Difference between revisions of "1992 IMO Problems/Problem 4"
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Let points <math>S</math>, <math>T</math>, and <math>U</math> be the points where lines <math>PQ</math>, <math>PR</math>, and <math>l</math> are tangent to circle <math>C</math> respectively. | Let points <math>S</math>, <math>T</math>, and <math>U</math> be the points where lines <math>PQ</math>, <math>PR</math>, and <math>l</math> are tangent to circle <math>C</math> respectively. | ||
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+ | First we get the coordinates for points <math>S</math> and <math>T</math>. | ||
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+ | Since the circle is the incenter we know the following properties: | ||
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+ | <math>\left| RU \right| = \left| RT \right|=(m+d)</math> | ||
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+ | and | ||
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+ | <math>\left| QU \right| = \left| QS \right|=(m-d)</math> | ||
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+ | Therefore, to get the coordinates of point <math>T=(T_{x},T_{y})</math>, we solve the following equations: | ||
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+ | <math>T_{x}^{2}+T_{y}^2=r^{2}</math> | ||
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+ | <math>\left| RT \right|^{2}=(m+d-T_{x})^{2}+(r+T_{y})^2</math> | ||
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+ | <math>(m+d)^{2}=(m+d-T_{x})^{2}+(r+T_{y})^2</math> | ||
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+ | After a lot of algebra, this solves to: | ||
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+ | <math>T_{x}=\frac{2r^{2}(m+d)}{(m+d)^{2}+r^{2}}</math> | ||
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+ | <math>T_{y}=\frac{r\left[ (m+d)^{2}-r^{2} \right]}{(m+d)^{2}+r^{2} }</math> | ||
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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 17:05, 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:
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.