Difference between revisions of "1976 USAMO Problems/Problem 2"
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− | WLOG, assume that the circle is the unit circle centered at the origin. Then the points <math>A</math> and <math>B</math> have coordinates <math>(-a,b)</math> and <math>(a,b)</math> respectively and <math>X</math> and <math>Y</math> have coordinates <math>(r,s)</math> and <math>(-r,-s)</math>. | + | WLOG, assume that the circle is the unit circle centered at the origin. Then the points <math>A</math> and <math>B</math> have coordinates <math>(-a,b)</math> and <math>(a,b)</math> respectively and <math>X</math> and <math>Y</math> have coordinates <math>(r,s)</math> and <math>(-r,-s)</math>. Note that these coordinates satisfy <math>a^2 + b^2 = 1</math> and <math>r^2 + s^2 = 1</math> since these points are on a unit circle. Now we can find equations for the lines: |
<cmath> \begin{align*} | <cmath> \begin{align*} | ||
AX \longrightarrow y &= \frac{(s-b)x+rb+sa}{r+a}\\ | AX \longrightarrow y &= \frac{(s-b)x+rb+sa}{r+a}\\ |
Revision as of 02:00, 27 February 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 solving 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.
See also
1976 USAMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |