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1981 IMO Problems/Problem 1: Difference between revisions

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== Problem ==
== Problem ==


<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
<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


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== Solution ==
== Solution ==


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]],
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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with equality exactly when <math> \displaystyle PD = PE = PF </math>, which occurs when <math> \displaystyle P </math> is the triangle's incenter, Q.E.D.
with equality exactly when <math>PD = PE = PF </math>, which occurs when <math>P </math> is the triangle's incenter, Q.E.D.


{{alternate solutions}}
{{alternate solutions}}


== Resources ==
{{IMO box|before=First question|num-a=2|year=1980}}
 
* [[1981 IMO Problems]]
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366638#p366638 Discussion on AoPS/MathLinks]
 


[[Category:Olympiad Geometry Problems]]
[[Category:Olympiad Geometry Problems]]

Revision as of 19:44, 25 October 2007

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, Q.E.D.

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

1980 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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