Difference between revisions of "1976 USAMO Problems/Problem 2"
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Now solving for <math>r</math> and <math>s</math> to get <math>r = \frac{1-by}{a}</math> and <math>s = \frac{bx}{a}</math> . Then since <math>r^s + s^2 = 1, \left(\frac{bx}{a}\right)^2 + \left(\frac{1-by}{a}\right)^2 = 1</math> which reduces to <math>x^2 + (y-1/b)^2 = \frac{a^2}{b^2}.</math> This equation defines a circle and is the locus of all intersection points <math>P</math>. In order to define this locus more generally, find the slope of this circle function using implicit differentiation: | Now solving for <math>r</math> and <math>s</math> to get <math>r = \frac{1-by}{a}</math> and <math>s = \frac{bx}{a}</math> . Then since <math>r^s + s^2 = 1, \left(\frac{bx}{a}\right)^2 + \left(\frac{1-by}{a}\right)^2 = 1</math> which reduces to <math>x^2 + (y-1/b)^2 = \frac{a^2}{b^2}.</math> This equation defines a circle and is the locus of all intersection points <math>P</math>. In order to define this locus more generally, find the slope of this circle function using implicit differentiation: | ||
<cmath> \begin{align*} | <cmath> \begin{align*} | ||
− | 2x + 2(y-1/b)y' | + | 2x + 2(y-1/b)y' = 0\\ |
− | (y-1/b)y' | + | (y-1/b)y' = -x\\ |
− | y' | + | y' = \frac{-x}{y-1/b}. |
\end{align*} </cmath> | \end{align*} </cmath> | ||
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, 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, 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. |
Revision as of 17:55, 22 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
. Then 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, 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 |