by Vo Duc Dien | by Vo Duc Dien | ||
− | Let E and F be the centers of the small and big circles, respectively, and r and R 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. |
− | Let M and N be the feet of E and F to AB, and | + | 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> |
We have: | We have: | ||
− | AP | + | <cmath>AP \times PB = 2r \cos{\alpha} \times 2R \cos{\epsilon} = 4 rR \cos{\alpha} \cos{\epsilon}</cmath> |
− | |||
− | |||
− | + | <math>AP\times PB</math> is maximum when the product <math>\cos{\alpha} \cos{\epsilon}</math> is a maximum. | |
− | + | We have <math>\cos{\alpha} \cos{\epsilon}= \frac{1}{2} [\cos(\alpha +\epsilon) + \cos(\alpha -\epsilon)]</math> | |
− | + | But <math>\alpha +\epsilon = 180^{\circ} - \angle EPF</math> and is fixed, so is <math>\cos(\alpha +\epsilon)</math>. | |
− | + | So its maximum depends on <math>cos(\alpha -\epsilon)</math> which occurs when <math>\alpha=\epsilon</math>. To draw the line <math>AB</math>: | |
− | + | 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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==See also== | ==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}} |
Revision as of 09:42, 13 May 2011
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
by Vo Duc Dien
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.
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 |