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Difference between revisions of "1976 USAMO Problems/Problem 2"
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==Solution==
 
==Solution==
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With loss of generality, we 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>. Then we can find equations for the lines:
+
WLOG, we 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>. Then 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 15:33, 22 February 2012

Problem

If $A$ and $B$ are fixed points on a given circle and $XY$ is a variable diameter of the same circle, determine the locus of the point of intersection of lines $AX$ and $BY$. You may assume that $AB$ 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]

Solution

WLOG, we assume that the circle is the unit circle centered at the origin. Then the points $A$ and $B$ have coordinates $(-a,b)$ and $(a,b)$ respectively and $X$ and $Y$ have coordinates $(r,s)$ and $(-r,-s)$. Then we can find equations for the lines: \begin{align*} AX \longrightarrow y &= \frac{(s-b)x+rb+sa}{r+a}\\ BY \longrightarrow y &= \frac{(s+b)x+rb-sa}{r+a}. \end{align*} Solving these simultaneous equations gives us coordinates for $P$ in terms of $a, b, r,$ and $s$: $P = (\frac{as}{b},1-ar)$. We can parametrize these coordinates as follows: \begin{align*} P_x &= \frac{as}{b}\\ P_y &= 1 - ar. \end{align*} Now since $x = \frac{as}{b},$ then $s = \frac{bx}{a}.$ Also, $r = \sqrt{1 - s^2} = \sqrt{1 - \left(\frac{bx}{a}\right)^2},$ so $y = 1 - a\sqrt{1 - \left(\frac{bx}{a}\right)^2} = 1 - \sqrt{a^2 - b^2 x^2}.$ This equation can be transformed to $\frac{x^2}{(a/b)^2} + \frac{(y-1)^2}{a^2} = 1$ 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
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