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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,
  
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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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Let <math>r</math> be the radius of the circle <math>C</math>.
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We define a cartesian coordinate system in two dimensions with the circle center at <math>(0,0)</math> and circle equation to be <math>x^{2}+y{2}=r^{2}</math>
  
 
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>
 
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>

Revision as of 17:49, 12 November 2023

Problem

In the plane let $C$ be a circle, $l$ a line tangent to the circle $C$, and $M$ a point on $l$. Find the locus of all points $P$ with the following property: there exists two points $Q$, $R$ on $l$ such that $M$ is the midpoint of $QR$ and $C$ is the inscribed circle of triangle $PQR$.

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 $r$ be the radius of the circle $C$.

We define a cartesian coordinate system in two dimensions with the circle center at $(0,0)$ and circle equation to be $x^{2}+y{2}=r^{2}$

We define the line $l$ by the equation $y=-r$, with point $M$ at a distance $m$ from the tangent and cartesian coordinates $(m,-r)$

In the plane let $C$ be a circle, $l$ a line tangent to the circle $C$, and $M$ a point on $l$. Find the locus of all points $P$ with the following property: there exists two points $Q$, $R$ on $l$ such that $M$ is the midpoint of $QR$ and $C$ is the inscribed circle of triangle $PQR$.

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

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