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Difference between revisions of "1981 IMO Problems/Problem 1"
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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]],
 
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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{(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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with equality exactly when <math>PD = PE = PF </math>, which occurs when <math>P </math> is the triangle's 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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[[Category:Olympiad Geometry Problems]]
 
[[Category:Olympiad Geometry Problems]]
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wait can u actually edit this..... if a mod sees this then it is public :O, mod delete it if you can see... I think this just shows up for me
 

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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