Difference between revisions of "1976 USAMO Problems/Problem 2"
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<cmath> \begin{align*} | <cmath> \begin{align*} | ||
2x + 2(y-1/b)y' &= 0\ | 2x + 2(y-1/b)y' &= 0\ | ||
− | (y-1/b)y' &= -x | + | (y-1/b)y' &= -x\ |
y' &= \frac{-x}{y-1/b}. | y' &= \frac{-x}{y-1/b}. | ||
\end{align*} </cmath> | \end{align*} </cmath> |
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 |