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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,
  
−
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>.  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>
+
Let <math>r</math> be the radius of the circle <math>C</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>.
+
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>
 +
 
 +
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 for <math>Q</math> are <math>(m-d,-r)</math>
 +
 
 +
Let points <math>S</math>, <math>T</math>, and <math>U</math> be the points where lines <math>PQ</math>, <math>PR</math>, and <math>l</math> are tangent to circle <math>C</math> respectively.
 +
 
 +
First we get the coordinates for points <math>S</math> and <math>T</math>.
 +
 
 +
Since the circle is the incenter we know the following properties:
 +
 
 +
<math>\left| RU \right| = \left| RT \right|=(m+d)</math>
 +
 
 +
and
 +
 
 +
<math>\left| QU \right| = \left| QS \right|=(m-d)</math>
 +
 
 +
Therefore, to get the coordinates of point <math>T=(T_{x},T_{y})</math>, we solve the following equations:
 +
 
 +
<math>T_{x}^{2}+T_{y}^2=r^{2}</math>
 +
 
 +
<math>\left| RT \right|^{2}=(m+d-T_{x})^{2}+(r+T_{y})^2</math>
 +
 
 +
<math>(m+d)^{2}=(m+d-T_{x})^{2}+(r+T_{y})^2</math>
 +
 
 +
After a lot of algebra, this solves to:
 +
 
 +
<math>T_{x}=\frac{2r^{2}(m+d)}{(m+d)^{2}+r^{2}}</math>
 +
 
 +
<math>T_{y}=\frac{r\left[ (m+d)^{2}-r^{2} \right]}{(m+d)^{2}+r^{2} }</math>
 +
 
 +
Now we calculate the slope of the line that passes through <math>PR</math> which is perpendicular to the line that passes from the center of the circle to point <math>T</math> as follows:
 +
 
 +
<math>Slope_{PR}=\frac{-T_{x}}{T_{y}}=\frac{-2r(m+d)}{(m+d)^{2}-r^2)}</math>
 +
 
 +
Then, the equation of the line that passes through <math>PR</math> is as follows:
 +
 
 +
<math>Line_{PR}\colon \; y+r=\frac{-2r(m+d)}{(m+d)^{2}-r^2)}\left( x-(m+d) \right)</math>  
 +
 
 +
Now we get the coordinates of point <math>S=(S_{x},S_{y})</math>, we solve the following equations:
 +
 
 +
<math>S_{x}^{2}+S_{y}^2=r^{2}</math>
 +
 
 +
<math>\left| QT \right|^{2}=(m-d-S_{x})^{2}+(r+S_{y})^2</math>
 +
 
 +
<math>(m-d)^{2}=(m-d-S_{x})^{2}+(r+S_{y})^2</math>
 +
 
 +
After a lot of algebra, this solves to:
 +
 
 +
<math>S_{x}=\frac{2r^{2}(m-d)}{(m-d)^{2}+r^{2}}</math>
 +
 
 +
<math>S_{y}=\frac{r\left[ (m-d)^{2}-r^{2} \right]}{(m-d)^{2}+r^{2} }</math>
 +
 
 +
Now we calculate the slope of the line that passes through <math>PQ</math> which is perpendicular to the line that passes from the center of the circle to point <math>S</math> as follows:
 +
 
 +
<math>Slope_{PQ}=\frac{-S_{x}}{S_{y}}=\frac{-2r(m-d)}{(m-d)^{2}-r^2)}</math>
 +
 
 +
Then, the equation of the line that passes through <math>PQ</math> is as follows:
 +
 
 +
<math>Line_{PQ}\colon \; y+r=\frac{-2r(m-d)}{(m-d)^{2}-r^2)}\left( x-(m-d) \right)</math>
 +
 
 +
Now we solve for the coordinates for point <math>P=(P_{x},P_{y})</math> by calculating the intersection of <math>Line_{PR}</math> and <math>Line_{PQ}</math> as follows:
 +
 
