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

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<math>Slope_{PR}=\frac{-T_{x}}{T_{y}}=\frac{-2r(m+d)}{(m+d)^{2}-r^2)}</math>
 
<math>Slope_{PR}=\frac{-T_{x}}{T_{y}}=\frac{-2r(m+d)}{(m+d)^{2}-r^2)}</math>
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Then, the equation of the line that passes through <math>PR</math> is as follows:
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<math>Line_{PR}\colon \; y+r=\frac{-2r(m+d)}{(m+d)^{2}-r^2)}\left( x-(m+d) \right)</math>
  
  

Revision as of 17:11, 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 for $Q$ are $(m-d,-r)$

Let points $S$, $T$, and $U$ be the points where lines $PQ$, $PR$, and $l$ are tangent to circle $C$ respectively.

First we get the coordinates for points $S$ and $T$.

Since the circle is the incenter we know the following properties:

$\left| RU \right| = \left| RT \right|=(m+d)$

and

$\left| QU \right| = \left| QS \right|=(m-d)$

Therefore, to get the coordinates of point $T=(T_{x},T_{y})$, we solve the following equations:

$T_{x}^{2}+T_{y}^2=r^{2}$

$\left| RT \right|^{2}=(m+d-T_{x})^{2}+(r+T_{y})^2$

$(m+d)^{2}=(m+d-T_{x})^{2}+(r+T_{y})^2$

After a lot of algebra, this solves to:

$T_{x}=\frac{2r^{2}(m+d)}{(m+d)^{2}+r^{2}}$

$T_{y}=\frac{r\left[ (m+d)^{2}-r^{2} \right]}{(m+d)^{2}+r^{2} }$

Now we calculate the slope of the line that passes through $PR$ which is perpendicular to the line that passes from the center of the circle to point $T$ as follows:

$Slope_{PR}=\frac{-T_{x}}{T_{y}}=\frac{-2r(m+d)}{(m+d)^{2}-r^2)}$

Then, the equation of the line that passes through $PR$ is as follows:

$Line_{PR}\colon \; y+r=\frac{-2r(m+d)}{(m+d)^{2}-r^2)}\left( x-(m+d) \right)$






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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