Difference between revisions of "1976 USAMO Problems/Problem 2"
m (→Solution) |
m (→Solution) |
||
| Line 33: | Line 33: | ||
y &= \frac{1 - ar}{b}. | y &= \frac{1 - ar}{b}. | ||
\end{align*} </cmath> | \end{align*} </cmath> | ||
| − | Now solve 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^ | + | Now solve 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^2 + 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' &= 0\\ | 2x + 2(y-1/b)y' &= 0\\ | ||
Revision as of 01:01, 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.
![[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
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 solve 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 | ||
