Difference between revisions of "1976 USAMO Problems/Problem 2"
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==Solution== | ==Solution== | ||
− | WLOG, | + | WLOG, 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}\\ | ||
BY \longrightarrow y &= \frac{(s+b)x+rb-sa}{r+a}. | BY \longrightarrow y &= \frac{(s+b)x+rb-sa}{r+a}. | ||
\end{align*} </cmath> | \end{align*} </cmath> | ||
− | Solving these simultaneous equations gives | + | Solving these simultaneous equations gives 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>. | + | <math>P = (\frac{as}{b},1-ar)</math>. These coordinates can be parametrized as follows: |
<cmath> \begin{align*} | <cmath> \begin{align*} | ||
P_x &= \frac{as}{b}\\ | P_x &= \frac{as}{b}\\ |
Revision as of 16:34, 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
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
. Then we can find equations for the lines:
Solving these simultaneous equations gives coordinates for
in terms of
and
:
. These coordinates can be parametrized 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 |