Difference between revisions of "1976 USAMO Problems/Problem 2"
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y &= \frac{1 - ar}{b}. | y &= \frac{1 - ar}{b}. | ||
\end{align*} </cmath> | \end{align*} </cmath> | ||
− | 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>. | + | 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*} | ||
+ | 2x + 2(y-1/b)y' &= 0\ | ||
+ | (y-1/b)y' &= -x | ||
+ | y' &= \frac{-x}{y-1/b}. | ||
+ | \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. | ||
==See also== | ==See also== |
Revision as of 16:55, 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 |