1992 IMO Problems/Problem 4

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 for $Q$ are $(m-d,-r)$

Let points $S$ and $T$ points where lines $PQ$ and $PR$ are tangent to circle $C$ respectively.

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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1992 IMO Problems/Problem 4

Problem

Video Solution

Solution