Difference between revisions of "1976 USAMO Problems/Problem 2"
(New page: ==Problem== If <math>A</math> and <math>B</math> are fixed points on a given circle and <math>XY</math> is a variable diameter of the same circle, determine the locus of the point of inter...) |
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==Solution== | ==Solution== | ||
− | {{ | + | 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: |
+ | <cmath> \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*} </cmath> | ||
+ | Solving these simultaneous equations gives us coordinates for <math>P</math> in terms of <math>a, b, r,</math> and <math>s</math>: | ||
+ | <math>P = (\frac{as}{b},1-ar)</math>. We can parametrize these coordinates as follows: | ||
+ | <cmath> \begin{align*} | ||
+ | P_x &= \frac{as}{b}\ | ||
+ | P_y &= 1 - ar. | ||
+ | \end{align*} </cmath> | ||
+ | Now since <math>x = \frac{as}{b},</math> then <math>s = \frac{bx}{a}.</math> Also, <math>r = \sqrt{1 - s^2} = \sqrt{1 - \left(\frac{bx}{a}\right)^2},</math> so <math>y = 1 - a\sqrt{1 - \left(\frac{bx}{a}\right)^2} = 1 - \sqrt{a^2 - b^2 x^2}.</math> | ||
+ | This equation can be transformed to <math>\frac{x^2}{(a/b)^2} + \frac{(y-1)^2}{a^2} = 1</math> which is the locus of an ellipse. | ||
==See also== | ==See also== |
Revision as of 15:32, 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
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 |