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

 
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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> \displaystyle PD = PE = PF </math>, which occurs when <math> \displaystyle P </math> is the triangle's incenter, Q.E.D.
  
 
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Revision as of 20:34, 24 March 2007

Problem

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

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

is least.

Solution

We note that $\displaystyle BC \cdot PD + CA \cdot PE + AB \cdot PF$ 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 $\displaystyle PD = PE = PF$, which occurs when $\displaystyle P$ 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.

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