Difference between revisions of "1976 USAMO Problems/Problem 2"
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Now note that at points <math>A</math> and <math>B</math>, this slope expression reduces to <math>y' = \frac{-b}{a}</math> and <math>y' = \frac{b}{a}</math> respectively, values which are identical to the slopes of lines <math>AO</math> and <math>BO</math>. Thus we conclude that the complete locus of intersection points is the circle tangent to lines <math>AO</math> and <math>BO</math> at points <math>A</math> and <math>B</math> respectively. | Now note that at points <math>A</math> and <math>B</math>, this slope expression reduces to <math>y' = \frac{-b}{a}</math> and <math>y' = \frac{b}{a}</math> respectively, values which are identical to the slopes of lines <math>AO</math> and <math>BO</math>. Thus we conclude that the complete locus of intersection points is the circle tangent to lines <math>AO</math> and <math>BO</math> at points <math>A</math> and <math>B</math> respectively. | ||
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==See Also== | ==See Also== |
Revision as of 15:07, 17 September 2012
Problem
If and
are fixed points on a given circle and
is a variable diameter of the same circle, determine the locus of the point of intersection of lines
and
. You may assume that
is not a diameter.
Solution
WLOG, assume that the circle is the unit circle centered at the origin. Then the points and
have coordinates
and
respectively and
and
have coordinates
and
. Note that these coordinates satisfy
and
since these points are on a unit circle. Now we can find equations for the lines:
Solving these simultaneous equations gives coordinates for
in terms of
and
:
. These coordinates can be parametrized in Cartesian variables as follows:
Now solve for
and
to get
and
. Then since
which reduces to
This equation defines a circle and is the locus of all intersection points
. In order to define this locus more generally, find the slope of this circle function using implicit differentiation:
Now note that at points
and
, this slope expression reduces to
and
respectively, values which are identical to the slopes of lines
and
. Thus we conclude that the complete locus of intersection points is the circle tangent to lines
and
at points
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
1976 USAMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |