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 16:32, 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 |