 +
<math>\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)</math>
 +
 
 +
Solving for <math>P_{x}</math> we get:
 +
 
 +
<math>P_{x}=\frac{2mr^{2}}{m^{2}+r^{2}-d^{2}}</math>
 +
 
 +
Solving for <math>P_{y}</math> we get:
 +
 
 +
<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>
 +
 
 +
<math>P_{y}=\frac{r(m^{2}-r^{2}-d^{2})}{m^{2}+r^{2}-d^{2}}</math>
 +
 
 +
Now we need to find the limit of <math>P_{x}</math> and <math>P_{y}</math> as <math>d</math> approaches infinity:
 +
 
 +
<math>P_{x_{d \to \infty}}=\lim_{d \to \infty} \frac{2mr^{2}}{m^{2}+r^{2}-d^{2}}=0</math>
 +
 
 +
<math>P_{y_{d \to \infty}}=\lim_{d \to \infty} \frac{r(m^{2}-r^{2}-d^{2})}{m^{2}+r^{2}-d^{2}}=r</math>
 +
 
 +
This means that the locus of <math>P</math> starts at point <math>(0,r)</math> on the circle <math>C</math> but that point is not included in the locus as that is the limit.
 +
 
 +
If we assume that the locus is a ray that starts at <math>(0,r)</math> let's calculate the slope of such ray:
 +
 
 +
<math>Slope_{locus}=\frac{P_{y}-r}{P_{x}}</math>
 +
 
 +
<math>Slope_{locus}=\frac{\frac{r(m^{2}-r^{2}-d^{2})}{m^{2}+r^{2}-d^{2}}-r}{\frac{2mr^{2}}{m^{2}+r^{2}-d^{2}}}</math>
 +
 
 +
<math>Slope_{locus}=\frac{-r^{2}}{2m}</math>
 +
 
 +
Since the calculated slope of such locus at any point <math>P</math> is not dependent on <math>d</math> and solely dependent on fixed <math>r</math> and <math>m</math>, then this proves the slope is fixed and thus the locus is a ray that starts at <math>(0,r)</math> excluding that point and with a slope of <math>\frac{-r^{2}}{2m}</math> in the cartesian coordinate system moving upwards to infinity.
 +
 
 +
We can also write the equation of the locus as: <math>y=\frac{-r^{2}}{2m}x+r,\;\;\forall y>r\;</math>and <math>x<0</math>
 +
 
 +
~Tomas Diaz. orders@tomasdiaz.com
  
 
{{alternate solutions}}
 
{{alternate solutions}}
 +
 +
==See Also==
 +
 +
{{IMO box|year=1992|num-b=3|num-a=5}}
 +
[[Category:Olympiad Geometry Problems]]
 +
[[Category:3D Geometry Problems]]

Latest revision as of 00:42, 17 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:

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

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

Now we need to find the limit of $P_{x}$ and $P_{y}$ as $d$ approaches infinity:

$P_{x_{d \to \infty}}=\lim_{d \to \infty} \frac{2mr^{2}}{m^{2}+r^{2}-d^{2}}=0$

$P_{y_{d \to \infty}}=\lim_{d \to \infty} \frac{r(m^{2}-r^{2}-d^{2})}{m^{2}+r^{2}-d^{2}}=r$

This means that the locus of $P$ starts at point $(0,r)$ on the circle $C$ but that point is not included in the locus as that is the limit.

If we assume that the locus is a ray that starts at $(0,r)$ let's calculate the slope of such ray:

$Slope_{locus}=\frac{P_{y}-r}{P_{x}}$

$Slope_{locus}=\frac{\frac{r(m^{2}-r^{2}-d^{2})}{m^{2}+r^{2}-d^{2}}-r}{\frac{2mr^{2}}{m^{2}+r^{2}-d^{2}}}$

$Slope_{locus}=\frac{-r^{2}}{2m}$

Since the calculated slope of such locus at any point $P$ is not dependent on $d$ and solely dependent on fixed $r$ and $m$, then this proves the slope is fixed and thus the locus is a ray that starts at $(0,r)$ excluding that point and with a slope of $\frac{-r^{2}}{2m}$ in the cartesian coordinate system moving upwards to infinity.

We can also write the equation of the locus as: $y=\frac{-r^{2}}{2m}x+r,\;\;\forall y>r\;$and $x<0$

~Tomas Diaz. orders@tomasdiaz.com

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

See Also

1992 IMO (Problems) • Resources
Preceded by
Problem 3 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 5
All IMO Problems and Solutions
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