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Difference between revisions of "1992 IMO Problems/Problem 4"

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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>
 
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>
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Let <math>d</math> be the distance from point <math>M</math> to point <math>R</math> such that the coordinates for <math>R</math> are <math>(m+d,-r)</math> and thus the coordinates from <math>Q</math> are <math>(m-d,-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 16:50, 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 $(m,-r)$

Let $d$ be the distance from point $M$ to point $R$ such that the coordinates for $R$ are $(m+d,-r)$ and thus the coordinates from $Q$ are $(m-d,-r)$

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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