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 01:00, 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 |