Difference between revisions of "1992 IMO Problems/Problem 4"
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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, | 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 <math>r</math> be the radius of the circle <math>C</math>. We define an <math>xy</math>-plane with the circle center in the origin of the plane and circle equation to be <math>x^{2}+y{2}=r^{2}</math>. | + | Let <math>r</math> be the radius of the circle <math>C</math>. We define an <math>xy</math>-plane with the circle center in the origin of the plane and circle equation to be <math>x^{2}+y{2}=r^{2}</math>. |
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| + | We define the line <math>l</math> by the equation <math>y=-r</math>, with point <math>M</math> at a distance <math>m</math> from the tangent and cartesian coordinates <math>(m,-r)</math> | ||
In the plane let <math>C</math> be a circle, <math>l</math> a line tangent to the circle <math>C</math>, and <math>M</math> a point on <math>l</math>. Find the locus of all points <math>P</math> with the following property: there exists two points <math>Q</math>, <math>R</math> on <math>l</math> such that <math>M</math> is the midpoint of <math>QR</math> and <math>C</math> is the inscribed circle of triangle <math>PQR</math>. | In the plane let <math>C</math> be a circle, <math>l</math> a line tangent to the circle <math>C</math>, and <math>M</math> a point on <math>l</math>. Find the locus of all points <math>P</math> with the following property: there exists two points <math>Q</math>, <math>R</math> on <math>l</math> such that <math>M</math> is the midpoint of <math>QR</math> and <math>C</math> is the inscribed circle of triangle <math>PQR</math>. | ||
{{alternate solutions}} | {{alternate solutions}} | ||
Revision as of 16:48, 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 an 
-plane with the circle center in the origin of the plane and circle equation to be 
.
We define the line by the equation 
, with point at a distance 
from the tangent and cartesian coordinates 
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.
