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Difference between revisions of "1981 IMO Problems/Problem 1"

 
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== Problem ==
 
== Problem ==
  
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<math>\displaystyle P</math> is a point inside a given triangle <math>\displaystyle ABC</math>.  <math>\displaystyle D, E, F</math> are the feet of the perpendiculars from <math>\displaystyle P</math> to the lines <math>\displaystyle BC, CA, AB</math>, respectively.  Find all <math>\displaystyle P</math> for which
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<math>P</math> is a point inside a given triangle <math>ABC</math>.  <math>D, E, F</math> are the feet of the perpendiculars from <math>P</math> to the lines <math>BC, CA, AB</math>, respectively.  Find all <math>P</math> for which
  
 
<center>
 
<center>
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== Solution ==
 
== Solution ==
  
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We note that <math>\displaystyle BC \cdot PD + CA \cdot PE + AB \cdot PF</math> is twice the triangle's area, i.e., constant.  By the [[Cauchy-Schwarz Inequality]],
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We note that <math>BC \cdot PD + CA \cdot PE + AB \cdot PF</math> is twice the triangle's area, i.e., constant.  By the [[Cauchy-Schwarz Inequality]],
  
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<center>
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<math>
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<cmath>
 
{(BC \cdot PD + CA \cdot PE + AB \cdot PF) \left(\frac{BC}{PD} + \frac{CA}{PE} + \frac{AB}{PF} \right) \ge ( BC + CA + AB )^2}
 
{(BC \cdot PD + CA \cdot PE + AB \cdot PF) \left(\frac{BC}{PD} + \frac{CA}{PE} + \frac{AB}{PF} \right) \ge ( BC + CA + AB )^2}
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</math>,
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</cmath>,
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</center>
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with equality exactly when <math> \displaystyle PD = PE = PF </math>, which occurs when <math> \displaystyle P </math> is the triangles incenter, Q.E.D.
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with equality exactly when <math>PD = PE = PF </math>, which occurs when <math>P </math> is the triangle's incenter or one of the three excenters. But since we know <math>P </math> is inside <math>\triangle ABC </math>, we can say <math>P </math> is the incenter. <math>\square </math>
  
 
{{alternate solutions}}
 
{{alternate solutions}}
  
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== Resources ==
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{{IMO box|before=First question|num-a=2|year=1981}}
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* [[1981 IMO Problems]]
 
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* [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366638#p366638 Discussion on AoPS/MathLinks]
 
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[[Category:Olympiad Geometry Problems]]
 
[[Category:Olympiad Geometry Problems]]

Latest revision as of 12:06, 26 August 2024

Problem

$P$ is a point inside a given triangle $ABC$. $D, E, F$ are the feet of the perpendiculars from $P$ to the lines $BC, CA, AB$, respectively. Find all $P$ for which

$\frac{BC}{PD} + \frac{CA}{PE} + \frac{AB}{PF}$

is least.

Solution

We note that $BC \cdot PD + CA \cdot PE + AB \cdot PF$ is twice the triangle's area, i.e., constant. By the Cauchy-Schwarz Inequality,


\[{(BC \cdot PD + CA \cdot PE + AB \cdot PF) \left(\frac{BC}{PD} + \frac{CA}{PE} + \frac{AB}{PF} \right) \ge ( BC + CA + AB )^2}\],


with equality exactly when $PD = PE = PF$, which occurs when $P$ is the triangle's incenter or one of the three excenters. But since we know $P$ is inside $\triangle ABC$, we can say $P$ is the incenter. $\square$

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

1981 IMO (Problems) • Resources
Preceded by
First question 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 2
All IMO Problems and Solutions
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