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

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<math>P_{x}=\frac{2mr^{2}}{m^{2}+r^{2}-d^{2}}</math>
 
<math>P_{x}=\frac{2mr^{2}}{m^{2}+r^{2}-d^{2}}</math>
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Solving for <math>P_{y}</math> we get:
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<math>P_{y}=\frac{-2r(m-d)}{(m-d)^{2}-r^2)}\left( \frac{2mr^{2}}{m^{2}+r^{2}-d^{2}}-(m-d) \right)-r</math>
  
  

Revision as of 17:25, 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:

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

Now we get the coordinates of point $S=(S_{x},S_{y})$, we solve the following equations:

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

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

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

After a lot of algebra, this solves to:

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

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

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

$Slope_{PQ}=\frac{-S_{x}}{S_{y}}=\frac{-2r(m-d)}{(m-d)^{2}-r^2)}$

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

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

Now we solve for the coordinates for point $P=(P_{x},P_{y})$ by calculating the intersection of $Line_{PR}$ and $Line_{PQ}$ as follows:

$\frac{-2r(m+d)}{(m+d)^{2}-r^2)}\left( P_{x}-(m+d) \right)=\frac{-2r(m-d)}{(m-d)^{2}-r^2)}\left( P_{x}-(m-d) \right)$

Solving for $P_{x}$ we get:

$P_{x}=\frac{2mr^{2}}{m^{2}+r^{2}-d^{2}}$

Solving for $P_{y}$ we get:

$P_{y}=\frac{-2r(m-d)}{(m-d)^{2}-r^2)}\left( \frac{2mr^{2}}{m^{2}+r^{2}-d^{2}}-(m-d) \right)-r$



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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