1992 IMO Problems/Problem 4
Contents
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:
Solving for we get:
Solving for we get:
Now we need to find the limit of and as approaches infinity:
This means that the locus of starts at point on the circle but that point is not included in the locus as that is the limit.
If we assume that the locus is a ray that starts at let's calculate the slope of such ray:
Since the calculated slope of such locus at any point is not dependent on and solely dependent on fixed and , then this proves the slope is fixed and thus the locus is a ray that starts at excluding that point and with a slope of in the cartesian coordinate system moving upwards to infinity.
We can also write the equation of the locus as: and
~Tomas Diaz. orders@tomasdiaz.com
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
1992 IMO (Problems) • Resources | ||
Preceded by Problem 3 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 5 |
All IMO Problems and Solutions |