1976 USAMO Problems/Problem 2
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.
![[asy] size(300); defaultpen(fontsize(8)); real r=10; picture pica, picb; pair A=r*expi(5*pi/6), B=r*expi(pi/6), X=r*expi(pi/3), X1=r*expi(-pi/12), Y=r*expi(4*pi/3), Y1=r*expi(11*pi/12), O=(0,0), P, P1; P = extension(A,X,B,Y);P1 = extension(A,X1,B,Y1); path circ1 = Circle((0,0),r); draw(pica, circ1);draw(pica, B--A--P--Y--X);dot(pica,P^^O); label(pica,"$A$",A,(-1,1));label(pica,"$B$",B,(1,0));label(pica,"$X$",X,(0,1));label(pica,"$Y$",Y,(0,-1));label(pica,"$P$",P,(1,1));label(pica,"$O$",O,(-1,1));label(pica,"(a)",O+(0,-13),(0,0)); draw(picb, circ1);draw(picb, B--A--X1--Y1--B);dot(picb,P1^^O); label(picb,"$A$",A,(-1,1));label(picb,"$B$",B,(1,1));label(picb,"$X$",X1,(1,-1));label(picb,"$Y$",Y1,(-1,0));label(picb,"$P'$",P1,(-1,-1));label(picb,"$O$",O,(-1,-1)); label(picb,"(b)",O+(0,-13),(0,0)); add(pica); add(shift(30*right)*picb); [/asy]](http://latex.artofproblemsolving.com/b/e/2/be2ee0dc899d7824f7648504e2e34196cd679264.png)
Solution
With loss of generality, we 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 us coordinates for 
in terms of 
and 
:
. We can parametrize these coordinates as follows:
Now since 
then 
Also, 
so 
This equation can be transformed to 
which is the locus of an ellipse.
See also
| 1976 USAMO (Problems • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 1 • 2 • 3 • 4 • 5 | ||
| All USAMO Problems and Solutions | ||
