Difference between revisions of "1992 IMO Problems/Problem 4"
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Now we solve for the coordinates for point <math>P=(P_{x},P_{y})</math> by calculating the intersection of <math>Line_{PR}</math> and <math>Line_{PQ}</math> as follows: | Now we solve for the coordinates for point <math>P=(P_{x},P_{y})</math> by calculating the intersection of <math>Line_{PR}</math> and <math>Line_{PQ}</math> as follows: | ||
+ | <math>\frac{-2r(m+d)}{(m+d)^{2}-r^2)}\left( P_{x}-(m+d) \right)=\frac{-2r(m-d)}{(m-d)^{2}-r^2)}\left( P_{x}-(m-d) \right)</math> | ||
Revision as of 18:21, 12 November 2023
Problem
In the plane let be a circle,
a line tangent to the circle
, and
a point on
. Find the locus of all points
with the following property: there exists two points
,
on
such that
is the midpoint of
and
is the inscribed circle of triangle
.
Video Solution
https://www.youtube.com/watch?v=ObCzaZwujGw
Solution
Note: This is an alternate method to what it is shown on the video. This alternate method is too long and too intensive in solving algebraic equations. A lot of steps have been shortened in this solution. The solution in the video provides a much faster solution,
Let be the radius of the circle
.
We define a cartesian coordinate system in two dimensions with the circle center at and circle equation to be
We define the line by the equation
, with point
at a distance
from the tangent and cartesian coordinates
Let be the distance from point
to point
such that the coordinates for
are
and thus the coordinates for
are
Let points ,
, and
be the points where lines
,
, and
are tangent to circle
respectively.
First we get the coordinates for points and
.
Since the circle is the incenter we know the following properties:
and
Therefore, to get the coordinates of point , we solve the following equations:
After a lot of algebra, this solves to:
Now we calculate the slope of the line that passes through which is perpendicular to the line that passes from the center of the circle to point
as follows:
Then, the equation of the line that passes through is as follows:
Now we get the coordinates of point , we solve the following equations:
After a lot of algebra, this solves to:
Now we calculate the slope of the line that passes through which is perpendicular to the line that passes from the center of the circle to point
as follows:
Then, the equation of the line that passes through is as follows:
Now we solve for the coordinates for point by calculating the intersection of
and
as follows:
In the plane let be a circle,
a line tangent to the circle
, and
a point on
. Find the locus of all points
with the following property: there exists two points
,
on
such that
is the midpoint of
and
is the inscribed circle of triangle
.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.